Help with a surface integral Let G(x, y, z) = $\ (1-x^{2}-y^{2})^{\frac{3}{2}}$
Evaluate the surface integral:
$\int\int_{S}^{}G(x, y,z)dS$
where S is the hemisphere:
z =$\ (1-x^{2}-y^{2})^{\frac{1}{2}}$
This is my work thus far:
$\int_{}^{}\int_{S}^{}G(x,y,z)dS=\int_{}^{}\int_{R}^{}(1-x^{2}-y^{2})^{\frac{3}{2}}\sqrt[]{1+\frac{x^{2}}{z^{2}}+\frac{y^{2}}{z^{2}}} \ dxdy$
Factoring 1/z^2 out of the square root we obtain:
= $\int_{}^{}\int_{R}^{}(1-x^{2}-y^{2})^{2}\ dxdy$
Converting to polar coordinates (x = cos$\theta$; y = sin$\theta$, dxdy = rdrd$\theta$):
$\int_{0}^{2\pi}\int_{0}^{1}(1-r^{2})^{2}r \ drd\theta =\frac{\pi}{3}$
The answer provided in the textbook is pi/2.
 A: Since,
$$\boxed{\iint_{S}f(x,y,z)\,{\rm d}S=\iint_{D}f(x,y,g(x,y))\sqrt{z_{x}^{2}+z_{y}^{2}+1}\,{\rm d}A}$$
So, we have
\begin{align*}
\iint_{S}(1-x^{2}-y^{2})^{3/2}\, {\rm d}S&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{z_{x}^{2}+z_{y}^{2}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\left(-\frac{x}{\sqrt{1-x^{2}-y^{2}}} \right)^{2}+\left(-\frac{y}{\sqrt{1-x^{2}-y^{2}}}\right)^{2}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}}{1-x^{2}-y^{2}}+\frac{y^{2}}{1-x^{2}-y^{2}}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}+y^{2}}{1-x^{2}-y^{2}}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}+y^{2}+1-x^{2}-y^{2}}{1-x^{2}-y^{2}}}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{1}{1-x^{2}-y^{2}}}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})\, {\rm d}A,\\
&=\int_{0}^{2\pi}\int_{0}^{1}(1-r^{2})r\, {\rm d}{\rm d}\theta,\\
&=\int_{0}^{2\pi} \frac{1}{4}\, {\rm d}\theta,\\
&=\frac{2\pi}{4},\\
&=\boxed{\frac{\pi}{2}}
\end{align*}
The change of variables $x=r\cos \theta$ and $y=r\sin \theta$ with $r\in \mathbb{R}^{*}_{+}$ and $\theta\in \left[0,2\pi\right[$ has Jacobian $r$ and the limits of integration is given because $S$ is given by $x^{2}+y^{2}+z^{2}=1$ in the top upper.
