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(\mathrm dx(t))/(\mathrm dt),(\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}`