$\int_R xy^2ds$ where $R$ is the upper half of the circle $x^2+y^2=25$. Is my procedure correct for the following problem?
Calculate
$$\int_R xy^2ds$$
where $R$ is the upper half of the circle $x^2+y^2=25$.
What I did was parametrize the circle which gives $g(t) = (5\cos(t), 5\sin(t))$ and $0 \leq t \leq \pi $. This ends in the integral
$$\int_0^{\pi} (125 \sin^2(t)\cos(t))(5)dt=0$$
Is it correct that it's $0$ or did I make an incorrect parametrization? Thanks.

It's unclear whether you want the surface integral over the region enclosed or the line integral over the boundary of the semicircle, but you can use a symmetry argument in any case.
Here is how you would do it for the surface integral. Let us split your region into two disjoint parts $A$ and $B$, where $A = \{(x, y) ~|~ x = 0\}$ and $B = \{(x, y) ~|~ x \neq 0\}$. Note that for every point $(x, y)$ with $x \neq 0$ in region $B$, the point $(-x, y)$ also lies in the region. Since the integrand is odd in $x$ (i.e. $xy^2 = -((-x)y^2$), and hence $\int_B xy^2 ds = 0$. In addition, the integrand is zero on every point in $A$, so it follows that $\int_A xy^2 ds = 0$. Hence,
$$\int_R xy^2 ds = \int_A xy^2 ds + \int_B xy^2 ds = 0$$
The symmetry argument for the line integral is similar.
