Compute the integral
$\int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
My try:
Put $x=2\cos \theta,y=2\sin \theta $
$\int (x-y) dx +xy dy$ becomes
$-4\int (\cos \theta -\sin \theta) \sin \theta d\theta+\int 8\sin \theta \cos^2 \theta d\theta$
$=2 \int 1-\cos 2\theta d\theta$
$=4\pi$
Is my answer correct?
Please help.
Seems good. You could have also applied Green's theorem: \begin{align} \oint_{x^2+y^2 = 4} (x-y)\, \mathrm dx + xy \, \mathrm dy &= \iint_{x^2+y^2 \leq 4} \frac{\partial}{\partial x} (xy) - \frac{\partial}{\partial y} (x-y) \, \mathrm dx \,\mathrm dy\\ &= \iint_{x^2+y^2 \leq 4} y+1 \, \mathrm dx \,\mathrm dy\\ &= \iint_{x^2+y^2 \leq 4} 1 \, \mathrm dx \,\mathrm dy = 4\pi, \end{align} since $$ \iint_{x^2+y^2 \leq 4} y \, \mathrm dx \,\mathrm dy = 0 $$ by symmetry.