Green's Theorem

**chella182** · May 6th 2009, 10:19 AM

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?

**NonCommAlg** · May 6th 2009, 10:22 AM

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 $2 \pi a^2.$

**chella182** · May 6th 2009, 10:24 AM

Thankyou

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