Question 1

Let $$R be $$the region bounded by $$y = (x-3)^2$, x = 0$, and y = 0$. Which of the following is the correct integral setup using the disk/washer method to compute the volume of the solid obtained by revolving R$ about the x$-axis?

Accepted Answer

$$\displaystyle \$$int_0$$^3 \pi \left[(x-3)^2$\right]^2 dx$$

Question 2

When revolving the region $$R$$ bounded by $$y = (x-3)^2$, x = 0$, and y = 0$ about the line y = 9$, what is the correct expression for the inner radius at a given x$?

Accepted Answer

$$9 - (x-3)^2$

Question 3

For the region $$R$$ bounded by $$y = (x-3)^2$, x = 0$, and y = 0$, which of the following is the correct integral setup for the volume when revolving R$ about the y$-axis?

Accepted Answer

$$\displaystyle \$$int_0$$^9 \pi \left[3 - \sqrt{y}\right]^2 dy$$

Question 4

Suppose you revolve the region $$R$$ bounded by $$y = (x-3)^2$, x = 0$, and y = 0$ about the line x = 4$. What is the inner radius in terms of y$?

Accepted Answer

$$1 + \sqrt{y}$$

Question 5

Given the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$, what are the $$x-$$values of intersection between $$y = x/2$$ and $$y = 15 - x$$?

Accepted Answer

$$x = 10$$

Question 6

When revolving the triangular region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$ about the line $$y = -1$$, what is the correct expression for the outer radius at a given $$x$$?

Accepted Answer

$$16 - x$$

Question 7

Which of the following integrals correctly represents the volume of the solid formed by revolving the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$ about the line $$y = 18$$?

Accepted Answer

$$\displaystyle \int_{-2}^{10} \pi \left[(18 - x/2)^2$ - (3 + x)^2$\right] dx$$

Question 8

For the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$, when revolving about $$x = 12$$, what is the correct expression for the inner radius for $$-1 \u003c y \u003c 5$$?

Accepted Answer

$$12 - 2y$$

Question 9

When setting up the volume integral for revolving the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$ about $$x = 12$$, why is the integral split into two parts?

Accepted Answer

Because the expressions for the inner radius change at $$y = 5$$ due to the intersection of the bounding lines.

Question 10

Suppose you are asked to find the volume of a solid formed by revolving a region about a horizontal line $$y = k. $$Which of the following best describes how to determine the radius for the washer method?

Accepted Answer

The radius is the vertical distance from $$y = k to $$the function at $$x.$$

Question 11

A region is bounded by $$y = (x-3)^2$, x = 0$, and y = 0$. If this region is revolved about the line y = 9$, what is the correct integral for the volume?

Accepted Answer

$$\displaystyle \$$int_0$$^3 \pi \left[9^2$ - (9 - (x-3)^2$)^2$$\right] dx$

Question 12

Given $$y = (x-3)^2$, solve for x$ in terms of y$ and select the correct expression for x$ that is used when setting up the integral in terms of y$ for the region described.

Accepted Answer

$$x = 3 - \sqrt{y}$$

Question 13

For the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$, what are the $$y-$$values of intersection between $$y = 15 - x$$ and $$y = 2y$$?

Accepted Answer

$$y = 5$$

Question 14

When revolving the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$ about $$x = 12$$, what is the correct integral setup for the volume for $$5 \u003c y \u003c 17$$?

Accepted Answer

$$\displaystyle \$$int_5$$^{17} \pi \left[14^2$ - (y-3)^2$\right] dy$$

Question 15

A chemical engineer wants to design a storage tank by revolving the region bounded by $$y = (x-3)^2$, x = 0$, and y = 0$ about the y$-axis. Which of the following integrals gives the correct volume?

Accepted Answer

$$\displaystyle \$$int_0$$^9 \pi \left[3 - \sqrt{y}\right]^2 dy$$

Question 16

A medical device is modeled by revolving the region bounded by $$y = x/2$$, $$y = 15 - x$$, and $$x = -2$$ about the line $$y = -1. $$What is the correct integral for the volume?

Accepted Answer

$$\displaystyle \int_{-2}^{10} \pi \left[(16 - x)^2$ - (x/2 + 1)^2$\right] dx$$

Applications of Integration: Volumes of Solids of Revolution (Disk and Washer Methods)

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