# Integrating with disks

• September 11th 2009, 05:34 PM
Hellacious D
Integrating with disks
The base of S is a circular disk with radius 4r. 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.
• September 11th 2009, 05:55 PM
skeeter
Quote:

Originally Posted by Hellacious D
The base of S is a circular disk with radius 4r. 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.

$x^2 + y^2 = (4r)^2
$

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

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

$dV = 4y^2 dx$

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

$V = 2\int_0^{4r} 4(16r^2-x^2) \, dx$
• September 11th 2009, 06:05 PM
Hellacious 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.
• September 11th 2009, 06:24 PM
skeeter
Quote:

Originally Posted by Hellacious 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.

show how you arrived at this "no deal" solution.
• September 11th 2009, 06:29 PM
Hellacious 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?
• September 11th 2009, 06:40 PM
skeeter
Quote:

Originally Posted by Hellacious 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?

http://www.mathhelpforum.com/math-help/latex-help/
• September 11th 2009, 06:59 PM
Hellacious 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
• September 11th 2009, 07:15 PM
skeeter
Quote:

Originally Posted by Hellacious 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 - ((x^3)/3)) evaluated from 0 to 4r

$16r^2$ is a constant ... antiderivative is $16r^2 \cdot x$
• September 11th 2009, 07:22 PM
Hellacious D
Thank you much, I don't know why I kept missing that.