Evaluate $ I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0 $ Evaluate $I$:
$$
I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0
$$
My attempt:
$$
\begin{aligned}
&z=-\sqrt{R^2-x^2-y^2}\Rightarrow
\begin{cases}
z'_x=\frac{x}{\sqrt{R^2-x^2-y^2}}\\
z'_y=\frac{y}{\sqrt{R^2-x^2-y^2}}
\end{cases}\\
&I=\iint\limits_{D(x,y)}\langle(x^5+z, 0, 0), (z'_x, z'_y,-1)\rangle\ dx\ dy=\iint\limits_{D(x,y)}\frac{x^6+xz}{\sqrt{R^2-x^2-y^2}}\ dx\ dy=\\
&=\iint\limits_{D(x,y)}\left(\frac{x^6}{\sqrt{R^2-x^2-y^2}}-x\right)\ dx\ dy=\int\limits_0^{2\pi}d\varphi\int\limits_0^R\left(\frac{r^6\cos^6\varphi}{\sqrt{R^2-r^2}}-r\cos\varphi\right)r\ dr=\\
&=\int\limits_0^{2\pi}\cos^6\varphi\ d\varphi\int\limits_0^R\frac{r^6}{\sqrt{R^2-r^2}}\ dr
\end{aligned}
$$
And then I got stuck because of the last integral. Perhaps I made a mistake somewhere in the beginning.
The answer should be the following:
$$
I=-\frac{2\pi R^7}{7}
$$
Could someone help me to solve this problem? I would appreciate it.

You forgot to multiply by another factor of $r$.  The integral should be
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr.
$$
Take $r = R\sin u$ then you obtain
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr \;\; =\;\; \int_0^{\frac{\pi}{2}} \frac{R^7\sin^7u}{R\sqrt{1-\sin^2u}}R\cos udu \;\; =\;\; \int_0^{\frac{\pi}{2}} R^7\sin^7udu.
$$
Expand $\sin^7u = \sin u\left (1-\cos^2u\right )^3$, and make the substitution $w = \cos u$.  Then your integral becomes
$$
-\int_1^0 R^7\left (1-w^2 \right )^3dw \;\; =\;\; R^7\int_0^1\left (1 - 3w^2 + 3w^4 - w^6\right )dw \;\; = \;\; R^7\left (1 - 1 + \frac{3}{5} - \frac{1}{7}\right ) \;\; =\;\; \frac{16R^7}{35}.
$$
