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Another Polar Double Integral

by G01
Tags: double, integral, polar
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Nov24-06, 07:03 PM
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P: 2,688
Hey everyone, My task this time is to derive the volume of a sphere using a polar double integral.

The sphere has radius [tex] a[/tex] we know that r goes from 0 to a in this integral.

The equation for a sphere is:

[tex] x^2 + y^2 +z^2 = r^2[/tex]
or [tex] f(x,y) = \sqrt{r^2 -x^2 -y^2}[/tex]
and it intersects the x-y plane in a circle:

[tex] x^2 + y^2 = r^2 [/tex]

So if we find the volume over this circle and under the positive half of the sphere and double it we should get the volume we want.

r goes from 0 to a, and [tex]\theta[/tex] goes from 0 to[tex]2\pi[/tex].

So we get:

[tex]2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta [/tex]

My problem is that whenever I fill in the polar conversions for x and y my sphere equation becomes z=0. I know I'm missing something simple but I can't figure it out. Any help would be appreciated. Thank you.
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Nov24-06, 07:18 PM
Sci Advisor
PF Gold
P: 4,500
Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere
Nov24-06, 07:21 PM
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P: 2,688
wow! i must be tired!! Thanks alot shredder

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