volume of the solid math question?

Question

find the volume of the solid generated by rotating the circle (x-10)^2 + y^2 = 36 about the y-axis.

basically all i can do is solve for y. i just don't know what to do afterwards.

y = (-x^2 + 20x - 64)^(1/2)

Answer 1

The first answerer gives the correct result, using the second theorem of Pappus. For those not familiar with that very useful theorem:

I would use the method of cylindrical shells. Leave the expression for y as
sqrt(36 - (x-10)²)

dV = 2πrh dr = 2π(x-0)[sqrt(36 - (x-10)²) - {-sqrt(36 - (x-10)²)}] dx
= 4πx sqrt(36 - (x-10)²) dx

and so

V = integral from 4 to 16 of 4πx sqrt(36 - (x-10)²) dx

The 0 in (x-0) in dV comes from the fact that the axis of revolution is the y-axis, whose equation is x=0.

Answer 2

It is a circle with a radius of 6 centered on (10,0).
If you rotate it around the y-axis you generate an donut shaped solid. called a torus.
The diameter in the middle of the donut is 10. The volume is the section of the tube (πr² with a radius of 6, or 36π) multiplied by it length, or in this case its circumference (2πR with a radius of 10, or 20π).
The volume is 720π²