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Evaluate $$\int\int z^2\,dS,$$ where $S$ is the part of outer surface of cylinder $x^2+y^2=4$ between the planes $z=0$ and $z=3$.

The answer given in book is $\pi$ but I am not getting this answer.

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Since your surface is a cylindar of radius $2$, the surface element can be written as $\mathrm{d}S=2\,\mathrm{d}\theta\,\mathrm{d}z$. Then your integral can be written as: $$I=2\iint\limits_{(\theta,z)\in[0,2\pi]\times[0,3]}z^2\,\mathrm{d}\theta\,\mathrm{d}z,$$ and using separation of variables: $$I=2\int\limits_{\theta\in[0,2\pi]}\mathrm{d}\theta\int\limits_{z\in[0,3]}z^2\mathrm{d}z=2\times2\pi\times\frac{3^3}3=36\pi.$$


The answer $\pi$ that you're given is certainly wrong for the following very simple reason: you're integrating a non-negative function on the cylinder; so the integral is non-less than the integral on the portion between $z=1$ and $z=3$; now, on this portion, $z^2\geq1$, hence the integral must be at least the surface area of the portion of the cylinder between $z=1$ and $z=3$ which is $2\times\pi\times2\times2=8\pi$.

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