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#### skatenerd

##### Active member

- Oct 3, 2012

- 114

So theres a sphere denoted as \(\Omega\):

$$x^2+y^2+z^2=R^2$$

and its cut by two planes \(z=a\) & \(z=b\) where \(0\leq{a}\leq{b}\leq{R}\). I have to find the area of the portion of \(\Omega\) cut by those two planes.

So I got to setting up this integral. Not so bad...

$$\int\int_{\Omega}^{ }\frac{|\nabla{F}|}{|\nabla{F}\bullet{p}|}\,dA$$

where $$\nabla{F}=2x\hat{i}+2y\hat{j}+2z\hat{k}$$

so now $$|\nabla{F}|=\sqrt{4x^2+4y^2+4z^2}=2\sqrt{x^2+y^2+z^2}=2R$$

and then $$\nabla{F}\bullet{p}$$ where \(p=\hat{k}\) (normal to the surface)

$$=(2x\hat{i}+2y\hat{j}+2z\hat{k})\bullet{\hat{k}}=2z$$

so now the integral looks like $$\int\int_{\Omega}^{ }\frac{2R}{2z}\,dA=R\int\int_{\Omega}^{ }\frac{1}{z}\,dA$$

This is where my problem comes up...I dont really have anything I can do with z...I have the two planes \(z=a\) & \(z=b\) but then how would I find my bound of integration? I feel like polar coordinates would be optimal but even so, what would the inner \(dr\) bounds be? The two connecting circles are

$$x^2+y^2=R^2-a^2$$ and $$x^2+y^2=R^2-b^2$$

Would integrating this integrand I found over this two circles make sense for finding the area?