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 $

Question

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.

Answer 1

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 $$ -\int_1^0 R^7\left (1-w^2 \right )^3dw \;\; =\;\; R^7\int_0^1\left (1 - 3w^2 + 3w^4 - w^6\right )dw \;\; = \;\; R^7\left (1 - 1 + \frac{3}{5} - \frac{1}{7}\right ) \;\; =\;\; \frac{16R^7}{35}. $$