Comprehensive Calculus Practice Exam Guidance

Q1(a). Calculate the volume of the solid obtained by rotating the region $0 \leq x \leq \tan\left(\frac{\pi}{4}y\right)$ from $x = 0$ to $x = 1$ about the $y$-axis.

Background

Topic: Volumes of solids of revolution (Calculus II)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $y$-axis, using integration.

Key Terms and Formulas

• Solid of revolution

• Disk/washer method

• Volume formula (about $y$-axis): $V = \int_{a}^{b} \pi [R(y)]^2 dy$

Step-by-Step Guidance

1. Identify the bounds for $y$. Since $x$ goes from $0$to $1$, solve $x = \tan\left(\frac{\pi}{4}y\right)$ for $y$ when $x = 0$ and $x = 1$.

2. Express $x$ in terms of $y$: $x = \tan\left(\frac{\pi}{4}y\right)$, so the region is bounded by $x = 0$ and $x = \tan\left(\frac{\pi}{4}y\right)$.

3. Set up the volume integral using the disk method about the $y$-axis: $V = \int_{y_1}^{y_2} \pi [x]^2 dy$.

4. Substitute $x = \tan\left(\frac{\pi}{4}y\right)$ into the formula: $V = \int_{y_1}^{y_2} \pi \left(\tan\left(\frac{\pi}{4}y\right)\right)^2 dy$.

5. Determine the limits $y_1$ and $y_2$ by solving for $y$ when $x = 0$ and $x = 1$.

Try solving on your own before revealing the answer!

Final Answer:

When $x = 0$, $y = 0$. When $x = 1$, $\frac{\pi}{4}y = \arctan(1) = \frac{\pi}{4}$, so $y = 1$.

The volume is $V = \int_{0}^{1} \pi \left(\tan\left(\frac{\pi}{4}y\right)\right)^2 dy$.

Evaluating this integral gives $V = \frac{\pi}{2}$.

This result comes from integrating the square of the tangent function over the interval $[0,1]$.

Q1(b). Calculate the volume of the solid obtained by rotating the region $0 \leq y \leq 1 - \frac{x^2}{4}$ from $x = 0$ to $x = 2$ about the $y$-axis.

Background

Topic: Volumes of solids of revolution (washer method)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $y$-axis.

Key Terms and Formulas

• Washer method

• Volume formula (about $y$-axis): $V = \int_{a}^{b} 2\pi x y dx$ (for shell method), or $V = \int_{c}^{d} \pi [R(y)]^2 dy$ (for disk/washer method)

Step-by-Step Guidance

1. Sketch the region: $y$ goes from $0$up to $1 - \frac{x^2}{4}$, with $x$ from $0$to $2$.

2. Since rotating about the $y$-axis, consider using the shell method: $V = \int_{a}^{b} 2\pi x h(x) dx$, where $h(x)$ is the height of the shell.

3. Here, $h(x) = 1 - \frac{x^2}{4}$.

4. Set up the integral: $V = \int_{0}^{2} 2\pi x \left(1 - \frac{x^2}{4}\right) dx$.

5. Expand and simplify the integrand before integrating.

Try solving on your own before revealing the answer!

Final Answer:

$V = \int_{0}^{2} 2\pi x \left(1 - \frac{x^2}{4}\right) dx = 2\pi \int_{0}^{2} \left(x - \frac{x^3}{4}\right) dx$

Evaluating gives $V = \frac{8\pi}{3}$.

This is the volume of the solid formed by rotating the given region about the $y$-axis.

Q1(c). Calculate the volume of the solid obtained by rotating the region $0 \leq y \leq \sin x \cos x$ from $x = 0$ to $x = \frac{\pi}{4}$ about the $x$-axis.

Background

Topic: Volumes of solids of revolution (disk method)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $x$-axis.

Key Terms and Formulas

• Disk method

• Volume formula (about $x$-axis): $V = \int_{a}^{b} \pi [f(x)]^2 dx$

Step-by-Step Guidance

1. Identify the bounds: $x$ goes from $0$to$\frac{\pi}{4}$.

2. The region is bounded above by $y = \sin x \cos x$ and below by $y = 0$.

3. Set up the volume integral: $V = \int_{0}^{\frac{\pi}{4}} \pi [\sin x \cos x]^2 dx$.

4. Simplify $[\sin x \cos x]^2$ using trigonometric identities.

5. Prepare to integrate the resulting expression over $[0, \frac{\pi}{4}]$.

Try solving on your own before revealing the answer!

