Question

Compute single integral of y^2 dx +x dy where C is the circle x^2+y^2=16 oriented positively

Answer 1

Parameterize $C$ by

$\vec r(t)=\langle x(t),y(t)\rangle=\langle4\cos t,4\sin t\rangle$

with $0\le t\le2\pi$. Then

$\displaystyle\int_Cy^2\,\mathrm dx+x\,\mathrm dy=\int_0^{2\pi}\langle y(t)^2,x(t)\rangle\cdot\left\langle\frac{\mathrm dx(t)}{\mathrm dt},\frac{\mathrm dy(t)}{\mathrm dt}\right\rangle\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\langle16\sin^2t,4\cos t\rangle\cdot\langle-4\sin t,4\cos t\rangle\,\mathrm dt$

$=\displaystyle16\int_0^{2\pi}(\cos^2t-4\sin^3t)\,\mathrm dt=\boxed{16\pi}$