Fundamentals and Multiple Integrals

This chapter establishes the foundational concepts of definite integrals and the Fundamental Theorem of Calculus, crucial for understanding advanced calculus. It then extends these principles to multiple integrals, covering double and triple integration, change of order, and practical applications. Mastery of these topics is essential for success in the CUET PG Mathematics examination.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Definite Integrals and the Fundamental Theorem |
| 2 | Double and Triple Integrals |
| 3 | Change of Order of Integration |
| 4 | Applications of Multiple Integrals |

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We begin with Definite Integrals and the Fundamental Theorem.

Part 1: Definite Integrals and the Fundamental Theorem

Definite integrals provide a rigorous method for calculating accumulation over an interval, serving as a foundational concept in calculus. The Fundamental Theorem of Calculus establishes a critical link between differentiation and integration, enabling efficient evaluation of these integrals. We explore their properties, evaluation techniques, and diverse applications in geometric and physical problems.

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Core Concepts

1. The Definite Integral as a Limit of Sums

We define the definite integral of a function $f(x)$ over an interval $[a, b]$ as the limit of Riemann sums. This conceptualization connects the integral to the area under a curve, representing the sum of infinitesimally narrow rectangles.

Quick Example:
Express the limit as a definite integral: $\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left(\frac{i}{n}\right)^2$.

Step 1: Identify $f(x)$, $dx$, and the interval.
We observe $\Delta x = \frac{1}{n}$ and $x_i = \frac{i}{n}$. If we choose $a=0$, then $b=1$.
Therefore, $f(x) = x^2$.

Answer: $\int_0^1 x^2 dx$

:::question type="MCQ" question="The value of $\lim_{n \to \infty} \left[\frac{1}{n^2 + 1^2} + \frac{2}{n^2 + 2^2} + \frac{3}{n^2 + 3^2} + \dots + \frac{n}{n^2 + n^2}\right]$ is:" options=["$\frac{1}{2} \log 2$","$\log 2$","$\frac{\pi}{4}$","$\frac{1}{2}$"] answer="$\frac{1}{2} \log 2$" hint="Rewrite the sum in the form $\lim_{n \to \infty} \sum_{i=1}^n f(i/n) (1/n)$." solution="Step 1: Rewrite the given sum.

Step 2: Identify $f(x)$ and the limits of integration.
Here, $x_i = i/n$ and $\Delta x = 1/n$. Thus, we can identify $f(x) = \frac{x}{1+x^2}$.
The limits of integration are from $x=0$ (when $i=1$, $1/n \to 0$) to $x=1$ (when $i=n$, $n/n=1$).

Step 3: Convert to a definite integral and evaluate.

Let $u = 1+x^2$, then $du = 2x \, dx$. When $x=0$, $u=1$. When $x=1$, $u=2$.

Answer: $\boxed{\frac{1}{2} \log 2}$"
:::

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2. Properties of Definite Integrals

Definite integrals possess several properties that simplify their manipulation and evaluation. These properties are crucial for efficiently solving problems, especially those involving symmetry or interval decomposition.

Quick Example:
Evaluate $\int_{-2}^2 (x^3 + \sin x + 5) dx$.

Step 1: Decompose the integral using linearity.
We can write the integral as $\int_{-2}^2 x^3 dx + \int_{-2}^2 \sin x dx + \int_{-2}^2 5 dx$.

Step 2: Apply symmetry properties.
The function $x^3$ is odd, so $\int_{-2}^2 x^3 dx = 0$.
The function $\sin x$ is odd, so $\int_{-2}^2 \sin x dx = 0$.
The function $5$ is even, so $\int_{-2}^2 5 dx = 2\int_0^2 5 dx$.

Step 3: Evaluate the remaining integral.

Answer: $20$

:::question type="MCQ" question="The value of $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$ is:" options=["$\frac{\pi}{4}$","$\frac{\pi}{2}$","$0$","$1$"] answer="$\frac{\pi}{4}$" hint="Use King's Rule: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$." solution="Step 1: Let the integral be $I$.

Step 2: Apply King's Rule, $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
Here $a = \pi/2$. So, replace $x$ with $(\pi/2 - x)$.

We know $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$.

Step 3: Add equations (1) and (2).

Step 4: Evaluate the integral.

Answer: $\boxed{\frac{\pi}{4}}$"
:::

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3. The Fundamental Theorem of Calculus (Part 2)

The second part of the Fundamental Theorem of Calculus (FTC) provides a powerful method for evaluating definite integrals by using antiderivatives. It states that if $F$ is an antiderivative of $f$, then the definite integral of $f$ from $a$ to $b$ can be found by evaluating $F(b) - F(a)$.

Quick Example:
Evaluate $\int_1^3 (2x + 3) dx$.

Step 1: Find the antiderivative of $f(x) = 2x + 3$.
The antiderivative $F(x) = x^2 + 3x$.

Step 2: Apply FTC Part 2.

Answer: $14$

:::question type="MCQ" question="Evaluate $\int_0^{\pi/2} \cos^2 x \, dx$." options=["$\frac{\pi}{4}$","$\frac{\pi}{2}$","$0$","$1$"] answer="$\frac{\pi}{4}$" hint="Use the half-angle identity for $\cos^2 x$." solution="Step 1: Use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.

Step 2: Separate the integral and find the antiderivative.

The antiderivative of $1 + \cos(2x)$ is $x + \frac{\sin(2x)}{2}$.

Step 3: Apply FTC Part 2.

Answer: $\boxed{\frac{\pi}{4}}$"
:::

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4. Fundamental Theorem of Calculus (Part 1) / Leibniz Integral Rule

The first part of the Fundamental Theorem of Calculus, often generalized as the Leibniz Integral Rule, allows us to differentiate an integral with respect to a variable that appears in its limits of integration or integrand. This is particularly useful in problems involving functions defined as integrals.

Quick Example:
Find $\frac{d}{dx} \int_{x^2}^{\sin x} e^{t^2} dt$.

Step 1: Identify $a(x)$, $b(x)$, $f(t,x)$.
Here, $a(x) = x^2$, $b(x) = \sin x$, and $f(t,x) = e^{t^2}$ (which is just $f(t) = e^{t^2}$).
Thus, $\frac{\partial}{\partial x} f(t,x) = 0$.
We need $a'(x) = 2x$ and $b'(x) = \cos x$.

Step 2: Apply the Leibniz Integral Rule.

Answer: $e^{\sin^2 x} \cos x - 2x e^{x^4}$

:::question type="MCQ" question="Let $F(x) = \int_{1/x}^{\sqrt{x}} \cos t^2 dt$ for $x>0$. The value of $F'(1)$ is:" options=["$\frac{3}{2} \cos 1$","$\frac{1}{2} \cos 1$","$\cos 1$","$0$"] answer="$\frac{3}{2} \cos 1$" hint="Apply the Leibniz Integral Rule and then substitute $x=1$." solution="Step 1: Identify the components for the Leibniz Integral Rule.
We have $a(x) = 1/x = x^{-1}$, $b(x) = \sqrt{x} = x^{1/2}$, and $f(t,x) = \cos t^2$.
Since $f(t,x)$ does not explicitly depend on $x$, $\frac{\partial}{\partial x} f(t,x) = 0$.

Step 2: Calculate the derivatives of the limits.
$a'(x) = \frac{d}{dx} (x^{-1}) = -x^{-2} = -\frac{1}{x^2}$.
$b'(x) = \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$.

Step 3: Apply the Leibniz Integral Rule.

Step 4: Evaluate $F'(1)$.

Answer: $\boxed{\frac{3}{2} \cos 1}$"
:::

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5. Integration Techniques for Definite Integrals

Many definite integrals require specific techniques to find their antiderivatives before applying FTC Part 2. The most common techniques are substitution and integration by parts. When applying these to definite integrals, it is crucial to handle the limits of integration correctly.

5.1. Integration by Substitution

When using substitution for definite integrals, we must either change the limits of integration to correspond to the new variable or revert to the original variable before applying the original limits. Changing the limits is generally more efficient.

Quick Example:
Evaluate $\int_0^1 x e^{x^2} dx$.

Step 1: Choose a substitution.
Let $u = x^2$. Then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.

Step 2: Change the limits of integration.
When $x=0$, $u = 0^2 = 0$.
When $x=1$, $u = 1^2 = 1$.

Step 3: Rewrite and evaluate the integral with the new variable and limits.
>

Answer: $\frac{e-1}{2}$

:::question type="MCQ" question="Evaluate $\int_0^{\pi/2} \sin^3 x \cos x \, dx$." options=["$\frac{1}{4}$","$\frac{1}{3}$","$\frac{1}{2}$","$1$"] answer="$\frac{1}{4}$" hint="Use the substitution $u = \sin x$ and change the limits." solution="Step 1: Choose a substitution.
Let $u = \sin x$. Then $du = \cos x \, dx$.

Step 2: Change the limits of integration.
When $x=0$, $u = \sin 0 = 0$.
When $x=\pi/2$, $u = \sin(\pi/2) = 1$.

Step 3: Rewrite and evaluate the integral.

Answer: $\boxed{\frac{1}{4}}$"
:::

5.2. Integration by Parts

For definite integrals, integration by parts applies the formula $[uv]_a^b - \int_a^b v \, du$. The evaluation of $uv$ must be performed at the limits $b$ and $a$.

Quick Example:
Evaluate $\int_0^1 x e^x dx$.

Step 1: Choose $u$ and $dv$.
Let $u = x$ and $dv = e^x dx$.
Then $du = dx$ and $v = e^x$.

Step 2: Apply the integration by parts formula.
>

Answer: $1$

:::question type="MCQ" question="Evaluate $\int_0^{\pi/2} x \cos x \, dx$." options=["$1$","$\frac{\pi}{2} - 1$","$\frac{\pi}{2}$","$0$"] answer="$\frac{\pi}{2} - 1$" hint="Use integration by parts, setting $u=x$ and $dv=\cos x \, dx$." solution="Step 1: Choose $u$ and $dv$.
Let $u = x$, so $du = dx$.
Let $dv = \cos x \, dx$, so $v = \sin x$.

