calculus
posted by Johnathon .
Find the volume V of the described solid S.
The base of S is a circular disk with radius 2r. Parallel crosssections perpendicular to the base are squares.

I may have this wrong, but here's how I see it:
Draw a circle of radius 2r. That's the base. Draw a chord perpendicular to the xaxis at distance x from the center. At this point,
x^2 + y^2 = 4r^2
Now, erect a square of height 2y on the base. That is a crosssection of the solid. It has area 4y^2
The volume is thus Int(4y^2 dx)[0,2r]
But, y^2 = 4r^2  x^2
v = 4*Int(4r^2  x^2) dx)[0,2r]
= 4*(4r^2 x  1/3 x^3)[0,2r]
= 4(8r^3  8r^3/3)
= 32(2/3 r^3)
= 64/3 r^3