I have the following problem where I'm trying to figure out how to convert a circle whose equation is $(x-1)^2 + (y+3)^2 = 25 $ traversed counterclockwise.
Here's the integral
$$ \oint_C 2ydx + 5xdy $$
From what I can see,
$$ x = 1 + 5cos\theta$$ $$ y = -3 + 5sin\theta$$
$$0 \leq \theta \leq 2\pi$$
where Green's theorem would yield $3$ when taking the partial derivates.
I know for a fact that my outer integral ranges from $0 \leq \theta \leq 2\pi$ but I cannot figure out how my $r$ ranges in this case. Does it have an affect or should my limits simply be:
$$ 3\int_o^{2\pi} \int_o^5 rdrd\theta$$
which doesn't seem right; I feel like it should be somewhere in the lines of
$$ 3\int_o^{2\pi} \int_{1 + 5cos\theta}^{??} rdrd\theta$$
Best Answer
From Green's Theorem we have
$$\oint_C(Pdx+Qdy)=\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$$
With $P=2y$, $Q=5x$, and $C$ is the boundary of $S$, a circular disk region with radius $5$ and center $(1,3)$, we have
$$\oint_C(2ydy+5xdx)=\iint_S 3 dxdy=3 (5)^2=75\pi$$
Direct Evaluation;
Let $x=1+5\cos t$ and $y=3+5\sin t$. Then, $dx=-5\sin tdt$ and $dy=5\cos tdt$. Thus,
$$\oint_C(2ydy+5xdx)=\int_0^{2\pi}\left(2(3+5\sin t)(-5\sin t)+5(1+5\cos t)(5\cos t)\right)dt=75\pi$$