Final Answer:

$[\sin x \cos x]^2 = \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x$

So $V = \int_{0}^{\frac{\pi}{4}} \frac{\pi}{4} \sin^2 2x dx$

Evaluating gives $V = \frac{\pi}{16}$

This is the volume of the solid formed by rotating the region about the $x$-axis.

Q1(d). Calculate the volume of the solid obtained by rotating the region $0 \leq x \leq y^2$ from $y = 0$ to $y = \sqrt{2}$ about the $x$-axis.

Background

Topic: Volumes of solids of revolution (shell method)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $x$-axis.

Key Terms and Formulas

• Shell method

• Volume formula (about $x$-axis): $V = \int_{a}^{b} 2\pi y h(y) dy$

Step-by-Step Guidance

1. Identify the bounds: $y$ goes from $0$to$\sqrt{2}$.

2. For each $y$, $x$ goes from $0$to$y^2$, so the width is$y^2$.

3. Set up the shell method integral: $V = \int_{0}^{\sqrt{2}} 2\pi y (y^2) dy$.

4. Simplify the integrand before integrating.

5. Prepare to integrate $2\pi y^3$ over $[0, \sqrt{2}]$.

Try solving on your own before revealing the answer!

Final Answer:

$V = \int_{0}^{\sqrt{2}} 2\pi y^3 dy = 2\pi \left[ \frac{y^4}{4} \right]_{0}^{\sqrt{2}} = 2\pi \left( \frac{(\sqrt{2})^4}{4} \right ) = 2\pi$

This is the volume of the solid formed by rotating the region about the $x$-axis.

Q1(e). Calculate the volume of the solid obtained by rotating the region $0 \leq y \leq \sqrt{\cos x}$ for $x = 0$ to $x = \frac{\pi}{2}$ about the $x$-axis.

Background

Topic: Volumes of solids of revolution (disk method)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $x$-axis.

Key Terms and Formulas

• Disk method

• Volume formula: $V = \int_{a}^{b} \pi [f(x)]^2 dx$

Step-by-Step Guidance

1. Identify the bounds: $x$ goes from $0$to$\frac{\pi}{2}$.

2. The region is bounded above by $y = \sqrt{\cos x}$ and below by $y = 0$.

3. Set up the volume integral: $V = \int_{0}^{\frac{\pi}{2}} \pi [\sqrt{\cos x}]^2 dx$.

4. Simplify $[\sqrt{\cos x}]^2$ to $\cos x$.

5. Prepare to integrate $\pi \cos x$ over $[0, \frac{\pi}{2}]$.

Try solving on your own before revealing the answer!

Final Answer:

$V = \int_{0}^{\frac{\pi}{2}} \pi \cos x dx = \pi [\sin x]_{0}^{\frac{\pi}{2}} = \pi$

This is the volume of the solid formed by rotating the region about the $x$-axis.

Q1(f). Calculate the volume of the solid obtained by rotating the region $0 \leq y \leq \frac{9x}{\sqrt{x^3 + 9}}$ from $x = 0$ to $x = 3$ about the $y$-axis.

Background

Topic: Volumes of solids of revolution (shell method)

This question tests your ability to set up and compute the volume of a solid formed by rotating a region about the $y$-axis.

Key Terms and Formulas

• Shell method

• Volume formula: $V = \int_{a}^{b} 2\pi x h(x) dx$

Step-by-Step Guidance

1. Identify the bounds: $x$ goes from $0$to $3$.

2. For each $x$, the height is $h(x) = \frac{9x}{\sqrt{x^3 + 9}}$.

3. Set up the shell method integral: $V = \int_{0}^{3} 2\pi x \frac{9x}{\sqrt{x^3 + 9}} dx$.

4. Simplify the integrand: $2\pi x \frac{9x}{\sqrt{x^3 + 9}} = 18\pi \frac{x^2}{\sqrt{x^3 + 9}}$.

5. Prepare to integrate $18\pi \frac{x^2}{\sqrt{x^3 + 9}}$ over $[0,3]$.

Try solving on your own before revealing the answer!

