Question

Evaluate the line integral ∫c3xy2ds, where c is the right half of the circle x2+y2=36.

Answer 1

We can choose $x=6\cos t$ and $y=6\cos t$, so that the path $C$ is traversed when $-\dfrac\pi2\le t\le\dfrac\pi2$.

Now

$\displaystyle\int_C 3xy^2\,\mathrm dS=3\int_{t=-\pi/2}^{t=\pi/2}(6\cos t)(6\sin t)^2\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt$
$=\displaystyle648\int_{t=-\pi/2}^{t=\pi/2}\cos t\sin t^2\sqrt{36\cos^2t+36\sin^2t}\,\mathrm dt$
$=\displaystyle3888\int_{t=-\pi/2}^{t=\pi/2}\cos t\sin t^2\,\mathrm dt$

Taking $u=\sin t$, we have

$=\displaystyle3888\int_{u=-1}^{u=1}u^2\,\mathrm du$
$=1296(1^3-(-1)^3)$
$=2592$