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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.

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$$F=(F_1(x,y),F_2(x,y))=(x^2,xy)$$ $$\oint\limits_{C}{F.dr}=\int{\int{\left( \frac{\partial {{F}_{2}}}{\partial x}-\frac{\partial {{F}_{1}}}{\partial y} \right)}}\,dA=\int\int y\,dA$$ let $x=2r\cos\theta$ and $y=3r\sin\theta$ we have $$\left| \frac{\partial (x,y)}{\partial (r,\theta )} \right|=6r$$ and $$\int\limits_{C}{F.dr}=\int_{0}^{2\pi}{\int_{0}^{1}} 18r^2\sin\theta\,drd\theta=\int_{0}^{2\pi}\sin \theta d\theta\times\int_{0}^{1}18r^2dr=0$$

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  • $\begingroup$ I got the same answer to the integral. Your setup is way faster though! $\endgroup$ – says Jun 10 '16 at 14:58
  • $\begingroup$ Please ,your solution is correct. good luck $\endgroup$ – Behrouz Maleki Jun 10 '16 at 15:00
  • $\begingroup$ how do you get x=2rcos and y=3rsin ? $\endgroup$ – says Jun 11 '16 at 15:09
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    $\begingroup$ $a=\sqrt{4}$ and $b=\sqrt{9}$ and $dxdy=a\,b\,r\,drd\theta$ $\endgroup$ – Behrouz Maleki Jun 11 '16 at 15:15
  • $\begingroup$ How come we write x=2rcos and y=3rsin. I can see where the 2 and 3 come from, they are the radius of the ellipse with respect to an axis, but why do we include the r in the equations? $\endgroup$ – says Jun 11 '16 at 15:19

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