Help with the line integral $\int x^2dS$ where C is a upper-half of a circle $x^2+y^2=r^2$ $$\int_C x^2dS$$ where C: $$x^2+y^2=r^2$$
So, $$ x=r\cos(\theta) $$ $$ y=r\sin(\theta) $$ $$ 0 \leq \theta \leq \pi $$
How would arc length of this curve go? $$ dS = \sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}d\theta $$ What should I put in $\left(\frac{dr}{d\theta}\right)^2$ and $r^2$?

$r$ is constant and $> 0$ in a circle. Therefore:
$
\frac{dr}{d\theta}=0
$
And
$
ds = \sqrt{(0)^2+(r)^2} d\theta = r d\theta
$
From there, your integral should be straightforward.

Another way to look at it:
\begin{align}
ds &= \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta \\
\frac{dx}{d\theta} &= -r\sin{\theta} \\
\frac{dy}{d\theta} &= r\cos{\theta} \\
\Rightarrow ds &= \sqrt{(-r\sin{\theta})^2 + (r\cos{\theta})^2}d\theta = r d\theta \\
\end{align}
