Calculus
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 yaxis.
b) Find the volume of the solid obtained by rotating the region bounded by y = 4x^2, x = 1, and y = 0 about the xaxis.
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 xaxis.
Thanks for your trouble.

Steve
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)
= 36*8π√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 )[0,6^{3/2}]
= 2π Int(x(x^1/3  x/6) dx )[0,6^{3/2}]
= 2π Int(x^{4/3}  x^{2}/6) dx )[0,6^{3/2}]
= 2π(3/7 x^{7/3}  x^{3}/18)[0,6^{3/2}]
= 2π(6^{7/2}  6^{9/2}/18)
= 2π*216√6(3/7  1/3)
= π*216√6(4/21)
= 36*8π√6/7 
Beth
Thank you!
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