Let γ be the ellipse $x^2 + 4y^ 2 = 4$, oriented anticlockwise. Compute $\int_c(4y − 3x)dx + (x − 4y)dy$

I used green theorem with P and Q. and got $$-3\int\int_\ dxdy$$

The answer is $-6\pi,$ so how do I get $2\pi$ from the double integral? Or am I just tackling the problem all wrong?

  • $\begingroup$ green is Green: always use a Capital for names. $\endgroup$ – Jean Marie Mar 22 '17 at 22:36

The double integral will give you the area of the ellipse.

Your ellipse is $$\left( \frac{x}{2}\right)^2+\left(\frac{y}{1}\right)^2 =1 $$ with $a=2$ and $b=1$ and has area $\pi a b= 2 \pi$

  • $\begingroup$ what would the bounds be though? isn't 2pi usually expressed in term of d(theta) not dx or dy? $\endgroup$ – user3427042 Mar 22 '17 at 22:04
  • $\begingroup$ dx dy is a 2d cartesian area element, and the double integral is over the entire ellipse so it is calculating the area of the ellipse. You don't need to change to another coordinate system. $\endgroup$ – PM. Mar 22 '17 at 22:09
  • $\begingroup$ denominator should be 4 right not 2? $\endgroup$ – user3427042 Mar 22 '17 at 22:11
  • $\begingroup$ A pleasure. You can see the details of the double integral, in both cartesian and polar coordinates, here mymathforum.com/calculus/29970-area-ellipse.html $\endgroup$ – PM. Mar 22 '17 at 22:16
  • $\begingroup$ Regarding denominator, 2^2 = 4. $\endgroup$ – PM. Mar 22 '17 at 22:23

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