Answer:
[tex]\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[/tex]
Step-by-step explanation:
We want to find the volume of the solid obtained by rotating the region between the two curves:
[tex]y=(x-2)^4\text{ and } 8x-y=16[/tex]
About the line x = 16.
Since our axis of revolution is vertical, we can use the washer method in terms of y.
[tex]\displaystyle V = \pi \int _c^d[R(y)]^2 -[r(y)}]^2\, dy[/tex]
Where R(y) is the outer radius and r(y) is the inner radius.
First, solve each equation in terms of y:
[tex]\displaystyle x_1 = \frac{1}{8}y+2\text{ and } x_2 = y^{{}^{1}\! /\! {}_{4}}+2[/tex]
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:
[tex]\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[/tex]
Thus, our volume is:
[tex]\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[/tex]
*I labeled the diagram incorrectly. Let R(x) be R(y) and r(x) be r(y).