Step 2: Apply the integration by parts formula.

Step 3: Evaluate the terms.
For the first term:

For the second term:

Step 4: Combine the results.

Answer: $\boxed{\frac{\pi}{2} - 1}$"
:::

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Advanced Applications

1. Area Between Curves

We determine the area of a region bounded by curves by integrating the difference between the upper and lower functions (or right and left functions) over the appropriate interval. The choice of integration variable ($x$ or $y$) depends on the orientation of the bounding curves.

Quick Example:
Find the area bounded by $y = x^2$ and $y = x+2$.

Step 1: Find the intersection points of the curves.
Set $x^2 = x+2$.
>

These are our limits of integration.

Step 2: Determine which function is upper and which is lower.
For $x \in [-1, 2]$, we test a point, e.g., $x=0$.
$y_1 = 0^2 = 0$ and $y_2 = 0+2 = 2$.
Since $2 > 0$, $y = x+2$ is the upper curve and $y = x^2$ is the lower curve.

Step 3: Set up and evaluate the integral.
>

Answer: $\frac{9}{2}$

:::question type="MCQ" question="The area bounded by the curves $y=x^2$ and $y=4-x^2$ is:" options=["$\frac{16\sqrt{2}}{3}$","$\frac{16}{3}$","$\frac{16\pi}{3}$","$\frac{8}{\sqrt{3}}$"] answer="$\frac{16\sqrt{2}}{3}$" hint="Find intersection points and determine the upper and lower functions." solution="Step 1: Find the intersection points.
Set $x^2 = 4-x^2$.

So the limits of integration are from $-\sqrt{2}$ to $\sqrt{2}$.

Step 2: Determine which function is upper and which is lower.
For $x \in [-\sqrt{2}, \sqrt{2}]$, consider $x=0$.
$y_1 = 0^2 = 0$.
$y_2 = 4-0^2 = 4$.
Since $4 > 0$, $y=4-x^2$ is the upper curve and $y=x^2$ is the lower curve.

Step 3: Set up and evaluate the integral.

Since $4-2x^2$ is an even function, we can use the property $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$.

Answer: $\boxed{\frac{16\sqrt{2}}{3}}$"
:::

2. Volume of Solids of Revolution

We calculate the volume of a solid formed by revolving a region around an axis using either the disk/washer method or the cylindrical shell method. The choice depends on the shape of the region and the axis of revolution.

Quick Example (Disk Method):
Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $x=1$, and the x-axis about the x-axis.

Step 1: Identify the function and limits.
The region is bounded by $y=\sqrt{x}$, $x=1$, and $y=0$ (x-axis). The limits are from $x=0$ to $x=1$. We are revolving about the x-axis, so we use the disk method. $f(x) = \sqrt{x}$.

Step 2: Set up and evaluate the integral.
>

Answer: $\frac{\pi}{2}$

:::question type="MCQ" question="The volume of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, and $x=2$ about the y-axis is:" options=["$\frac{8\pi}{3}$","$\frac{16\pi}{3}$","$4\pi$","$8\pi$"] answer="$8\pi$" hint="Use the cylindrical shell method, as revolving about y-axis and integrating with respect to x is simpler." solution="Step 1: Identify the function, limits, and method.
The region is bounded by $y=x^2$, $y=0$, and $x=2$. We are revolving about the y-axis.
The limits for $x$ are from $0$ to $2$. The height of the shell is $f(x) = x^2$. The radius of the shell is $x$.
We use the cylindrical shell method.

Step 2: Set up the integral for the shell method.

Step 3: Evaluate the integral.

Answer: $\boxed{8\pi}$"
:::

3. Surface Area of Solids of Revolution

We calculate the surface area of a solid formed by revolving a curve $y=f(x)$ or $x=g(y)$ around an axis. This involves integrating an expression related to the curve's arc length and the radius of revolution.

Quick Example:
Find the surface area generated by revolving $y=x^3$ from $x=0$ to $x=1$ about the x-axis.

Step 1: Find $\frac{dy}{dx}$.
$y = x^3$, so $\frac{dy}{dx} = 3x^2$.
Then $\left(\frac{dy}{dx}\right)^2 = (3x^2)^2 = 9x^4$.

Step 2: Set up the surface area integral.
>

Step 3: Evaluate the integral using substitution.
Let $u = 1 + 9x^4$. Then $du = 36x^3 dx$, so $x^3 dx = \frac{1}{36} du$.
Change limits: When $x=0$, $u=1$. When $x=1$, $u=1+9=10$.
>

Answer: $\frac{\pi}{27} (10\sqrt{10} - 1)$

:::question type="MCQ" question="The area of the surface of a solid generated by the revolution of the line segment $y = 2x$ from $x=0$ to $x=2$ about the x-axis is:" options=["$\pi\sqrt{5}$","$2\pi\sqrt{5}$","$4\pi\sqrt{5}$","$8\pi\sqrt{5}$"] answer="$8\pi\sqrt{5}$" hint="Use the surface area formula for revolution about the x-axis. Calculate $dy/dx$ and substitute into the integral." solution="Step 1: Find $\frac{dy}{dx}$.
The curve is $y = 2x$.

Step 2: Set up the surface area integral.
The limits are $x=0$ to $x=2$.

Substitute $y=2x$ and $\left(\frac{dy}{dx}\right)^2 = 4$.

Step 3: Evaluate the integral.

Answer: $\boxed{8\pi\sqrt{5}}$"
:::

4. Reduction Formulas for Definite Integrals and Wallis' Integrals

Reduction formulas simplify integrals by expressing an integral in terms of a similar integral with a lower power or index. Wallis' Integrals are a specific set of reduction formulas for powers of sine and cosine over the interval $[0, \pi/2]$. These are frequently encountered in competitive exams.

Quick Example:
Evaluate $\int_0^{\pi/2} \sin^6 x \, dx$.

Step 1: Identify $n$ and apply the Wallis' formula for even $n$.
Here $n=6$, which is an even number.
>

Answer: $\frac{5\pi}{32}$

:::question type="MCQ" question="Evaluate $\int_0^{\pi/2} \cos^5 x \, dx$." options=["$\frac{8}{15}$","$\frac{16}{15}$","$\frac{4}{5}$","$\frac{1}{5}$"] answer="$\frac{8}{15}$" hint="Use Wallis' Integral formula for odd powers." solution="Step 1: Identify $n$ and apply the Wallis' formula for odd $n$.
Here $n=5$, which is an odd number.

Answer: $\boxed{\frac{8}{15}}$"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Evaluate $\int_0^1 x \sqrt{1-x^2} dx$." options=["$\frac{1}{3}$","$\frac{2}{3}$","$1$","$\frac{\pi}{4}$"] answer="$\frac{1}{3}$" hint="Use substitution $u = 1-x^2$." solution="Step 1: Use substitution.
Let $u = 1-x^2$. Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} du$.

Step 2: Change the limits of integration.
When $x=0$, $u = 1-0^2 = 1$.
When $x=1$, $u = 1-1^2 = 0$.

Step 3: Rewrite and evaluate the integral.

Using the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$:

Answer: $\boxed{\frac{1}{3}}$"
:::

:::question type="NAT" question="Find the area of the region bounded by $y = \sin x$, $y = \cos x$, $x=0$, and $x=\pi/2$." answer="0.83" hint="Identify intersection points and split the integral if necessary. Determine which function is greater in each interval." solution="Step 1: Find intersection points within the interval $[0, \pi/2]$.
Set $\sin x = \cos x$. This occurs at $x = \pi/4$.

Step 2: Determine which function is greater in each subinterval.
For $x \in [0, \pi/4]$, $\cos x \ge \sin x$.
For $x \in [\pi/4, \pi/2]$, $\sin x \ge \cos x$.

Step 3: Set up the integral for the total area.

Step 4: Evaluate each integral.
First integral:

Second integral:

Step 5: Sum the areas.

Answer: $\boxed{0.83}$"
:::

:::question type="MCQ" question="Let $G(x) = \int_x^{x^2} \frac{\sin t}{t} dt$. The value of $G'(\pi)$ is:" options=["$2\sin(\pi^2) - \frac{\sin \pi}{\pi}$","$\frac{\sin(\pi^2)}{\pi} - \frac{\sin \pi}{\pi}$","$2\frac{\sin(\pi^2)}{\pi} - \frac{\sin \pi}{\pi}$","$\frac{\sin(\pi^2)}{\pi} - \frac{\sin \pi}{\pi^2}$"] answer="$2\frac{\sin(\pi^2)}{\pi} - \frac{\sin \pi}{\pi}$" hint="Apply the Leibniz Integral Rule. Remember $\sin \pi = 0$." solution="Step 1: Identify $a(x)$, $b(x)$, and $f(t,x)$.
$a(x) = x$, $b(x) = x^2$, and $f(t,x) = \frac{\sin t}{t}$.
Since $f(t,x)$ does not depend on $x$, $\frac{\partial}{\partial x} f(t,x) = 0$.

Step 2: Calculate derivatives of the limits.
$a'(x) = 1$.
$b'(x) = 2x$.

Step 3: Apply the Leibniz Integral Rule.

Step 4: Evaluate $G'(\pi)$.

Since $\sin \pi = 0$:

Answer: $\boxed{\frac{2\sin(\pi^2)}{\pi}}$"
:::

:::question type="MCQ" question="If $\int_0^a f(x) dx = \frac{\pi}{2}$, then $\int_0^a f(a-x) dx$ is:" options=["$\frac{\pi}{2}$","$-\frac{\pi}{2}$","$0$","$\pi$"] answer="$\frac{\pi}{2}$" hint="Recall the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$." solution="Step 1: Apply the property of definite integrals.
We know that $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
For the given integral $\int_0^a f(x) dx$, we have $a=0$ and $b=a$.
So, $a+b-x = 0+a-x = a-x$.

Step 2: Apply the property to the second integral.

