Question 1

Which of the following is the correct setup for the arc length of the curve $$x = \cos(y)$$ for $$0 \leq x \leq \frac{1}{2}$$ using $$dx$$?

Accepted Answer

$$\int_{0}^{\frac{1}{2}} \sqrt{\frac{2 - x^2$}{1 - x^2$}} \, dx$$

Question 2

Given $$x = \cos(y)$$, what are the correct limits of integration for $$y if$$ $$x$$ goes from $$0 to$$ $$\frac{1}{2}$$?

Accepted Answer

$$\frac{\pi}{3} \leq y \leq \frac{\pi}{2}$$

Question 3

Which of the following integrals correctly represents the arc length of $$x = \cos(y)$$ for $$y$$ from $$\frac{\pi}{3} to$$ $$\frac{\pi}{2}$$ using $$dy$$?

Accepted Answer

$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{1 + \sin^2$(y)} \, dy$$

Question 4

Suppose you are asked to find the surface area generated by revolving $$x = \sqrt{y + 5}$$, $$\sqrt{5} \leq x \leq 3$$, about the $$y-$$axis using $$dx. $$Which of the following is the correct setup?

Accepted Answer

$$\int_{\sqrt{5}}^{3} 2\pi x \sqrt{1 + 4x^2$} \, dx$$

Question 5

For the same surface area problem, what is the correct setup using $$dy$$?

Accepted Answer

$$\int_{0}^{4} \pi \frac{4y + 21}{4(y + 5)} \, dy$$

Question 6

Given $$y = x^2$ - 5$$, what is $$\frac{dy}{dx}$$ and how does it appear in the arc length formula?

Accepted Answer

$$\frac{dy}{dx} = 2x$$, so the formula is $$\sqrt{1 + 4x^2$} \, dx$$

Question 7

If $$y = 6x^2$ + 2$$, what is the correct surface area integral for revolving about the $$y-$$axis from $$x = 1 to$$ $$x = 2$$?

Accepted Answer

$$\int_{1}^{2} 2\pi x \sqrt{1 + 36x^2$} \, dx$$

Question 8

Which of the following best describes the process for finding the limits of integration when switching from $$x to$$ $$y$$ for $$x = \cos(y)$$, $$0 \leq x \leq \frac{1}{2}$$?

Accepted Answer

Solve $$y = \arccos(x)$$ and substitute $$x = 0$$ and $$x = \frac{1}{2} to $$get $$y$$ limits.

Question 9

Given $$\frac{dx}{dy} = -\sin(y)$$ for $$x = \cos(y)$$, what is the correct integrand for arc length in terms of $$y$$?

Accepted Answer

$$\sqrt{1 + \sin^2$(y)}$$

Question 10

A region bounded by $$y = x - 1$$, $$y = \sqrt{2x + 6}$$, and the $$x-$$axis is revolved about the $$x-$$axis. Which method can be used to set up the volume integral?

Accepted Answer

Washer method with respect to $$x$$ and $$y$$

Question 11

For the region in the previous question, what is the correct setup for the volume using the washer method with respect to $$x$$?

Accepted Answer

$$\pi \int_{-3}^{5} \left[ (\sqrt{2x + 6})^2$ - (x - 1)^2$ \right] dx$$

Question 12

If the base of a solid is the region between $$y = x^2$ and y = x + 2$, and cross-sections perpendicular to the x$-axis are semicircles, what is the correct setup for the volume?

Accepted Answer

$$\int_{a}^{b} \frac{\pi}{8} (x + 2 - x^2$)^2 dx$

Question 13

For the surface area of revolution about the $$y-$$axis, which formula is generally used for $$y = f(x)$$?

Accepted Answer

$$SA = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\$$right)^2$$} dx$$

Question 14

If $$\frac{dy}{dx} = 6x$$, what is the integrand for the arc length formula?

Accepted Answer

$$\sqrt{1 + 36x^2$}$$

Question 15

Which of the following is the correct formula for the surface area of revolution about the $$y-$$axis for $$x = \sqrt{y + 5}$$, $$0 \leq y \leq 4$$, using $$dy$$?

Accepted Answer

$$\int_{0}^{4} \pi \frac{4y + 21}{4(y + 5)} dy$$

Arc Length and Surface Area Integrals in Calculus

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