# Green's Theorem with change of variables

Evaluate $$\int_C F dr$$ $$F =< x2, xy >$$ $$C: x^2/4^2 + y^2/9 = 1$$

With y ≥ 0 positively oriented.

For the circle

$$u^2 + v^2 = 1$$

$$u=x/a$$

$$v=y/b$$

$$x=au$$

$$y=bv$$

For the ellipse

$$(x/a)^2 + (y/b)^2 = 1$$

Computing the jacobian, I get 6. So, using greens theorem and switching to polar I get:

$$\int \int (6rsinθ) rdrdθ$$

Just want someone to see if I've completed the changing of variables correctly. Computing integrals isn't all that difficult but I'm having a bit of trouble with the setup still.

$$F=(F_1(x,y),F_2(x,y))=(x^2,xy)$$ $$\oint\limits_{C}{F.dr}=\int{\int{\left( \frac{\partial {{F}_{2}}}{\partial x}-\frac{\partial {{F}_{1}}}{\partial y} \right)}}\,dA=\int\int y\,dA$$ let $x=2r\cos\theta$ and $y=3r\sin\theta$ we have $$\left| \frac{\partial (x,y)}{\partial (r,\theta )} \right|=6r$$ and $$\int\limits_{C}{F.dr}=\int_{0}^{2\pi}{\int_{0}^{1}} 18r^2\sin\theta\,drd\theta=\int_{0}^{2\pi}\sin \theta d\theta\times\int_{0}^{1}18r^2dr=0$$
• $a=\sqrt{4}$ and $b=\sqrt{9}$ and $dxdy=a\,b\,r\,drd\theta$ – Behrouz Maleki Jun 11 '16 at 15:15