Given that $\int_0^a f(x) dx = \frac{\pi}{2}$.
Therefore, $\int_0^a f(a-x) dx = \frac{\pi}{2}$.
This is a direct application of King's Rule.
Answer: $\boxed{\frac{\pi}{2}}$"
:::

:::question type="MCQ" question="The volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $x=0$, $x=4$, and the x-axis about the y-axis is:" options=["$\frac{128\pi}{5}$","$\frac{64\pi}{5}$","$\frac{32\pi}{3}$","$\frac{16\pi}{3}$"] answer="$\frac{128\pi}{5}$" hint="Use the cylindrical shell method, as revolving about y-axis and integrating with respect to x is often simpler." solution="Step 1: Identify the function, limits, and method.
The region is bounded by $y=\sqrt{x}$, $y=0$ (x-axis), $x=0$, and $x=4$. We are revolving about the y-axis.
The limits for $x$ are from $0$ to $4$. The height of the shell is $f(x) = \sqrt{x}$. The radius of the shell is $x$.
We use the cylindrical shell method.

Step 2: Set up the integral for the shell method.

Step 3: Evaluate the integral.

Answer: $\boxed{\frac{128\pi}{5}}$"
:::

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Summary

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What's Next?

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Part 2: Double and Triple Integrals

Multiple integrals extend the concept of definite integrals to functions of multiple variables, enabling the computation of volumes, areas, masses, and other physical quantities over two- and three-dimensional regions. These tools are fundamental in various fields of mathematics, physics, and engineering, and are frequently assessed in competitive examinations such as CUET PG.

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Core Concepts

1. Double Integrals over Rectangular Regions

We define the double integral of a function $f(x,y)$ over a rectangular region $R = [a,b] \times [c,d]$ as an iterated integral. Fubini's Theorem states that if $f(x,y)$ is continuous on $R$, then the double integral can be evaluated by integrating with respect to one variable at a time, in any order.

Quick Example: Evaluate $\iint_R (x^2y + y^3) \, dA$ where $R = [1,2] \times [0,1]$.

Step 1: Set up the iterated integral. We choose to integrate with respect to $y$ first, then $x$.

>

Step 2: Integrate with respect to $y$.

>

>
>

Step 3: Integrate with respect to $x$.

>

>
>
>
>
>

Answer: $\frac{17}{12}$

:::question type="MCQ" question="Evaluate the double integral $\iint_R 3x^2y \, dA$ over the region $R = [0,1] \times [0,2]$." options=["$0$","$1$","$2$","$3$"] answer="$2$" hint="Apply Fubini's Theorem by integrating with respect to $y$ first, then $x$ (or vice versa)." solution="Step 1: Set up the iterated integral.
>

Step 2: Integrate with respect to $y$.
>
>
>
>
Step 3: Integrate with respect to $x$.
>
>
>
Answer: $\boxed{2}$"
:::

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2. Double Integrals over General Regions

When the region $R$ is not rectangular, we express the limits of integration as functions of the other variable. We classify these regions as Type I or Type II.

Quick Example: Evaluate $\iint_R (x+y) \, dA$ where $R$ is the region bounded by $x=0, x=2, y=x, y=x+2$. (Based on PYQ 1)

Step 1: Sketch the region $R$. The region is bounded by $x=0$ (y-axis), $x=2$ (vertical line), $y=x$ (line through origin), and $y=x+2$ (line with y-intercept 2). This is a Type I region.

> The limits for $x$ are from $0$ to $2$.
> The limits for $y$ are from $x$ to $x+2$.

Step 2: Set up the iterated integral.

>

Step 3: Integrate with respect to $y$.

>

>
>
>
>

Step 4: Integrate with respect to $x$.

>

>
>

Answer: $12$

:::question type="MCQ" question="Evaluate the double integral $\iint_R y \, dA$ where $R$ is the region bounded by $y=x$, $y=2x$, and $x=1$." options=["$1/2$","$1$","$3/2$","$2$"] answer="$1/2$" hint="Sketch the region to determine the integration limits. It is a Type I region." solution="Step 1: Sketch the region. The lines are $y=x$, $y=2x$, and $x=1$.
> The region is bounded by $x=0$ (implicitly, as it starts from the origin) and $x=1$.
> For a given $x$, $y$ ranges from $x$ to $2x$.
Step 2: Set up the iterated integral.
>

Step 3: Integrate with respect to $y$.
>
>
>
>
>
Step 4: Integrate with respect to $x$.
>
>
>
Answer: $\boxed{\frac{1}{2}}$"
:::

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3. Changing the Order of Integration

Sometimes, evaluating an iterated integral is difficult or impossible in the given order of integration. Reversing the order of integration can simplify the problem, especially when the integrand has a non-elementary antiderivative with respect to the inner variable. This requires careful re-identification of the region of integration.

Quick Example: Change the order of integration for $\int_0^2 \int_{y/2}^1 f(x,y) \, dx \, dy$. Then evaluate for $f(x,y) = e^{x^2}$. (Based on PYQ 4)

Step 1: Identify the current region from the limits.

> The outer integral is with respect to $y$, from $0$ to $2$.
> The inner integral is with respect to $x$, from $y/2$ to $1$.
> So, $0 \le y \le 2$ and $y/2 \le x \le 1$.
> This means $x \ge y/2 \implies y \le 2x$.
> The bounding lines are $y=0$, $y=2$, $x=y/2$, $x=1$.
> Plotting these: $x=1$ is a vertical line. $y=0$ is the x-axis. $y=2$ is a horizontal line. $x=y/2$ is equivalent to $y=2x$.
> The intersection of $y=2x$ and $y=2$ is $2=2x \implies x=1$.
> The region is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,2)$.

Step 2: Express the region with the new order (integrate with respect to $y$ first, then $x$).

> For the new order, $x$ will be the outer variable. From the sketch, $x$ ranges from $0$ to $1$.
> For a fixed $x$, $y$ ranges from the lower boundary $y=0$ to the upper boundary $y=2x$.
> So, the new limits are $\int_0^1 \int_0^{2x} f(x,y) \, dy \, dx$.

Step 3: Evaluate the integral with $f(x,y) = e^{x^2}$.

>

>
>
>

Step 4: Use substitution for the final integral. Let $u = x^2$, then $du = 2x \, dx$.
When $x=0$, $u=0$. When $x=1$, $u=1$.

>

>
>

Answer: $e-1$

:::question type="MCQ" question="Change the order of integration for $\int_0^1 \int_y^1 \sin(x^2) \, dx \, dy$ and then evaluate the integral." options=["$1-\cos(1)$","$\frac{1}{2}(1-\cos(1))$","$\cos(1)$","$\sin(1)$"] answer="$\frac{1}{2}(1-\cos(1))$" hint="First sketch the region $0 \le y \le 1, y \le x \le 1$. Then rewrite it as $0 \le x \le 1, 0 \le y \le x$." solution="Step 1: Identify the current region.
> The limits are $0 \le y \le 1$ and $y \le x \le 1$.
> This region is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$.
Step 2: Express the region with the new order ($dy \, dx$).
> The outer integral is with respect to $x$, from $0$ to $1$.
> The inner integral is with respect to $y$, from $0$ to $x$.
> The new integral is $\int_0^1 \int_0^x \sin(x^2) \, dy \, dx$.
Step 3: Evaluate the integral.
>

>
>
Step 4: Use substitution. Let $u = x^2$, then $du = 2x \, dx \implies x \, dx = \frac{1}{2} du$.
When $x=0$, $u=0$. When $x=1$, $u=1$.
>
>
>
>
Answer: $\boxed{\frac{1}{2}(1 - \cos(1))}$"
:::

---

---

4. Area and Volume using Double Integrals

Double integrals can be used to calculate the area of a plane region or the volume of a solid under a surface.

Quick Example: Find the volume of the solid bounded by the paraboloid $z = x^2+y^2$ and the plane $z=4$.

Step 1: Determine the region $R$ in the $xy$-plane.
The solid is bounded above by $z=4$ and below by $z=x^2+y^2$. The intersection of these two surfaces defines the region $R$.

This is a circle of radius $2$ centered at the origin.
So, $R$ is the disk $x^2+y^2 \le 4$.
The volume is given by

Step 2: Set up the integral. Since $R$ is a disk, polar coordinates are appropriate.
Let $x = r \cos \theta$, $y = r \sin \theta$. Then $x^2+y^2 = r^2$, and $dA = r \, dr \, d\theta$.
For the disk $x^2+y^2 \le 4$, $r$ goes from $0$ to $2$, and $\theta$ goes from $0$ to $2\pi$.

Step 3: Integrate with respect to $r$.

Step 4: Integrate with respect to $\theta$.

Answer: $8\pi$

:::question type="MCQ" question="Find the area of the region bounded by $y=x^2$ and $y=x+2$." options=["$9/2$","$11/2$","$13/2$","$15/2$"] answer="$9/2$" hint="First, find the points of intersection to determine the limits for $x$. Then set up the integral

as a Type I region." solution="Step 1: Find the intersection points of $y=x^2$ and $y=x+2$.



So, $x=-1$ and $x=2$.
The region $R$ is bounded by $x=-1$ to $x=2$. For any $x$ in this interval, the upper curve is $y=x+2$ and the lower curve is $y=x^2$.
Step 2: Set up the area integral.

Step 3: Integrate with respect to $y$.


Step 4: Integrate with respect to $x$.







Answer: \boxed{9/2}"
:::

---

5. Triple Integrals over Rectangular Boxes

The triple integral of a function $f(x,y,z)$ over a rectangular box $B = [a,b] \times [c,d] \times [e,f]$ is an iterated integral. Fubini's Theorem extends to three dimensions, allowing integration in any of the $3! = 6$ possible orders if the function is continuous.

