Use Green's Theorem to evalutae
$$ \int <2y^2 + \sqrt{1+x^5} , (5x-e^y)> dr $$
$$ C: x^2+y^2=4 $$
C is positively orientated
$$ \int \int (dN/dx - dM/dy) dA $$
$$ = (5 - 4y) dA $$
$$ \int \int_0^2 (5 - 4y)) r drd(\theta) $$
$$ y=r\sin(\theta)$$
$$ \int_0^{2\pi} \int_0^2 (5r - 8\sin(\theta)r)) drd(\theta) $$
Evaluating inside integral =
$$ \int (10-16\sin(\theta)) d(\theta) $$
with a region between $0$ and $2\pi$
Evaluating this outside integral $= 20 \pi $
I was just hoping if someone had the time they could check if my setup and integration was correct, particularly in switching to polar coordinates. Thanks :)
Everything is right, but you missed an extra $r$ on the second last integral. It should be \begin{align*} & \int_0^{2\pi}\int_0^2 5r - 4r^2\sin(\theta)\, drd\theta\\ & = \int_0^{2\pi} \dfrac{5}{2}r^2-\dfrac{4}{3}r^3\sin(\theta)\bigg|_0^2 \,d\theta\\\ & = \int_0^{2\pi} 10-\dfrac{32}{3}\sin(\theta)\, d\theta\\ & = 20\pi. \end{align*}