1. ## [SOLVED] Line integral

I must calculate $\int _{C} xy^4 ds$ where $C$ is the right side of $x^2+y^2=16$.
My attempt : It's my first exercise involving such integrals and I don't understand well something.
So $C$ is the right side of a circle whose radius is $4$.
$\vec{r}(t)=(4\cos t, 4\sin t)$
If $f(x,y)=xy^4$ then $\int _{C} xy^4 ds=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt$.
How do I interpret $f(\vec{r}(t))$?
Is it $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4\cos t \cdot (4\sin(t))^4) \cdot 4 dt$?

2. Yes, that is correct.

Another way to write your parametrization is

x = 4cost
y = 4sint