Problem:
Evaluate $I$: $$ I=\iint\limits_S yz^2\ dx\ dz\ \ \text{where}\ S\ \text{is the inner side of a cylinder}\ x^2+y^2=r^2,\ y\leqslant 0,\ 0\leqslant z\leqslant r $$
My attempt №1: $$ \begin{aligned} &y\leqslant0\Rightarrow y=-\sqrt{r^2-x^2},\\ &\text{The normal that goes through the inner side of $S$ makes an acute angle with $Oy+$, therefore we get:}\\ &I=-\iint\limits_{D(x,z)}\sqrt{r^2-x^2}z^2\ dx\ dz=-\int\limits_0^r z^2\ dz\int\limits_{-r}^{r}\sqrt{r^2-x^2}\ dx=-\frac{2r^3}{3}\int\limits_0^r\sqrt{r^2-x^2}\ dx \end{aligned} $$ And then I realized that something went wrong. I also tried a different approach.
My attempt №2 $$ \begin{aligned} &\begin{cases} x=u\cos v\\ y=u\sin v\\ z=u \end{cases},\ r=r(x(u,v),y(u,v),z(u,v))\Rightarrow \begin{cases} r'_u=(\cos v, \sin v, 1)\\ r'_v=(-u\sin v, u\cos v, 0) \end{cases}\Rightarrow\\ &\Rightarrow r'_u\times r'_v=(-u\cos v, -u\sin v, u)\\ &I=\iint\limits_{D(u,v)}\langle(0,yz^2,0),(u\cos v, -u\sin v, u)\rangle\ du\ dv=-\iint\limits_{D(u,v)} u^4\sin^2v\ du\ dv=\\ &=-\int\limits_0^r u^4\ du\int\limits_{\pi/2}^{3\pi/2}\sin^2 v\ dv=-\frac{r^5}{5}\cdot\frac{1}{2}\cdot\pi=-\frac{\pi r^5}{10} \end{aligned} $$ However, the correct answer is this: $$I=-\frac{\pi r^5}{6}$$
Could someone explain what is wrong with my attempts?
The half-cylinder $C$ is parametrised by $$ {\bf r}(\theta,z)=(r\cos\theta,r\sin\theta,z) $$ where $y\leq 0$ requires $\pi\leq\theta\leq 2\pi$ and the height range is $0\leq z\leq r$. The scalar field we are integrating over $C$ is given by $$ f(x,y,z)=yz^2 $$ We need to evaluate $$ I=\iint_D f({\bf r}(\theta,z))\|{\bf r}_{\theta}\times {\bf r}_z\|d\theta dz $$ where $D=\{(\theta,z): \pi\leq\theta\leq 2\pi, 0\leq z\leq r\}$.
Well, ${\bf r}_{\theta}=(-r\sin\theta, r\cos\theta,0)$ and ${\bf r}_z=(0,0,1)$. Hence, ${\bf r}_{\theta}\times {\bf r}_z=(r\cos\theta,r\sin\theta,0)$ which means that $\|{\bf r}_{\theta}\times {\bf r}_z\|=r$ and hence the infinitesimal area element is $dS = \|{\bf r}_{\theta}\times {\bf r}_z\|d\theta dz = rd\theta dz$, as intuition confirms it should be: $r d\theta$ is the sector length (or width) of the area element and $dz$ is the vertical height of the area element, so the area is $r d\theta\times dz$.
Putting all the pieces together, using that $y=r\sin\theta$ and $z$ is itself, we get \begin{eqnarray*} I &=& \int_{\theta=\pi}^{2\pi}\int_{z=0}^{z=r} (r\sin\theta)(z^2)rd\theta dz\\ &=& r^2\int_{\theta=\pi}^{2\pi}\sin\theta d\theta\int_{z=0}^{z=r}z^2dz\\ &=& r^2\Big(-\cos\theta|_{\theta=\pi}^{2\pi}\Big)\Big(\frac{1}{3}z^3|_{z=0}^{z=r}\Big)\\ &=& r^2\Big(-2\Big)\Big(\frac{1}{3}r^3\Big)\\ &=& -\frac{2}{3}r^5 \end{eqnarray*}
I don't think the question is correctly formulated since $dxdz$ does not describe the infinitesimal area $dS$ of $C$. In the original coordinates $(x,y,z)$ the infinitesimal area element is given by $$ \sqrt{(dx)^2+(dy)^2}dz = \sqrt{1+\Big(\frac{dy}{dx}\Big)^2}dxdz = \frac{r}{\sqrt{r^2-x^2}}dxdz = r d\theta dz $$ where you use $y=-\sqrt{r^2-x^2}$ then substitute in $x=r\cos\theta$ and $dx=-r\sin\theta d\theta$. Your attempt No 1. does integrate to $-\frac{\pi}{6}r^5$ but I don't believe this answer is the correct answer for the surface integral of $f(x,y,z)=yz^2$ over $C$. Your attempt No 2. is an attempt to integrate a vector field over $C$ when what you want is to do is integrate a scalar field over $C$, as I have done above. It would be helpful if you could double check that the statement of the problem is correct.