Calculus Practice Exam Guidance: Areas, Volumes, Arc Length, and Surface Area

Q1. Sketch the region bounded by $y = 6 - 3x^2$ and $y = 6x - 3$. Then show that the area of this region is 32.

Background

Topic: Area Between Curves

This question tests your ability to find the area between two curves by integrating the difference of their equations over the appropriate interval.

Key Terms and Formulas

• Area between curves: $A = \int_a^b [f(x) - g(x)]\,dx$ where $f(x)$ is the upper function and $g(x)$ is the lower function.

Step-by-Step Guidance

1. Set the two equations equal to each other to find the points of intersection: $6 - 3x^2 = 6x - 3$.

2. Solve for $x$ to determine the limits of integration.

3. Determine which function is on top (greater $y$ value) between the intersection points.

4. Set up the definite integral for the area: $A = \int_{x_1}^{x_2} [(\text{top function}) - (\text{bottom function})] \, dx$.

Try solving on your own before revealing the answer!

Q2. Sketch the region bounded by $y = 3\cos x$ and $y = 3\cos 2x$ for $0 \leq x \leq \pi$. Then show that the area of this region is $\frac{9\sqrt{3}}{2}$.

Background

Topic: Area Between Curves (Trigonometric Functions)

This question involves finding the area between two trigonometric curves over a specified interval.

Key Terms and Formulas

• Area between curves: $A = \int_a^b [f(x) - g(x)]\,dx$

• Trigonometric identities may be useful for simplifying the integrand.

Step-by-Step Guidance

1. Find the points where $3\cos x = 3\cos 2x$ within $[0, \pi]$ to determine where the curves cross.

2. Decide which function is on top in each subinterval.

3. Set up the integral(s) for the area, possibly splitting the interval if the top/bottom function changes.

4. Simplify the integrand as much as possible before integrating.

Try solving on your own before revealing the answer!

Q3. Sketch the region bounded by $y = 2 - x$, $y = 2 - 2x$, and $y = 0$. Then use the washer method to show that the volume of the solid is $4\pi$ when this region is revolved about $x = 3$.

Background

Topic: Volumes of Solids of Revolution (Washer Method)

This question tests your ability to use the washer method to find the volume of a solid generated by revolving a region about a vertical line.

Key Terms and Formulas

• Washer method: $V = \pi \int_a^b \left([R_{\text{outer}}(x)]^2 - [R_{\text{inner}}(x)]^2\right) dx$

• For revolution about $x = 3$, the radius is the horizontal distance from $x$ to $x = 3$.

Step-by-Step Guidance

1. Find the $x$-values where $y = 2 - x$ and $y = 2 - 2x$ intersect and where they meet $y = 0$ to determine the limits of integration.

2. Express the outer and inner radii in terms of $x$ as the distance from $x$ to $x = 3$.

3. Set up the washer method integral using these radii and the appropriate limits.

4. Simplify the integrand before integrating.

Try solving on your own before revealing the answer!

Q4. (a) Sketch the region in the first quadrant bounded by $f(x) = \begin{cases} 1 & 0 \leq x \leq 1 \\ 2 - \sqrt{x} & 1 < x \leq 4 \end{cases}$

Background

Topic: Piecewise Functions and Area

This part asks you to visualize a region defined by a piecewise function, which is important for setting up integrals in later parts.

Key Terms and Formulas

• Piecewise function: A function defined by different expressions over different intervals.

Step-by-Step Guidance

1. Plot $y = 1$ for $0 \leq x \leq 1$ and $y = 2 - \sqrt{x}$ for $1 < x \leq 4$ in the first quadrant.

2. Shade the region under the curve from $x = 0$ to $x = 4$.

Try sketching the region before moving on!

Q4. (b) Use the disk method to show that the volume of the solid is $\frac{11\pi}{6}$ when the region is revolved about the $x$-axis.

Background

Topic: Disk Method for Volumes of Revolution

This part asks you to set up and evaluate the volume of a solid generated by revolving a region about the $x$-axis using the disk method.

Key Terms and Formulas

• Disk method: $V = \pi \int_a^b [f(x)]^2 dx$

• For piecewise functions, split the integral at the point where the formula changes.

Step-by-Step Guidance

1. Set up two separate integrals: one for $0 \leq x \leq 1$ with $f(x) = 1$, and one for $1 < x \leq 4$ with $f(x) = 2 - \sqrt{x}$.

2. Write the total volume as $V = \pi \int_0^1 [1]^2 dx + \pi \int_1^4 [2 - \sqrt{x}]^2 dx$.

3. Expand $[2 - \sqrt{x}]^2$ to simplify the second integral before integrating.

Try setting up and simplifying the integrals before calculating!

Q4. (c) Use the shell method to show that the volume of the solid is $\frac{31\pi}{5}$ when the region is revolved about the $y$-axis.

Background

Topic: Shell Method for Volumes of Revolution

This part asks you to use the shell method to find the volume when revolving about the $y$-axis.

Key Terms and Formulas

• Shell method: $V = 2\pi \int_a^b x f(x) dx$

• For piecewise functions, split the integral at the point where the formula changes.

Step-by-Step Guidance

1. Set up two integrals: $2\pi \int_0^1 x \cdot 1 dx$ and $2\pi \int_1^4 x (2 - \sqrt{x}) dx$.

2. Expand $x (2 - \sqrt{x})$ to simplify the second integral before integrating.

Try setting up and simplifying the integrals before calculating!

Q5. Show that the length of the curve $y = \frac{x^3}{6} + \frac{1}{2x}$ on $[1, 2]$ is $\frac{17}{12}$.

Background

Topic: Arc Length of a Curve

This question tests your ability to use the arc length formula for a function $y = f(x)$ over a given interval.

Key Terms and Formulas

• Arc length: $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$

Step-by-Step Guidance

1. Find $f'(x)$ for $y = \frac{x^3}{6} + \frac{1}{2x}$.

2. Compute $[f'(x)]^2$ and add 1 inside the square root.

3. Set up the arc length integral $L = \int_1^2 \sqrt{1 + [f'(x)]^2} dx$.

4. Simplify the expression under the square root as much as possible before integrating.

Try simplifying the integrand before integrating!

Q6. Show that the area of the surface generated when the curve $y = \frac{1}{4}(e^{2x} + e^{-2x})$ is revolved about the $x$-axis over $[-2, 2]$ is $\frac{\pi}{8}(e^8 + 16 - e^{-8})$.

Background

Topic: Surface Area of Revolution

This question tests your ability to use the surface area formula for a curve revolved about the $x$-axis.

Key Terms and Formulas

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

Step-by-Step Guidance

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

2. Compute $[y'(x)]^2$ and add 1 inside the square root.

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

4. Simplify the integrand as much as possible before integrating.

Try simplifying the integrand before integrating!