The base ofSis a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares.

Find the volumeVof this solid.

I'd write out my work on this problem, but I don't know how to use the thing that lets you do that.

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- Sep 11th 2009, 04:34 PMHellacious DIntegrating with disks
The base of

*S*is a circular disk with radius 4*r*. Parallel cross-sections perpendicular to the base are squares.

Find the volume*V*of this solid.

I'd write out my work on this problem, but I don't know how to use the thing that lets you do that. - Sep 11th 2009, 04:55 PMskeeter
$\displaystyle x^2 + y^2 = (4r)^2

$

side of a square cross-section perpendicular to the x-axis = $\displaystyle 2y$

area of a square cross-section = $\displaystyle 4y^2$

$\displaystyle dV = 4y^2 dx$

$\displaystyle dV = 4(16r^2-x^2) \, dx$

$\displaystyle V = 2\int_0^{4r} 4(16r^2-x^2) \, dx$ - Sep 11th 2009, 05:05 PMHellacious D
I'm getting (-512r^3)/3 which is what I've gotten, but webassign says no deal. I've tried (-512x^3)/3 and -512/3 with no success.

- Sep 11th 2009, 05:24 PMskeeter
- Sep 11th 2009, 05:29 PMHellacious D
How do I write out my work? I can do it in text, but what's the program you guys use on here to make it look nice?

- Sep 11th 2009, 05:40 PMskeeter
- Sep 11th 2009, 05:59 PMHellacious D
eh, no time to learn that today.

2*(the integral from 0 to 4r) of 4(16(r^2)-(x^2))dx

2*(4(16(r^2)-((x^3)/3)) evaluated from 0 to 4r

8(16(r^2)-(64r^3)/3) - 8(16r^2)

(128r^2)-((512r^3)/3)-128r^2

(-512r^3)/3 - Sep 11th 2009, 06:15 PMskeeter
- Sep 11th 2009, 06:22 PMHellacious D
Thank you much, I don't know why I kept missing that.