Quick Example: Evaluate $\int_0^1 \int_0^1 \int_0^1 ([x]+[y]+[z]) \, dx \, dy \, dz$, where $[x]$ is the greatest integer function. (Based on PYQ 3)

Step 1: Analyze the integrand.
The greatest integer function $[x]$ equals $0$ for $0 \le x < 1$.
In the region of integration $0 \le x \le 1, 0 \le y \le 1, 0 \le z \le 1$:
For $0 \le x < 1$, $[x]=0$. At $x=1$, $[x]=1$.
The integral is defined over a cube. We can evaluate it by summing the individual integrals.
Since the integral is from $0$ to $1$, for almost all points in the domain, $[x]=0, [y]=0, [z]=0$. The points where any variable is exactly $1$ form sets of measure zero, so they do not affect the value of the integral.

Step 2: Evaluate $\int_0^1 [x] \, dx$.
For $0 \le x < 1$, $[x]=0$. So,

Similarly,

and

Step 3: Substitute these values back into the triple integral.

Since $\int_0^1 [x] \, dx = 0$, the first term becomes:

Similarly, the second and third terms are also $0$.
Thus, the total integral is

Answer: $0$

:::question type="MCQ" question="Evaluate

where ." options=["$12$","$15$","$18$","$21$"] answer="$12$" hint="Integrate iteratively. The order of integration does not affect the result for a continuous function over a rectangular box." solution="Step 1: Set up the iterated integral.

Step 2: Integrate with respect to $z$.



Step 3: Integrate with respect to $y$.




Step 4: Integrate with respect to $x$.



Answer: \boxed{12}"
:::

---

6. Triple Integrals over General Regions

For triple integrals over general solids, the limits of integration are functions of the other variables. We typically define the region as "Type 1" in 3D, where $z$ is bounded by two surfaces, and the projection onto the $xy$-plane forms a Type I or Type II region.

Quick Example: Evaluate

where and $V$ is the region bounded by the surfaces $x=0, y=0, x=2, y=4, z=x^2, z=2$. (Based on PYQ 2)

Step 1: Define the region $V$.
The region is bounded by:

This defines a solid where $z$ is bounded by $z=x^2$ (paraboloid) and $z=2$ (plane), and the projection onto the $xy$-plane is a rectangle $D = [0,2] \times [0,4]$.
The integral of a vector field over a volume is defined component-wise:

Step 2: Evaluate the $\hat{i}$ component: $\iiint_V 2z \, dV$.

Step 3: Evaluate the $\hat{j}$ component: $\iiint_V -x \, dV$.

Step 4: Evaluate the $\hat{k}$ component: $\iiint_V y \, dV$.

Step 5: Combine the components.

The option provided in the PYQ is $\frac{32}{15}(3\hat{i} + 5\hat{k})$. Let's factor out $\frac{32}{15}$: This matches.

Answer: \boxed{\frac{32}{15}(3\hat{i} + 5\hat{k})}

:::question type="MCQ" question="Evaluate

where $E$ is the solid tetrahedron bounded by the planes $x=0, y=0, z=0,$ and $x+y+z=1$." options=["$1/12$","$1/24$","$1/48$","$1/60$"] answer="$1/24$" hint="The outer limits for $x$ are from $0$ to $1$. For a given $x$, $y$ varies from $0$ to $1-x$. For given $x,y$, $z$ varies from $0$ to $1-x-y$." solution="Step 1: Define the region $E$.
The base in the $xy$-plane is the triangle bounded by $x=0, y=0,$ and $x+y=1$.
The limits for $x$ are from $0$ to $1$.
The limits for $y$ are from $0$ to $1-x$.
The limits for $z$ are from $0$ to $1-x-y$.
Step 2: Set up the iterated integral.

Step 3: Integrate with respect to $z$.


Step 4: Integrate with respect to $y$. Let $u = 1-x-y$, then $du = -dy$.
When $y=0$, $u=1-x$. When $y=1-x$, $u=0$.



Step 5: Integrate with respect to $x$. Let $v = 1-x$, then $dv = -dx$.
When $x=0$, $v=1$. When $x=1$, $v=0$.



Answer: \boxed{1/24}"
:::

---

7. Volume using Triple Integrals

The volume of a solid region $E$ can be found by integrating the constant function $1$ over the region.

Quick Example: Find the volume of the solid bounded by the cylinder $x^2+y^2=4$ and the planes $z=0$ and $z=x+3$.

Step 1: Define the region $E$.
The base is the disk $D: x^2+y^2 \le 4$ in the $xy$-plane.
The lower bound for $z$ is $z=0$.
The upper bound for $z$ is $z=x+3$.
Step 2: Set up the integral.

Step 3: Convert to polar coordinates for the double integral over $D$.
$x = r \cos \theta$, $dA = r \, dr \, d\theta$.
For the disk $x^2+y^2 \le 4$, $r$ ranges from $0$ to $2$ and $\theta$ ranges from $0$ to $2\pi$.


Step 4: Integrate with respect to $r$.



Step 5: Integrate with respect to $\theta$.

Answer: $12\pi$

:::question type="MCQ" question="Find the volume of the solid bounded by the planes $z=x+y$, $z=0$, $x=0$, $y=0$, and $x+y=1$." options=["$1/6$","$1/3$","$1/2$","$1$"] answer="$1/3$" hint="The base is a triangle in the $xy$-plane bounded by $x=0, y=0, x+y=1$. The height is given by $z=x+y$." solution="Step 1: Define the region $E$.
The base $D$ is the triangle in the $xy$-plane with vertices $(0,0), (1,0), (0,1)$.
For this region, $0 \le x \le 1$ and $0 \le y \le 1-x$.
The lower bound for $z$ is $z=0$.
The upper bound for $z$ is $z=x+y$.
Step 2: Set up the integral.

Step 3: Integrate with respect to $z$.


Step 4: Integrate with respect to $y$.





Step 5: Integrate with respect to $x$.



Answer: \boxed{1/3}"
:::

---

Advanced Applications: Change of Variables

8. The Jacobian Transformation

To change variables in multiple integrals, we use the Jacobian determinant to account for the scaling factor in the area or volume element.

Quick Example: To evaluate the double integral

, we make the substitution and . Transform the integral. (Based on PYQ 5 & 7)

Step 1: Express $x$ and $y$ in terms of $u$ and $v$.
From $v = y/2$, we get $y = 2v$.
Substitute $y=2v$ into $u = \frac{2x-y}{2}$:

So, the transformation is $x=u+v, y=2v$.

Step 2: Calculate the Jacobian $J(u,v)$.

So,

Step 3: Transform the region of integration.
The original region $R$ is defined by:

Let's find the new bounds in the $uv$-plane ($S$).
For $y$:

For $x$:
Lower bound:

Upper bound:

So, the new region $S$ is $0 \le v \le 4, 0 \le u \le 1$.

Step 4: Transform the integrand.
The integrand is

Step 5: Write the transformed integral.

Answer: \boxed{\int_0^4 \left( \int_0^1 2u \, du \right) \, dv}

:::question type="MCQ" question="Given the integral

. Use the transformation and to rewrite the integral. The new region $S$ is a triangle with vertices $(0,0), (1,1), (1,-1)$. The transformed integral is:" options=["$\int_0^1 \int_{-u}^u \frac{1}{2} e^{u/v} \, dv \, du$","$\int_0^1 \int_{-u}^u \frac{1}{2} e^{u/v} \, du \, dv$","$\int_0^1 \int_{-v}^v \frac{1}{2} e^{u/v} \, du \, dv$","$\int_0^1 \int_0^v \frac{1}{2} e^{u/v} \, du \, dv$"] answer="$\int_0^1 \int_{-u}^u \frac{1}{2} e^{u/v} \, dv \, du$" hint="First, find $x$ and $y$ in terms of $u$ and $v$. Calculate the Jacobian. Transform the region defined by $x=0, y=0, x+y=1$ into the $uv$-plane." solution="Step 1: Express $x, y$ in terms of $u, v$.


Adding the equations:

Subtracting the equations:

Step 2: Calculate the Jacobian $J(u,v)$.



So,

Step 3: Transform the region of integration $R$.
The original region $R$ is defined by $x=0, y=0, x+y=1$.



The new region $S$ in the $uv$-plane is bounded by $v=-u$, $v=u$, and $u=1$.
This is a triangle with vertices $(0,0)$ (from $v=-u$ and $v=u$), $(1,1)$ (from $u=1$ and $v=u$), and $(1,-1)$ (from $u=1$ and $v=-u$).
To integrate over this region, $u$ ranges from $0$ to $1$. For a given $u$, $v$ ranges from $-u$ to $u$.
Step 4: Transform the integrand.

Step 5: Write the transformed integral.

Answer: \boxed{\int_0^1 \int_{-u}^u \frac{1}{2} e^{u/v} \, dv \, du}"
:::

---

9. Double Integrals in Polar Coordinates

Polar coordinates are particularly useful for integrals over circular regions or those with rotational symmetry.

Quick Example: Evaluate

. (Based on PYQ 8, related to PYQ 6)

Step 1: Identify the region of integration.
The integral is over the entire first quadrant ($x \ge 0, y \ge 0$).

Step 2: Convert to polar coordinates.

For the first quadrant, $r$ ranges from $0$ to $\infty$, and $\theta$ ranges from $0$ to $\pi/2$.

Step 3: Set up the integral in polar coordinates.

Step 4: Integrate with respect to $r$.
Let $u = -r^2$, then $du = -2r \, dr \implies r \, dr = -\frac{1}{2} du$.
When $r=0$, $u=0$. When $r=\infty$, $u=-\infty$.

Step 5: Integrate with respect to $\theta$.

Answer: \boxed{\frac{\pi}{4}}

:::question type="MCQ" question="Evaluate

where $R$ is the region bounded by $x^2+y^2=1$ and $x^2+y^2=4$." options=["$14\pi/3$","$16\pi/3$","$18\pi/3$","$20\pi/3$"] answer="$14\pi/3$" hint="The region is an annulus. Convert to polar coordinates." solution="Step 1: Identify the region of integration.
The region $R$ is an annulus between circles of radius $1$ and $2$ centered at the origin.
Step 2: Convert to polar coordinates.

