1. ## Green's Theorem

I only have a minor problem with this really... the integral is...
$\oint_{\Gamma}\{(y\exp{[x^2y^2]}-y)dx+(x\exp{[x^2y^2]}+x)dy\}$
...where $\Gamma$, mapped counter-clockwise, is the circle of radius $a$ with centre at the origin.

So I identified $P=y\exp{[x^2y^2]}-y$ and $Q=x\exp{[x^2y^2]}+x$ and similarly $\frac{\partial P}{\partial y}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}-1$ and $\frac{\partial Q}{\partial x}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}+1$.

Putting that into the double integral you end up with $2\iint_{A}dA$, but I'm not sure where to proceed from here. Is $\iint_{A}dA$ the area of the circle?

2. Originally Posted by chella182
I only have a minor problem with this really... the integral is...
$\oint_{\Gamma}\{(y\exp{[x^2y^2]}-y)dx+(x\exp{[x^2y^2]}+x)dy\}$
...where $\Gamma$, mapped counter-clockwise, is the circle of radius $a$ with centre at the origin.

So I identified $P=y\exp{[x^2y^2]}-y$ and $Q=x\exp{[x^2y^2]}+x$ and similarly $\frac{\partial P}{\partial y}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}-1$ and $\frac{\partial Q}{\partial x}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}+1$.

Putting that into the double integral you end up with $2\iint_{A}dA$, but I'm not sure where to proceed from here. Is $\iint_{A}dA$ the area of the circle?
yes, it's the area of the circle. so your integral is equal to $2 \pi a^2.$

3. Thankyou