<s>[INST] Finding $\iint_S {z \:ds}$ for some $S$ $$\iint_S {z \:ds}$$
In this double integral above, $S$ is the part of a sphere, $x^2+y^2+z^2=1$, which lies above the cone, $z=\sqrt{x^2+y^2}$. How can I calculate the above double integral.
Can someone help me to solve this? I got $\frac{\pi}{3\sqrt{2}}$ as my answer. Can someone verify this
 [/INST] This problem is actually best done in cylindrical coordinates.  Here when we let $x = r\cos\theta$ and $y = r\sin\theta$ we can rewrite our $z$-limits as $z = r$ and $r^2  +z^2 = 1$.  Note that when $z =1$ we have $r = 0$, and similarly when $z = r$ we get that $r^2 + r^2 = 1$ implying that $r = \frac{1}{\sqrt{2}}$.  Therefore our integral can be written
\begin{eqnarray*}
\int_0^{2\pi} d\theta \int_0^{1/\sqrt{2}} \int_r^{\sqrt{1-r^2}}rzdzdr & = & 2\pi\int_0^{1/\sqrt{2}}\frac{r}{2}(1-r^2 - r^2)dr \\
& = & \pi\int_0^{1/\sqrt{2}}r - 2r^3 dr \\
& = & \pi\left (\frac{1}{2}\cdot \frac{1}{2} - \frac{1}{2}\cdot \frac{1}{4} \right )\\
& = & \pi\left (\frac{2}{8} - \frac{1}{8} \right ) \\
& = & \frac{\pi}{8}.
\end{eqnarray*}
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