How to integrate this integral using Green's theorem?

Question

$\int_{L} (x+2xy)dy + y^{2}dx$ in the circle $L=\{(x,y)|x^2+y^2=y\}$. I already know that $P_y'=2y$ and that $Q_x'=1+2y$ where $P$ is the function multiplied by $dx$ and $Q$ is the function multiplied by $dy$. So I figured that since the given circle has $x$ in $[-0.5,0.5]$ and $y$ is in $[0,1]$ then I should integrate twice $\int_{0}^{1}(\int_{-0.5}^{0.5}1+4ydx)dy$. Is this correct?

Answer 1

It'll look cluttered in a comment, so I'll try to address the main points here.

You're correct in deducing that $L$ traces out a circle with radius $1/2$ and center $(0,1/2)$. Green's theorem can be used for your integral. There is a sign error in your expression for the curl of your vector field $\mathbf{F}(x,y) = (y^2, x+2xy)$. It should be

$$ \text{curl}(\mathbf{F}) = \left( \dfrac{\partial Q}{\partial x} \ {\color{red}-} \ \dfrac{\partial P}{\partial y} \right) = 1+2y-2y = 1 $$

Let $R$ denote the region enclosed by $L$. By Green's theorem, we can express our integral as

$$ \oint_{L} \mathbf{F} \cdot \text{d}s = \iint_{R} \text{curl}(\mathbf{F}) \text{ d}A $$

For the purpose of this question, it should be easy to see that $ \iint_{R} \text{d}A = \text{area}(R) $.

If you wanted to actually compute such an integral, the bounds you've selected, namely $x \in [-1/2,1/2]$ and $y \in [0,1]$, actually trace out a square in $\mathbb{R}^{2}$. You can try parameterising a circular region with center $(0,0)$ and radius $1$ using Cartesian coordinates. The you can just shift and scale this region to $R$.)