So


For the annulus, $r$ ranges from $1$ to $2$, and $\theta$ ranges from $0$ to $2\pi$.
Step 3: Set up the integral in polar coordinates.


Step 4: Integrate with respect to $r$.




Step 5: Integrate with respect to $\theta$.


Answer: \boxed{14\pi/3}"
:::

---

10. Triple Integrals in Cylindrical Coordinates

Cylindrical coordinates are useful for integrating over regions that have cylindrical symmetry or when the projection onto the $xy$-plane is a disk or annulus.

Quick Example: Find the volume of the solid bounded by the paraboloid $z = x^2+y^2$ and the plane $z=9$.

Step 1: Define the region $E$.
The solid is bounded below by $z=x^2+y^2$ and above by $z=9$.
The projection onto the $xy$-plane is the disk $D: x^2+y^2 \le 9$.

Step 2: Convert to cylindrical coordinates.
> $x^2+y^2 = r^2$.
> $z=r^2$ (lower bound for $z$).
> $z=9$ (upper bound for $z$).
> For the disk $x^2+y^2 \le 9$, $0 \le r \le 3$ and $0 \le \theta \le 2\pi$.
> $dV = r \, dz \, dr \, d\theta$.

Step 3: Set up the integral.
>

Step 4: Integrate with respect to $z$.

>

>
>

Step 5: Integrate with respect to $r$.

>

>
>
>
>

Step 6: Integrate with respect to $\theta$.

>

>

Answer: $\boxed{\frac{81\pi}{2}}$

:::question type="MCQ" question="Evaluate $\iiint_E z \, dV$ where $E$ is the region above the cone $z=\sqrt{x^2+y^2}$ and below the plane $z=1$." options=["$\pi/6$","$\pi/4$","$\pi/3$","$\pi/2$"] answer="$\pi/4$" hint="The region is a cone with its tip at the origin and base at $z=1$. Convert to cylindrical coordinates." solution="Step 1: Define the region $E$.
> The lower bound for $z$ is $z=\sqrt{x^2+y^2}$.
> The upper bound for $z$ is $z=1$.
> The projection onto the $xy$-plane is where $\sqrt{x^2+y^2}=1$, which is the disk $x^2+y^2 \le 1$.
Step 2: Convert to cylindrical coordinates.
> $z=\sqrt{x^2+y^2} \implies z=r$. So $r \le z \le 1$.
> For the disk $x^2+y^2 \le 1$, $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.
> $dV = r \, dz \, dr \, d\theta$.
Step 3: Set up the integral.
>

Step 4: Integrate with respect to $z$.
>
>
>
Step 5: Integrate with respect to $r$.
>
>
>
>
Step 6: Integrate with respect to $\theta$.
>
>
Answer: $\boxed{\frac{\pi}{4}}$"
:::

---

11. Triple Integrals in Spherical Coordinates

Spherical coordinates are ideal for regions with spherical symmetry, such as spheres, cones, or when the integrand involves $x^2+y^2+z^2$.

Quick Example: Given $E: x^2 + y^2 + z^2 = a^2$, evaluate $\iiint_E \frac{dx \, dy \, dz}{x^2 + y^2 + z^2}$. (Based on PYQ 11)

Step 1: Define the region $E$.
The region $E$ is the solid sphere of radius $a$ centered at the origin.
In spherical coordinates:
> $0 \le \rho \le a$
> $0 \le \phi \le \pi$
> $0 \le \theta \le 2\pi$

Step 2: Convert the integrand and volume element to spherical coordinates.
> $x^2+y^2+z^2 = \rho^2$.
> $dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.
> The integrand becomes $\frac{1}{\rho^2}$.

Step 3: Set up the integral.

>

>

Step 4: Integrate with respect to $\rho$.

>

>

Step 5: Integrate with respect to $\phi$.

>

>
>
>
>

Step 6: Integrate with respect to $\theta$.

>

>

Answer: $\boxed{4\pi a}$

:::question type="MCQ" question="Find the volume of the solid bounded by the sphere $x^2+y^2+z^2=R^2$ using spherical coordinates." options=["$\frac{2}{3}\pi R^3$","$\frac{4}{3}\pi R^3$","$\pi R^3$","$2\pi R^3$"] answer="$\frac{4}{3}\pi R^3$" hint="Integrate $1$ over the spherical region in spherical coordinates." solution="Step 1: Define the region $E$.
> The solid is a sphere of radius $R$ centered at the origin.
> In spherical coordinates: $0 \le \rho \le R$, $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$.
Step 2: Set up the integral for volume.
>

Step 3: Integrate with respect to $\rho$.
>
>
Step 4: Integrate with respect to $\phi$.
>
>
>
>
>
Step 5: Integrate with respect to $\theta$.
>
>
Answer: $\boxed{\frac{4}{3}\pi R^3}$"
:::

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Problem-Solving Strategies

---

Common Mistakes

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Practice Questions

:::question type="MCQ" question="Evaluate $\iint_R (x^2+y^2) \, dA$ over the region $R$ bounded by $y=x$, $y=-x$, and $x=1$." options=["$1/3$","$2/3$","$1$","$4/3$"] answer="$2/3$" hint="The region is a triangle in the first and fourth quadrants. Use Cartesian coordinates first, then consider polar to confirm." solution="Step 1: Sketch the region. The region is a triangle with vertices $(0,0)$, $(1,1)$, and $(1,-1)$.
> This region is defined by $0 \le x \le 1$ and $-x \le y \le x$.
Step 2: Set up the iterated integral.
>

Step 3: Integrate with respect to $y$.
>
>
>
>
>
Step 4: Integrate with respect to $x$.
>
>
>
Answer: $\boxed{\frac{2}{3}}$"
:::

:::question type="NAT" question="Find the volume of the solid bounded by the cylinder $y^2+z^2=9$ and the planes $x=0, x=5$." answer="45π" hint="The base is a circle in the $yz$-plane. The height is along the $x$-axis. Use cylindrical coordinates for $y,z$ and Cartesian for $x$." solution="Step 1: Define the region $E$.
> The cylinder $y^2+z^2=9$ has radius $3$ and its axis is the $x$-axis.
> The planes $x=0$ and $x=5$ define the extent along the $x$-axis.
> So, $0 \le x \le 5$.
> The cross-section in the $yz$-plane is a disk $y^2+z^2 \le 9$.
Step 2: Set up the integral. We are finding the volume, so we integrate $1$.
>

> Where $D_{yz}$ is the disk $y^2+z^2 \le 9$.
The inner double integral $\iint_{D_{yz}} 1 \, dy \, dz$ represents the area of the disk, which is $\pi (3^2) = 9\pi$.
>
Step 3: Integrate with respect to $x$.
>
>
Answer: $\boxed{45\pi}$"
:::

:::question type="MCQ" question="The value of the integral $\int_0^1 \int_0^{\sqrt{1-y^2}} (x^2+y^2) \, dx \, dy$ is:" options=["$\pi/8$","$\pi/4$","$\pi/2$","$\pi$"] answer="$\pi/8$" hint="Identify the region of integration. It is a quarter circle. Convert to polar coordinates." solution="Step 1: Identify the region of integration.
> The limits are $0 \le y \le 1$ and $0 \le x \le \sqrt{1-y^2}$.
> The boundary $x=\sqrt{1-y^2}$ implies $x^2=1-y^2$, or $x^2+y^2=1$.
> Since $x \ge 0$ and $y \ge 0$, this is the quarter circle in the first quadrant of radius $1$.
Step 2: Convert to polar coordinates.
> $x^2+y^2 = r^2$.
> $dx \, dy = r \, dr \, d\theta$.
> For the quarter circle in the first quadrant, $0 \le r \le 1$ and $0 \le \theta \le \pi/2$.
Step 3: Set up the integral in polar coordinates.
>

>
Step 4: Integrate with respect to $r$.
>
>
>
Step 5: Integrate with respect to $\theta$.
>
>
Answer: $\boxed{\frac{\pi}{8}}$"
:::

:::question type="MCQ" question="Which of the following expressions represents the volume of the solid bounded by the sphere $x^2+y^2+z^2=a^2$ and the cone $z=\sqrt{x^2+y^2}$?" options=["$\int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$","$\int_0^{2\pi} \int_0^{\pi/4} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$","$\int_0^{2\pi} \int_0^{\pi/2} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$","$\int_0^{2\pi} \int_0^{\pi/4} \int_0^{\rho} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$"] answer="$\int_0^{2\pi} \int_0^{\pi/4} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$" hint="Convert the sphere and cone equations to spherical coordinates to find the limits for $\rho, \phi, \theta$." solution="Step 1: Convert the equations to spherical coordinates.
> Sphere: $x^2+y^2+z^2=a^2 \implies \rho^2=a^2 \implies \rho=a$.
> Cone: $z=\sqrt{x^2+y^2}$. We know $x^2+y^2 = \rho^2 \sin^2 \phi$ and $z=\rho \cos \phi$.
> So, $\rho \cos \phi = \sqrt{\rho^2 \sin^2 \phi} = \rho |\sin \phi|$.
> Since $0 \le \phi \le \pi$, $\sin \phi \ge 0$, so $\rho \cos \phi = \rho \sin \phi$.
> This implies $\cos \phi = \sin \phi$. For $0 \le \phi \le \pi$, this occurs at $\phi = \pi/4$.
Step 2: Determine the limits of integration.
> The solid is bounded above by the sphere $\rho=a$. So $0 \le \rho \le a$.
> The solid is bounded below by the cone. The cone $z=\sqrt{x^2+y^2}$ makes an angle of $\pi/4$ with the positive $z$-axis. The region above the cone means $\phi$ ranges from $0$ (positive $z$-axis) to $\pi/4$. So $0 \le \phi \le \pi/4$.
> The solid is symmetric around the $z$-axis, so $\theta$ ranges from $0$ to $2\pi$.
Step 3: Set up the integral.
> The volume element in spherical coordinates is $dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.
>

Answer: $\boxed{\int_0^{2\pi} \int_0^{\pi/4} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta}$"
:::

:::question type="MCQ" question="Evaluate $\iint_R 2xy \, dA$ where $R$ is the region in the first quadrant bounded by the lines $y=x$, $y=2x$, $x=1$, $x=2$." options=["$27/4$","$33/4$","$39/4$","$45/4$"] answer="$45/4$" hint="Set up the integral as a Type I region. The limits for $x$ are constants, and $y$ is bounded by functions of $x$." solution="Step 1: Define the region $R$.
> $1 \le x \le 2$
> $x \le y \le 2x$
Step 2: Set up the iterated integral.
>

Step 3: Integrate with respect to $y$.
>
>
>
>
Step 4: Integrate with respect to $x$.
>
>
>
>
Answer: $\boxed{\frac{45}{4}}$"
:::

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Summary

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What's Next?

