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 $$\int\limits_{C}{F.dr}=\int_{0}^{2\pi}{\int_{0}^{1}} 18r^2\sin\theta\,drd\theta=\int_{0}^{2\pi}\sin \theta d\theta\times\int_{0}^{1}18r^2dr=0$$