posted by Beth
a) Find the volume formed by rotating the region enclosed by x = 6y and y^3 = x with y greater than, equal to 0 about the y-axis.
b) Find the volume of the solid obtained by rotating the region bounded by y = 4x^2, x = 1, and y = 0 about the x-axis.
c) Find the volume of the solid obtained by rotating the region bounded by y = 1/x^6, y = 0, x = 3, and x = 9, about the x-axis.
Thanks for your trouble.
I'll do (a)
First, determine where the curves intersect:
6y = y^3 at the point (6√6,√6)
You can do this using discs or shells. Using discs, or washers, you have a stack of washers. The area of a disc of outer radius R and inner hole of radis r is
m(R^2 - r^2)
Since 6y > y^3 on the interval desired
The thickness of each disc is dy. So, the volume of the stack of discs is the integral
π Int(R^2 - r^2 dy)[0,√6]
= π Int(36y^2 - (y^3)^2) dy)[0,√6]
= π Int(36y^2 - y^6 dy)[0,√6]
= π (12y^3 - 1/7 y^7)[0,√6]
= π (12*6√6 - 1/7 * 216√6)
= 36π√6(2 - 6/7)
Or, if you want to calculate using shells, the volume of a shell is 2πrh
r = x, h = x^(1/3) - x/6
So, the volume of a ring of shells is
2π Int(x(x^1/3 - x/6) dx )[0,63/2]
= 2π Int(x(x^1/3 - x/6) dx )[0,63/2]
= 2π Int(x4/3 - x2/6) dx )[0,63/2]
= 2π(3/7 x7/3 - x3/18)[0,63/2]
= 2π(67/2 - 69/2/18)
= 2π*216√6(3/7 - 1/3)