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Question: Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.

The solution given to us by our instructor: $$\int_{0} ^{2\pi}\int _{0} ^{1} \int _{-\sqrt3} ^{\sqrt3} r\,dz\,dr\,d\theta$$

I think this solution is incorrect since the top and bottom surfaces in the $z$ direction are clearly not $-\sqrt3$ and $\sqrt3$ and that the radius within the cylinder can't possibly be a constant like 1 if we are starting from the origin. (I could be wrong though).

Can someone please show me how to correctly set up a triple integral to solve this problem or explain to me why the equation above is correct?

Thanks in advance!

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  • $\begingroup$ Your instinct is spot on. The top and bottom surfaces should be the sphere, not planes. $\endgroup$
    – Ninad Munshi
    Oct 26, 2020 at 9:19

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You are right. That's the volume of a cylinder, and therefore it cannot possibly be the right integral. It should be$$\int_0^{2\pi}\int_0^1\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$

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