I must calculate $\displaystyle \int _{C} xy^4 ds$ where $\displaystyle C$ is the right side of $\displaystyle x^2+y^2=16$.

My attempt : It's my first exercise involving such integrals and I don't understand well something.

So $\displaystyle C$ is the right side of a circle whose radius is $\displaystyle 4$.

$\displaystyle \vec{r}(t)=(4\cos t, 4\sin t)$

If $\displaystyle f(x,y)=xy^4$ then $\displaystyle \int _{C} xy^4 ds=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt$.

How do I interpret $\displaystyle f(\vec{r}(t))$?

Is it $\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4\cos t \cdot (4\sin(t))^4) \cdot 4 dt$?