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

Step-by-step explanation:

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

About the line x = 16.

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

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

First, solve each equation in terms of y:

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:

Thus, our volume is:

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

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

step-by-step explanation: