[SOLVED] Line integral

**arbolis** · June 14th 2009, 06:15 PM

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

**Random Variable** · June 14th 2009, 06:27 PM

Yes, that is correct.

Another way to write your parametrization is

x = 4cost
y = 4sint

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