Calc 2 help please...?

Question

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves
x=0, y=1, x=y^12, about the line y=1

Answer 1

First, let Y = y - 1. In the x-Y coordinate system, the curves are: x = 0, Y = 0, x = (Y + 1)^12, about the line Y = 0.
We may think that the solid has been sliced into many thin disks perpendicular to the x-axis with thickness dx. The circular disks have the radius Y = x^(1/12) - 1, and thus the area pi*Y^2 = pi* {x^(1/6) - 2x^(1/12) + 1}. So the volume is:
∫[0 to 1] pi*Y^2 dx
= pi* ∫[0 to 1] {x^(1/6) - 2x^(1/12) + 1}dx
= pi [ (6/7)x^(7/6) - (24/13)x^(13/12) + x] [0 to 1]
= pi*(6/7 - 24/13 + 1)
= pi*(6/7 - 11/13)
= pi/91