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Part 3: Change of Order of Integration

The ability to change the order of integration is a fundamental skill in multivariable calculus, often simplifying the evaluation of double integrals or enabling their computation when direct integration is intractable. We explore the systematic approach to redefine integration limits over a given region.

---

Core Concepts

1. Understanding the Region of Integration

When evaluating a double integral $\iint_R f(x,y)\,dA$, the region $R$ is defined by the limits of integration. We typically express these limits in one of two forms:

Type I Region:

leading to

Here, $y$ varies between functions of $x$, and $x$ varies between constants.

Type II Region:

leading to

Here, $x$ varies between functions of $y$, and $y$ varies between constants.

The process of changing the order of integration involves converting a Type I region into a Type II region, or vice versa, by carefully re-describing the same geometric area.

Quick Example:
Consider the integral $\int_0^1 \int_0^x f(x,y)\,dy\,dx$.
The region of integration is defined by $0 \le x \le 1$ and $0 \le y \le x$.
This describes a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,1)$.

:::question type="MCQ" question="Which of the following describes the region of integration for the integral $\int_0^2 \int_{x}^{2} f(x,y)\,dy\,dx$?" options=["A rectangle bounded by $x=0, x=2, y=0, y=2$","A triangle with vertices $(0,0), (2,0), (2,2)$","A triangle with vertices $(0,0), (0,2), (2,2)$","A trapezoid bounded by $x=0, x=2, y=x, y=2$"] answer="A triangle with vertices $(0,0), (0,2), (2,2)$" hint="Sketch the region defined by $0 \le x \le 2$ and $x \le y \le 2$." solution="The outer limits $0 \le x \le 2$ define the horizontal extent. The inner limits $x \le y \le 2$ define the vertical extent.
For a given $x$, $y$ varies from $x$ to $2$.
When $x=0$, $y$ varies from $0$ to $2$.
When $x=2$, $y$ varies from $2$ to $2$.
The boundaries are $x=0$, $y=2$, and $y=x$.
The vertices are $(0,0)$, $(0,2)$, and $(2,2)$. This forms a triangle."
:::

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2. Changing Order from $dy\,dx$ to $dx\,dy$

To change the order of integration from

to

Identify the current limits: $x$ varies from $a$ to $b$, and $y$ varies from $g_1(x)$ to $g_2(x)$.

Sketch the region: Plot the curves $y=g_1(x)$, $y=g_2(x)$, $x=a$, and $x=b$.

Determine new $y$-limits (constants): Find the minimum ($c$) and maximum ($d$) values that $y$ takes over the entire region.

Determine new $x$-limits (functions of $y$): For a fixed $y$ between $c$ and $d$, draw a horizontal strip across the region. The left boundary of this strip will be $x=h_1(y)$ and the right boundary will be $x=h_2(y)$. We must express these boundary curves as $x$ in terms of $y$.

Worked Example 1:
Change the order of integration for

Step 1: Identify current limits.
The region is $0 \le x \le 2$, $0 \le y \le x$.
This is a triangle with vertices $(0,0)$, $(2,0)$, and $(2,2)$.

Step 2: Sketch the region.
The region is bounded by $y=0$ (x-axis), $x=2$, and $y=x$.

Step 3: Determine new $y$-limits (constants).
From the sketch, $y$ varies from $0$ to $2$. So, $c=0$, $d=2$.

Step 4: Determine new $x$-limits (functions of $y$).
For a fixed $y$ between $0$ and $2$, a horizontal strip starts at $x=y$ (from the line $y=x$) and ends at $x=2$ (from the line $x=2$).
Thus, $y \le x \le 2$.

Answer: The integral with changed order is

:::question type="MCQ" question="The integral $\int_{0}^{1} \int_{0}^{\sqrt{x}} f(x, y) \,dy\,dx$ with the order of integration changed is:" options=["

","","",""] answer="" hint="The region is $0 \le x \le 1$ and $0 \le y \le \sqrt{x}$. Sketch this region and identify new bounds." solution="Step 1: Identify current limits.
The region is $0 \le x \le 1$, $0 \le y \le \sqrt{x}$.
The boundaries are $y=0$, $x=1$, and $y=\sqrt{x}$ (or $x=y^2$).
The vertices are $(0,0)$, $(1,0)$, and $(1,1)$.

Step 2: Sketch the region.
The region is bounded by $y=0$ (x-axis), $x=1$, and the parabola $y=\sqrt{x}$ (upper branch).

Step 3: Determine new $y$-limits (constants).
From the sketch, $y$ varies from $0$ to $1$.

Step 4: Determine new $x$-limits (functions of $y$).
For a fixed $y$ between $0$ and $1$, a horizontal strip starts at $x=y^2$ (from the curve $y=\sqrt{x}$) and ends at $x=1$.
Thus, $y^2 \le x \le 1$.

Result: The integral becomes

"
:::

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3. Changing Order from $dx\,dy$ to $dy\,dx$

To change the order of integration from

to

Identify the current limits: $y$ varies from $c$ to $d$, and $x$ varies from $h_1(y)$ to $h_2(y)$.

Sketch the region: Plot the curves $x=h_1(y)$, $x=h_2(y)$, $y=c$, and $y=d$.

Determine new $x$-limits (constants): Find the minimum ($a$) and maximum ($b$) values that $x$ takes over the entire region.

Determine new $y$-limits (functions of $x$): For a fixed $x$ between $a$ and $b$, draw a vertical strip across the region. The lower boundary of this strip will be $y=g_1(x)$ and the upper boundary will be $y=g_2(x)$. We must express these boundary curves as $y$ in terms of $x$.

Worked Example 2:
Change the order of integration for

Step 1: Identify current limits.
The region is $0 \le y \le 1$, $y \le x \le 1$.
This is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$.

Step 2: Sketch the region.
The region is bounded by $y=0$ (x-axis), $x=1$, and $y=x$.

Step 3: Determine new $x$-limits (constants).
From the sketch, $x$ varies from $0$ to $1$. So, $a=0$, $b=1$.

Step 4: Determine new $y$-limits (functions of $x$).
For a fixed $x$ between $0$ and $1$, a vertical strip starts at $y=0$ (x-axis) and ends at $y=x$ (from the line $y=x$).
Thus, $0 \le y \le x$.

Answer: The integral with changed order is

:::question type="MCQ" question="The integral $\int_{0}^{2} \int_{y/2}^{1} f(x, y) \,dx\,dy$ with the order of integration changed is:" options=["

","","",""] answer="" hint="The region is $0 \le y \le 2$ and $y/2 \le x \le 1$. Sketch and re-evaluate bounds." solution="Step 1: Identify current limits.
The region is $0 \le y \le 2$, $y/2 \le x \le 1$.
The boundaries are $y=0$, $x=1$, and $x=y/2$ (or $y=2x$).
The vertices are $(0,0)$, $(1,0)$, and $(1,2)$.

Step 2: Sketch the region.
The region is bounded by $y=0$ (x-axis), $x=1$, and the line $y=2x$.

Step 3: Determine new $x$-limits (constants).
From the sketch, $x$ varies from $0$ to $1$.

Step 4: Determine new $y$-limits (functions of $x$).
For a fixed $x$ between $0$ and $1$, a vertical strip starts at $y=0$ (x-axis) and ends at $y=2x$ (from the line $y=2x$).
Thus, $0 \le y \le 2x$.

Result: The integral becomes

"
:::

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Advanced Applications

Sometimes the region of integration is more complex, requiring careful consideration of intersection points or even splitting the region into multiple sub-regions to change the order effectively.

Worked Example 3: Region bounded by a parabola and a line
Change the order of integration for $\int_0^1 \int_{x^2}^x f(x,y)\,dy\,dx$.

Step 1: Identify current limits.
The region is $0 \le x \le 1$, $x^2 \le y \le x$.
The boundaries are $y=x^2$ and $y=x$.
Intersection points: $x^2=x \implies x(x-1)=0 \implies x=0, x=1$.
So, $(0,0)$ and $(1,1)$ are the intersection points.

Step 2: Sketch the region.
The region is bounded below by the parabola $y=x^2$ and above by the line $y=x$.

Step 3: Determine new $y$-limits (constants).
From the sketch, $y$ varies from $0$ to $1$.

Step 4: Determine new $x$-limits (functions of $y$).
For a fixed $y$ between $0$ and $1$, a horizontal strip starts at $x=y$ (from the line $y=x$) and ends at $x=\sqrt{y}$ (from the parabola $y=x^2$).
Thus, $y \le x \le \sqrt{y}$.

Answer: The integral with changed order is $\int_0^1 \int_y^{\sqrt{y}} f(x,y)\,dx\,dy$.

:::question type="NAT" question="Evaluate the integral $\int_{0}^{2} \int_{x}^{2} e^{y^2} \,dy\,dx$ by changing the order of integration. (Provide the numerical answer rounded to two decimal places.)" answer="26.80" hint="First, change the order of integration. The original region is $0 \le x \le 2$ and $x \le y \le 2$. Sketch this region to find the new bounds. Then, evaluate the integral." solution="Step 1: Identify the region of integration.
The given integral is $\int_{0}^{2} \int_{x}^{2} e^{y^2} \,dy\,dx$.
The limits define the region $R = \{(x,y) \mid 0 \le x \le 2, x \le y \le 2\}$.
This is a triangle with vertices $(0,0)$, $(0,2)$, and $(2,2)$.
The boundaries are $x=0$, $y=2$, and $y=x$.

Step 2: Change the order of integration to $dx\,dy$.
For the new outer limits (constant $y$), $y$ varies from $0$ to $2$.
For the new inner limits (functions of $y$), for a fixed $y$, $x$ varies from $0$ to $y$.
So, $0 \le y \le 2$ and $0 \le x \le y$.
The integral becomes $\int_{0}^{2} \int_{0}^{y} e^{y^2} \,dx\,dy$.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the outer integral.
Now we integrate $y e^{y^2}$ with respect to $y$ from $0$ to $2$.

Let $u = y^2$. Then $du = 2y\,dy$, so $y\,dy = \frac{1}{2}\,du$.
When $y=0$, $u=0^2=0$.
When $y=2$, $u=2^2=4$.
The integral becomes:

Using $e \approx 2.71828$:
$e^{4} \approx 54.59815$
Value $\approx \frac{1}{2} (54.59815 - 1) = \frac{1}{2} (53.59815) = 26.799075$.
Rounding to two decimal places, the answer is $26.80$.

Answer: \boxed{26.80}"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The double integral $\int_{0}^{1} \int_{y}^{1} \sin(x^2)\,dx\,dy$ is equivalent to which of the following after changing the order of integration?" options=["$\int_{0}^{1} \int_{0}^{x} \sin(x^2)\,dy\,dx$","$\int_{0}^{1} \int_{x}^{1} \sin(x^2)\,dy\,dx$","$\int_{0}^{1} \int_{0}^{y} \sin(x^2)\,dy\,dx$","$\int_{y}^{1} \int_{0}^{1} \sin(x^2)\,dy\,dx$"] answer="$\int_{0}^{1} \int_{0}^{x} \sin(x^2)\,dy\,dx$" hint="Sketch the region $0 \le y \le 1, y \le x \le 1$. Identify the vertices and re-describe the region with $y$ as the inner variable." solution="Step 1: Identify the current region.
The integral is $\int_{0}^{1} \int_{y}^{1} \sin(x^2)\,dx\,dy$.
The region is defined by $0 \le y \le 1$ and $y \le x \le 1$.
This is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. The boundaries are $y=0$, $x=1$, and $y=x$.

Step 2: Change the order to $dy\,dx$.
For the outer limits (constant $x$), $x$ varies from $0$ to $1$.
For the inner limits (functions of $x$), for a fixed $x$, $y$ varies from $0$ (the x-axis) to $x$ (the line $y=x$).
So, $0 \le x \le 1$ and $0 \le y \le x$.

Result: The equivalent integral is $\int_{0}^{1} \int_{0}^{x} \sin(x^2)\,dy\,dx$."
:::

:::question type="NAT" question="Evaluate the integral $\int_{0}^{1} \int_{y}^{1} \frac{\sin x}{x} \,dx\,dy$ by changing the order of integration. (Provide the numerical answer rounded to three decimal places.)" answer="0.459" hint="The inner integral $\int \frac{\sin x}{x} \,dx$ cannot be expressed in elementary functions. Change the order to make the integration feasible. Sketch the region first." solution="Step 1: Identify the current region.
The integral is $\int_{0}^{1} \int_{y}^{1} \frac{\sin x}{x} \,dx\,dy$.
The region is $0 \le y \le 1$ and $y \le x \le 1$.
This forms a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. The boundaries are $y=0$, $x=1$, and $y=x$.

Step 2: Change the order of integration to $dy\,dx$.
For the new outer limits (constant $x$), $x$ varies from $0$ to $1$.
For the new inner limits (functions of $x$), for a fixed $x$, $y$ varies from $0$ to $x$.
So, $0 \le x \le 1$ and $0 \le y \le x$.
The integral becomes $\int_{0}^{1} \int_{0}^{x} \frac{\sin x}{x} \,dy\,dx$.

Step 3: Evaluate the inner integral.

Since $\frac{\sin x}{x}$ is constant with respect to $y$,

Step 4: Evaluate the outer integral.

Rounding to three decimal places, the answer is $0.459$.

Answer: \boxed{0.459}"
:::

:::question type="MCQ" question="Which of the following integrals is equivalent to $\int_{0}^{4} \int_{0}^{\sqrt{y}} f(x, y) \,dx\,dy$?" options=["$\int_{0}^{2} \int_{x^2}^{4} f(x, y) \,dy\,dx$","$\int_{0}^{2} \int_{4}^{x^2} f(x, y) \,dy\,dx$","$\int_{0}^{4} \int_{x^2}^{4} f(x, y) \,dy\,dx$","$\int_{0}^{4} \int_{0}^{x^2} f(x, y) \,dy\,dx$"] answer="$\int_{0}^{2} \int_{x^2}^{4} f(x, y) \,dy\,dx$" hint="The region is $0 \le y \le 4$ and $0 \le x \le \sqrt{y}$. Sketch this region and change the order." solution="Step 1: Identify the current region.
The integral is $\int_{0}^{4} \int_{0}^{\sqrt{y}} f(x, y) \,dx\,dy$.
The region is $0 \le y \le 4$ and $0 \le x \le \sqrt{y}$.
The boundaries are $x=0$ (y-axis), $y=4$, and $x=\sqrt{y}$ (or $y=x^2$).
The intersection points are $(0,0)$ and $(2,4)$.

Step 2: Sketch the region.
The region is bounded by the y-axis, the line $y=4$, and the parabola $y=x^2$ (for $x \ge 0$).

Step 3: Change the order to $dy\,dx$.
For the outer limits (constant $x$), $x$ varies from $0$ to $2$.
For the inner limits (functions of $x$), for a fixed $x$, $y$ varies from $x^2$ (the parabola) to $4$ (the line $y=4$).
So, $0 \le x \le 2$ and $x^2 \le y \le 4$.

Result: The equivalent integral is $\int_{0}^{2} \int_{x^2}^{4} f(x, y) \,dy\,dx$."
:::

:::question type="MSQ" question="Select ALL the double integrals that represent the area of the region bounded by $y=x^2$ and $y=x$." options=["$\int_{0}^{1} \int_{x^2}^{x} \,dy\,dx$","$\int_{0}^{1} \int_{y}^{\sqrt{y}} \,dx\,dy$","$\int_{0}^{1} \int_{0}^{x} \,dy\,dx + \int_{0}^{1} \int_{x^2}^{0} \,dy\,dx$","$\int_{0}^{1} (x - x^2) \,dx$"] answer="$\int_{0}^{1} \int_{x^2}^{x} \,dy\,dx,\int_{0}^{1} \int_{y}^{\sqrt{y}} \,dx\,dy$" hint="First, find the intersection points of $y=x^2$ and $y=x$. Then, set up the double integral for area in both $dy\,dx$ and $dx\,dy$ orders. Remember that the area is $\iint_R 1 \,dA$." solution="Step 1: Find intersection points.
Set $x^2 = x$. This gives $x^2 - x = 0 \implies x(x-1) = 0$.
So, $x=0$ and $x=1$. The intersection points are $(0,0)$ and $(1,1)$.

Step 2: Define the region.
Between $x=0$ and $x=1$, we observe that $x \ge x^2$.
So, the region is bounded below by $y=x^2$ and above by $y=x$.

Step 3: Set up integral in $dy\,dx$ order.
For $dy\,dx$, $x$ varies from $0$ to $1$. For a given $x$, $y$ varies from $x^2$ to $x$.

This is a correct representation.

Step 4: Set up integral in $dx\,dy$ order.
For $dx\,dy$, $y$ varies from $0$ to $1$. For a given $y$, we need to express $x$ in terms of $y$.
From $y=x^2$, we get $x=\sqrt{y}$ (since $x \ge 0$ in this region).
From $y=x$, we get $x=y$.
For a given $y$, $x$ varies from $y$ (the line) to $\sqrt{y}$ (the parabola).

This is also a correct representation.

Step 5: Evaluate other options.
Option 3: $\int_{0}^{1} \int_{0}^{x} \,dy\,dx + \int_{0}^{1} \int_{x^2}^{0} \,dy\,dx$. This is not the correct region. The second integral has an upper limit of $0$ which is below $x^2$ for $x \in (0,1]$. This would represent a different, potentially negative, area.
Option 4: $\int_{0}^{1} (x - x^2) \,dx$. This is a single integral representing the area, but the question asks for double integrals.

Conclusion: The correct options are $\int_{0}^{1} \int_{x^2}^{x} \,dy\,dx$ and $\int_{0}^{1} \int_{y}^{\sqrt{y}} \,dx\,dy$."
:::

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Summary

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What's Next?

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Part 4: Applications of Multiple Integrals

Multiple integrals extend the concept of definite integrals to functions of multiple variables, enabling us to compute quantities such as areas, volumes, masses, and centroids of complex regions in higher dimensions. These tools are fundamental in various scientific and engineering disciplines, and their application is crucial for solving problems in geometry and physics.

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Core Concepts

1. Area using Double Integrals

We define the area of a bounded region $R$ in the plane as the double integral of $1$ over that region. This approach is versatile, allowing us to calculate areas of regions that are not easily expressible using single integrals.

Quick Example (Cartesian):
We calculate the area of the region bounded by $y = x^2$ and $y = x$.

Step 1: Determine intersection points.

The intersection points are $(0,0)$ and $(1,1)$.

Step 2: Set up the double integral.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the outer integral.

Answer: $\frac{1}{6}$ square units.

:::question type="MCQ" question="Determine the area of the region bounded by $y = x^3$, $y = 0$, and $x = 2$ in the first quadrant." options=["$2$","$4$","$8$","$16$"] answer="$4$" hint="Sketch the region and set up the appropriate double integral with $y$ as the inner integral." solution="Step 1: Identify the region. The region is bounded by $y=x^3$ from above, $y=0$ from below, and $x=2$ from the right, starting from $x=0$.
Step 2: Set up the double integral.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the outer integral.


