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October 1, 2014

October 1, 2014

Posted by **Johnathon** on Thursday, November 10, 2011 at 12:09pm.

The base of S is a circular disk with radius 2r. Parallel cross-sections perpendicular to the base are squares.

- calculus -
**Steve**, Thursday, November 10, 2011 at 12:54pmI 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 x-axis 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 cross-section 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

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