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The sphere $x^2 + y^2 + z^2 = 4$ is cut by the plane $z = 1/2$. How do you calculate the volume of two parts of the sphere using integrals? Thank you!

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The standard setup is $$ \begin{align} \int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}\int_{-\sqrt{4-z^2-y^2}}^\sqrt{4-z^2-y^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z &=\int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}2\sqrt{4-z^2-y^2}\;\mathrm{d}y\;\mathrm{d}z\\ &=\int_\frac{1}{2}^2\pi(4-z^2)\;\mathrm{d}z\tag{1} \end{align} $$ and $$ \begin{align} \int_{-2}^\frac{1}{2}\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}\int_{-\sqrt{4-z^2-y^2}}^\sqrt{4-z^2-y^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z &=\int_{-2}^\frac{1}{2}\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}2\sqrt{4-z^2-y^2}\;\mathrm{d}y\;\mathrm{d}z\\ &=\int_{-2}^\frac{1}{2}\pi(4-z^2)\;\mathrm{d}z\tag{2} \end{align} $$ You might only need the last integral of each, but I started at ground-zero.

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thank you, that's what I wasn't sure about – Eva K. Dec 4 '11 at 19:34

I'll provide only a skeleton answer since this must be homework, which it should be tagged as. btw.

There should be an example in your calculus textbook where they compute the volume of a sphere or related three dimensional object using a triple integral. You could modify that argument by simply changing the limits of integration for $z$.

Alternatively, you could simply integrate the area formula for a circle between the relevant bounds for $z$.

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Yeah, I understand both methods, I'm just not sure about the "lower" part of the sphere, since at that part z is negative and then positive, so I have z = -sqrt(4 - x^2 - y^2) and then z = +sqrt (4 - x^2 - y^2). Not sure what to do with limits of z... – Eva K. Dec 4 '11 at 18:41
I just substracted the volume of the upper part from 4/3pi*r^2, but what if I didn't know the formula? – Eva K. Dec 4 '11 at 19:05

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