Final Answer:

$V = \int_{0}^{3} 18\pi \frac{x^2}{\sqrt{x^3 + 9}} dx$

Let $u = x^3 + 9$, $du = 3x^2 dx$, so $x^2 dx = \frac{du}{3}$.

Substitute and evaluate: $V = 6\pi \int_{9}^{36} u^{-1/2} du = 6\pi [2u^{1/2}]_{9}^{36} = 12\pi (6 - 3) = 36\pi$

This is the volume of the solid formed by rotating the region about the $y$-axis.

Q2(a). Find the arclength of the curve $y = \frac{1}{3}(x^2 + 2)^{3/2}$ from $x = 0$ to $x = 3$.

Background

Topic: Arc length of a curve (Calculus II)

This question tests your ability to compute the length of a curve using the arc length formula.

Key Terms and Formulas

• Arc length formula: $L = \int_{a}^{b} \sqrt{1 + [y'(x)]^2} dx$

Step-by-Step Guidance

1. Find $y'(x)$ for $y = \frac{1}{3}(x^2 + 2)^{3/2}$.

2. Compute $[y'(x)]^2$ and add $1$ to get the expression under the square root.

3. Set up the arc length integral: $L = \int_{0}^{3} \sqrt{1 + [y'(x)]^2} dx$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[0,3]$.

Try solving on your own before revealing the answer!

Final Answer:

$y'(x) = (x^2 + 2)^{1/2} x$

$[y'(x)]^2 = x^2 (x^2 + 2)$

$1 + [y'(x)]^2 = 1 + x^2 (x^2 + 2) = x^4 + 2x^2 + 1 = (x^2 + 1)^2$

So $L = \int_{0}^{3} |x^2 + 1| dx = \int_{0}^{3} (x^2 + 1) dx = \left[ \frac{x^3}{3} + x \right]_{0}^{3} = 12$

The arc length is $12$ units.

Q2(b). Find the arclength of the curve $x = \frac{y^3}{6} + \frac{1}{2}y$ from $y = 2$ to $y = 3$.

Background

Topic: Arc length of a curve (parametric form)

This question tests your ability to compute the length of a curve given $x$ as a function of $y$.

Key Terms and Formulas

• Arc length formula: $L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$

Step-by-Step Guidance

1. Find $\frac{dx}{dy}$ for $x = \frac{y^3}{6} + \frac{1}{2}y$.

2. Compute $\left(\frac{dx}{dy}\right)^2$ and add $1$ to get the expression under the square root.

3. Set up the arc length integral: $L = \int_{2}^{3} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[2,3]$.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{dx}{dy} = \frac{y^2}{2} + \frac{1}{2}$

$\left(\frac{dx}{dy}\right)^2 = \left(\frac{y^2}{2} + \frac{1}{2}\right)^2$

$1 + \left(\frac{dx}{dy}\right)^2 = 1 + \frac{y^4}{4} + y^2 + \frac{1}{4}$

Set up and evaluate $L = \int_{2}^{3} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$ numerically or symbolically.

The exact value is $\int_{2}^{3} \sqrt{\frac{y^4}{4} + y^2 + \frac{5}{4}} dy$.

Q2(c). Find the arclength of the curve $y = \frac{3}{4}x^{4/3} - \frac{3}{8}x^{2/3} + 5$ from $x = 1$ to $x = 8$.

Background

Topic: Arc length of a curve (Calculus II)

This question tests your ability to compute the length of a curve using the arc length formula.

Key Terms and Formulas

• Arc length formula: $L = \int_{a}^{b} \sqrt{1 + [y'(x)]^2} dx$

Step-by-Step Guidance

1. Find $y'(x)$ for $y = \frac{3}{4}x^{4/3} - \frac{3}{8}x^{2/3} + 5$.

2. Compute $[y'(x)]^2$ and add $1$ to get the expression under the square root.

3. Set up the arc length integral: $L = \int_{1}^{8} \sqrt{1 + [y'(x)]^2} dx$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[1,8]$.

