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Another Polar Double Integral 
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#1
Nov2406, 07:03 PM

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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 xy 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. 


#2
Nov2406, 07:18 PM

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Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere



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