Question

Which of the following integrals represents the volume of the solid obtained by rotating the region bounded by the curves y = (x - 2)^4 and 8x - y =16 about the line x= 10?
A. Pi integral^4_2 {[10 - (1/8 y + 2)^2] - [10 - (2 + ^4 squareroot y)^2]} dy
B. Pi integral^16_0 {[10 - (1/8 y + 2)] - [10 - (2 + ^4 Squareroot)]}^2 dy
C. Pi integral^4_2 {[10 - (1/8 y + 2)] - [10 - 2 + ^4 squareroot y)]}^2 dy
D.Pi integral^16_0 {[10 - (1/8 y + 2)]^2 - [10 - 2 + ^4 squareroot y)]^2} dy
E. Pi integral^16_0 {[10 - (1/8 y + 2)^2] - [10 - 2 + ^4 squareroot y)^2]} dy
F. Pi integral^4_2 {[10 - (1/8 y + 2)]^2 - [10 - 2 + ^4 squareroot y)]^2} dy

Answer 1

Answer:

$\displaystyle V = \pi \int _0^{16}\left[10-\left(\frac{1}{8}y-2\right)\right] ^2 - \left[10 - \left(2+y^{{}^{1}\!/\!{}_{4}}\right)\right]^2\, dy$

Step-by-step explanation:

We want to find the volume of the solid obtained by rotating the region between the two curves:

$y=(x-2)^4\text{ and } 8x-y=16$

About the line x = 16.

Since our axis of revolution is vertical, we can use the washer method in terms of y.

$\displaystyle V = \pi \int _c^d[R(y)]^2 -[r(y)}]^2\, dy$

Where R(y) is the outer radius and r(y) is the inner radius.

First, solve each equation in terms of y:

$\displaystyle x_1 = \frac{1}{8}y+2\text{ and } x_2 = y^{{}^{1}\! /\! {}_{4}}+2$

From the diagram below, we can see that the outer radius R(y) is (10 - x₁) and that the inner radius r(y) is (10 - x₂). The limits of integration will be from y = 0 to y = 16. Substitute:

$\displaystyle V = \pi \int_0^{16}\left[\underbrace{10-\left(\frac{1}{8}y+2\right)}_{R(y)}\right]^2 - \left[\underbrace{10-\left(y^{{}^{1}\!/\!{}_{4}}+2\right)}_{r(y)}\right]^2\, dy$

Thus, our volume is:

$\displaystyle V = \pi \int _0^{16}\left[10-\left(\frac{1}{8}y-2\right)\right] ^2 - \left[10 - \left(2+y^{{}^{1}\!/\!{}_{4}}\right)\right]^2\, dy$

*I labeled the diagram incorrectly. Let R(x) be R(y) and r(x) be r(y).