Try solving on your own before revealing the answer!

Final Answer:

$y'(x) = x^{1/3} - \frac{1}{4}x^{-1/3}$

$[y'(x)]^2 = x^{2/3} - \frac{1}{2} + \frac{1}{16}x^{-2/3}$

$1 + [y'(x)]^2 = x^{2/3} + \frac{1}{2} + \frac{1}{16}x^{-2/3}$

Set up and evaluate $L = \int_{1}^{8} \sqrt{x^{2/3} + \frac{1}{2} + \frac{1}{16}x^{-2/3}} dx$.

Q3(a). Find the surface area of the surface of revolution $y = \frac{x^3}{9}$, $0 \leq x \leq 2$ revolved around the $x$-axis.

Background

Topic: Surface area of revolution (Calculus II)

This question tests your ability to compute the surface area of a solid formed by revolving a curve about the $x$-axis.

Key Terms and Formulas

• Surface area formula: $S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx$

Step-by-Step Guidance

1. Find $f'(x)$ for $f(x) = \frac{x^3}{9}$.

2. Compute $[f'(x)]^2$ and add $1$ to get the expression under the square root.

3. Set up the surface area integral: $S = \int_{0}^{2} 2\pi \frac{x^3}{9} \sqrt{1 + [f'(x)]^2} dx$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[0,2]$.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = \frac{1}{3}x^2$

$[f'(x)]^2 = \frac{1}{9}x^4$

$S = \int_{0}^{2} 2\pi \frac{x^3}{9} \sqrt{1 + \frac{1}{9}x^4} dx$

This integral gives the surface area of the solid formed by revolving the curve about the $x$-axis.

Q3(b). Find the surface area of the surface of revolution $y = \sqrt{x}$, $\frac{3}{4} \leq x \leq \frac{15}{4}$ revolved around the $x$-axis.

Background

Topic: Surface area of revolution (Calculus II)

This question tests your ability to compute the surface area of a solid formed by revolving a curve about the $x$-axis.

Key Terms and Formulas

• Surface area formula: $S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx$

Step-by-Step Guidance

1. Find $f'(x)$ for $f(x) = \sqrt{x}$.

2. Compute $[f'(x)]^2$ and add $1$ to get the expression under the square root.

3. Set up the surface area integral: $S = \int_{3/4}^{15/4} 2\pi \sqrt{x} \sqrt{1 + [f'(x)]^2} dx$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[\frac{3}{4}, \frac{15}{4}]$.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = \frac{1}{2}x^{-1/2}$

$[f'(x)]^2 = \frac{1}{4}x^{-1}$

$S = \int_{3/4}^{15/4} 2\pi \sqrt{x} \sqrt{1 + \frac{1}{4}x^{-1}} dx$

This integral gives the surface area of the solid formed by revolving the curve about the $x$-axis.

Q3(c). Find the surface area of the surface of revolution $x = \frac{y^4}{8} + \frac{1}{8}y^2$, $1 \leq y \leq 2$ revolved around the $x$-axis.

Background

Topic: Surface area of revolution (parametric form)

This question tests your ability to compute the surface area of a solid formed by revolving a curve about the $x$-axis, given $x$ as a function of $y$.

Key Terms and Formulas

• Surface area formula: $S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$

Step-by-Step Guidance

1. Find $\frac{dx}{dy}$ for $x = \frac{y^4}{8} + \frac{1}{8}y^2$.

2. Compute $\left(\frac{dx}{dy}\right)^2$ and add $1$ to get the expression under the square root.

3. Set up the surface area integral: $S = \int_{1}^{2} 2\pi y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$.

4. Simplify the integrand as much as possible.

5. Prepare to integrate over $[1,2]$.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{dx}{dy} = \frac{1}{2}y^3 + \frac{1}{4}y$

$\left(\frac{dx}{dy}\right)^2 = \frac{1}{4}y^6 + \frac{1}{2}y^4 + \frac{1}{16}y^2$

$S = \int_{1}^{2} 2\pi y \sqrt{1 + \frac{1}{4}y^6 + \frac{1}{2}y^4 + \frac{1}{16}y^2} dy$

This integral gives the surface area of the solid formed by revolving the curve about the $x$-axis.