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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 a we know that r goes from 0 to a in this integral.
The equation for a sphere is:
x^2 + y^2 +z^2 = r^2
or f(x,y) = \sqrt{r^2 -x^2 -y^2}
and it intersects the x-y plane in a circle:
x^2 + y^2 = r^2So 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 \theta goes from 0 to2\pi.
So we get:
2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta
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.
The sphere has radius a we know that r goes from 0 to a in this integral.
The equation for a sphere is:
x^2 + y^2 +z^2 = r^2
or f(x,y) = \sqrt{r^2 -x^2 -y^2}
and it intersects the x-y plane in a circle:
x^2 + y^2 = r^2So 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 \theta goes from 0 to2\pi.
So we get:
2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta
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.