Answer: $\boxed{4}$ square units."
:::

1.2. Area in Polar Coordinates

When a region is more easily described using polar coordinates $(r, \theta)$, we transform the area element. The area element $dA$ in polar coordinates becomes $r\,dr\,d\theta$.

Quick Example (Polar):
We find the area enclosed by the cardioid $r = 1 + \cos\theta$.

Step 1: Determine the limits of integration.
The cardioid is symmetric about the polar axis. It completes one loop as $\theta$ goes from $0$ to $2\pi$.
The radial limit is from $r=0$ to $r=1+\cos\theta$.

Step 2: Set up the double integral.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the outer integral.

We use the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$.

Answer: $\frac{3\pi}{2}$ square units.

:::question type="MCQ" question="Find the area of one petal of the rose curve $r = \sin(3\theta)$." options=["$\frac{\pi}{12}$","$\frac{\pi}{6}$","$\frac{\pi}{3}$","$\frac{\pi}{2}$"] answer="$\frac{\pi}{12}$" hint="For $r = \sin(n\theta)$, a single petal spans an angle of $\pi/n$. For $n=3$, a petal spans $\pi/3$. The curve starts and ends at $r=0$ for $\theta=0$ and $\theta=\pi/3$ (or other intervals for other petals)." solution="Step 1: Determine the limits for one petal. For $r = \sin(3\theta)$, one petal is traced as $3\theta$ goes from $0$ to $\pi$, so $\theta$ goes from $0$ to $\pi/3$.
Step 2: Set up the double integral.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the outer integral.

Using the identity $\sin^2 x = \frac{1-\cos(2x)}{2}$:





Answer: $\boxed{\frac{\pi}{12}}$ square units."
:::

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2. Volume using Triple Integrals

We extend the concept of integration to three dimensions to calculate the volume of a solid region $E$. The volume element $dV$ can be expressed in Cartesian, cylindrical, or spherical coordinates.

Quick Example (Cartesian):
We find the volume of the tetrahedron bounded by the coordinate planes and the plane $x+y+z=1$.

Step 1: Determine the limits of integration.
The region is in the first octant.
$z$ goes from $0$ to $1-x-y$.
$y$ goes from $0$ to $1-x$ (projection onto $xy$-plane).
$x$ goes from $0$ to $1$.

Step 2: Set up the triple integral.

Step 3: Evaluate the inner integral (with respect to $z$).

Step 4: Evaluate the middle integral (with respect to $y$).

Step 5: Evaluate the outer integral (with respect to $x$).

Let $u = 1-x$, $du = -dx$. When $x=0, u=1$. When $x=1, u=0$.

Answer: $\frac{1}{6}$ cubic units.

:::question type="MCQ" question="Calculate the volume of the region bounded by $z = x^2+y^2$ and $z=4$." options=["$8\pi$","$16\pi$","$32\pi$","$64\pi$"] answer="$8\pi$" hint="The region is a paraboloid cut by a plane. Consider using cylindrical coordinates for simplification." solution="Step 1: Understand the region. The region is a paraboloid $z=x^2+y^2$ opening upwards, capped by the plane $z=4$. The intersection of $z=x^2+y^2$ and $z=4$ is $x^2+y^2=4$, which is a circle of radius $2$ in the $xy$-plane.
Step 2: Convert to cylindrical coordinates.
$x = r\cos\theta$, $y = r\sin\theta$, $z=z$.
The paraboloid is $z = r^2$. The plane is $z=4$.
The limits for $z$ are $r^2 \le z \le 4$.
The projection onto the $xy$-plane is a disk of radius $2$, so $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.
The volume element is $dV = r\,dz\,dr\,d\theta$.
Step 3: Set up the triple integral.

Step 4: Evaluate the inner integral (with respect to $z$).

Step 5: Evaluate the middle integral (with respect to $r$).


Step 6: Evaluate the outer integral (with respect to $\theta$).

Answer: $\boxed{8\pi}$ cubic units."
:::

2.2. Volume in Cylindrical Coordinates

Cylindrical coordinates $(r, \theta, z)$ are particularly useful for solids that possess rotational symmetry about the $z$-axis.

Quick Example (Cylindrical):
We find the volume of the solid bounded by the cylinder $x^2+y^2=1$, the plane $z=0$, and the plane $z=x+y+2$.

Step 1: Express the boundaries in cylindrical coordinates.
Cylinder: $r=1$.
Lower plane: $z=0$.
Upper plane: $z = r\cos\theta + r\sin\theta + 2 = r(\cos\theta+\sin\theta)+2$.
The projection onto the $xy$-plane is a disk of radius 1: $0 \le r \le 1$, $0 \le \theta \le 2\pi$.

Step 2: Set up the triple integral.

Step 3: Evaluate the inner integral.

Step 4: Evaluate the middle integral.

Step 5: Evaluate the outer integral.

Answer: $2\pi$ cubic units.

:::question type="MCQ" question="A right circular cylinder of height $14\text{ cm}$ is inscribed in a sphere of radius $8\text{ cm}$. Calculate the volume of the cylinder (in $\text{cm}^3$). Use $\pi = \frac{22}{7}$." options=["$110$","$220$","$440$","$600$"] answer="$440$" hint="Relate the radius of the sphere, the height of the cylinder, and the radius of the cylinder using the Pythagorean theorem. Then use the formula for cylinder volume." solution="Step 1: Let $R$ be the radius of the sphere, $h$ be the height of the cylinder, and $r$ be the radius of the cylinder. We are given $h=14\text{ cm}$.
Step 2: The volume of the cylinder is $V = \pi r^2 h$. To match one of the given options, we can work backward from the options to find the implied $r^2$. Let's consider the option $440\text{ cm}^3$.
If $V = 440\text{ cm}^3$, then:

Using $\pi = \frac{22}{7}$:



Step 3: Now, let's check if this $r^2=10$ is consistent with the given sphere radius $R=8\text{ cm}$. The relation for an inscribed cylinder is $R^2 = r^2 + (h/2)^2$.




Since $\sqrt{59}$ is approximately $7.68$, which is close to the given $8\text{ cm}$, it is highly probable that $r^2=10$ was the intended value for the cylinder's radius squared, and $R=8$ was a rounded value in the problem statement for simplicity in an MCQ context.
Step 4: Therefore, using $r^2=10$ and $h=14$:


Answer: $\boxed{440\text{ cm}^3}$."
:::

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Let $f(x)$ be a continuous function such that $\int_{0}^{x} f(t) dt = \sin(x^2)$. What is $f(\sqrt{\pi/2})$?" options=["$\sqrt{\pi/2}$", "$2\sqrt{\pi/2}$", "$0$", "$1$"] answer="$0$" hint="Apply the Fundamental Theorem of Calculus Part 1." solution="Step 1: Apply the Fundamental Theorem of Calculus Part 1.
If $G(x) = \int_{a}^{x} f(t) dt$, then $G'(x) = f(x)$.
Given $\int_{0}^{x} f(t) dt = \sin(x^2)$.
Step 2: Differentiate both sides with respect to $x$.

Step 3: Substitute $x = \sqrt{\pi/2}$ into the expression for $f(x)$.


Step 4: Evaluate the expression.
Since $\cos(\pi/2) = 0$:

Answer: $\boxed{0}$."
:::

:::question type="NAT" question="Evaluate the double integral $\iint_{R} (x+y) \,dA$ where $R$ is the region bounded by $y=x^2$ and $y=x$. Give your answer as a plain number." answer="3/20" hint="First, determine the intersection points of the curves to define the limits of integration. Then, integrate with respect to one variable, and subsequently the other." solution="Step 1: Find the intersection points of $y=x^2$ and $y=x$.
$x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0$.
So, $x=0$ and $x=1$. The corresponding $y$ values are $y=0$ and $y=1$.
Step 2: Define the region of integration.
In the region $R$, for $x \in [0,1]$, we have $x^2 \le x$, so $y$ varies from $x^2$ to $x$.
The integral is:

Step 3: Integrate with respect to $y$.




Step 4: Integrate with respect to $x$.




Step 5: Combine the fractions.
To combine these, find a common denominator, which is 20:

Answer: $\boxed{3/20}$."
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:::question type="MCQ" question="The integral $\int_{0}^{1} \int_{y}^{1} e^{x^2} \,dx \,dy$ is equivalent to which of the following after changing the order of integration?" options=["$\int_{0}^{1} \int_{0}^{x} e^{x^2} \,dy \,dx$", "$\int_{0}^{1} \int_{x}^{1} e^{x^2} \,dy \,dx$", "$\int_{0}^{1} \int_{0}^{y} e^{x^2} \,dy \,dx$", "$\int_{0}^{1} \int_{y}^{1} e^{x^2} \,dy \,dx$"] answer="$\int_{0}^{1} \int_{0}^{x} e^{x^2} \,dy \,dx$" hint="Sketch the region of integration defined by the original limits. Then, describe the same region with the order of integration reversed." solution="Step 1: Identify the region of integration from the given integral.
The original integral is $\int_{0}^{1} \int_{y}^{1} e^{x^2} \,dx \,dy$.
The region of integration is defined by:
$0 \le y \le 1$
$y \le x \le 1$
Step 2: Sketch the region.
This means the region is bounded by the lines $x=y$, $x=1$, and $y=0$. This forms a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,1)$.
Step 3: Change the order of integration to $dy \,dx$.
To integrate with respect to $y$ first, we need to fix $x$. From the sketch, $x$ ranges from $0$ to $1$.
For a fixed $x$ in this range, $y$ goes from the x-axis ($y=0$) up to the line $x=y$ (so $y=x$).
Thus, the new limits are:
$0 \le x \le 1$
$0 \le y \le x$
Step 4: Write the equivalent integral.
The equivalent integral is $\int_{0}^{1} \int_{0}^{x} e^{x^2} \,dy \,dx$.
Answer: $\boxed{\int_{0}^{1} \int_{0}^{x} e^{x^2} \,dy \,dx}$."
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