It is parameterized as follows: $x = 2\cos(\theta)$ and $y = 3\sin(\theta)$.

This is my double integral. It is not evaluating to the right answer. Why?

$$\int_0^{2\pi} \int_0^{\sqrt{4\cos^2(\theta) + 9\sin^2(\theta)}} dr\,d\theta$$

I remember from calculus 3 that to get the area for polar coordinates, I just evaluate $dr$ and $d\theta$. I don't see what's wrong with the limits of integration. The radius is from 0 to the formula and the radians are a full revolution.

  • 3
    $\begingroup$ The area element in polar coordinates is $r \, dr \, d\theta$. $\endgroup$
    – Martin R
    Aug 21 '16 at 2:23
  • $\begingroup$ ahh i see sorry my mistake $\endgroup$ Aug 21 '16 at 2:27
  • 4
    $\begingroup$ Worse, there is a serious problem. The $\theta$ that appears in this parametrization is NOT the polar coordinates $\theta$. You can see this quite easily if you think about stretching a circle to make the ellipse. $\endgroup$ Aug 21 '16 at 4:27
  • $\begingroup$ The corrected integral is giving me $13\pi/2$, and computing the area using Green's theorem is giving me the correct $6\pi$. Is there still a mistake here? $\endgroup$
    – JAustin
    Aug 21 '16 at 4:29
  • 1
    $\begingroup$ The area of an ellipse of major axis $2a$ and minor axis $2b$ is certainly $\pi ab$. $\endgroup$
    – Lubin
    Aug 21 '16 at 4:37

In order to include a diagram, I'm turning my comment into an answer. As I said, the $\theta$ that appears in this parametrization is NOT the polar coordinates $\theta$. You can see this quite easily if you think about stretching a circle to make the ellipse. I've substituted $t$ in the parametrization and indicated the polar coordinates $\theta$ as well.


To do this integral correctly in polar coordinates you must get the polar coordinates equation of the ellipse for starters: $$\frac{x^2}4 + \frac{y^2}9 = 1 \implies \frac{(r\cos\theta)^2}4 + \frac{(r\sin\theta)^2}9 = 1 \implies r = \frac1{\sqrt{\frac{\cos^2\theta}4+\frac{\sin^2\theta}9}}.$$ Ugh! This will give us the integral $$\frac12\int_0^{2\pi} \frac1{\frac{\cos^2\theta}4+\frac{\sin^2\theta}9}\,d\theta,$$ which can be done, but this is not the right way to do this problem!

  • $\begingroup$ Thank you for the figure. However, how is the fact that the thetas don't equal each other when the x1=x2 a problem? It is an ellipse, but that should be taken into consideration from the limit of integration of this particular integral that I wrote in my question-$\int_0^{\sqrt(4*cos...)}dr$ The radius limit is not constant, thus the radius changes based on the formula. $\endgroup$ Aug 22 '16 at 2:32
  • $\begingroup$ And then the theta tells it how far to traverse. The ellipse is still 2pi radians $\endgroup$ Aug 22 '16 at 2:34
  • $\begingroup$ Also, the constants 2 and 3 stretch out the x and y coordinate respectively. Since you drew the line to the ellipse at a different y coordinate so it is not proportional to the original circle of 2/1 x and 3/1 y, so of course the degrees aren't the same, but I don't understand the relation between evaluating the area. $\endgroup$ Aug 22 '16 at 2:37
  • $\begingroup$ When you compute the area by doing $$\int_0^{2\pi}\int_0^{r(\theta)} r\,dr\,d\theta,$$ the function $r(\theta)$ must actually be giving $r$ in terms of the polar coordinate $\theta$. That is wrong with your parametrization, as I've tried to explain. The vector to the point $(2\cos t,3\sin t)$ makes angle $\theta=\tan^{-1}\big(\frac32\tan t\big) \ne t$ with the positive $x$-axis. $\endgroup$ Aug 22 '16 at 3:45
  • $\begingroup$ I am beginning to understand, but I've thought about this for a while and I still have some intuitive questions. Say that I merely substitute $d\theta$ for $dt$. I know this is mathematically incorrect because of the formula you've given relating theta and t, but I don't see how this is wrong intuitively. I keep the upper limit of the integral at my mistaken $r(t)$ rather than $r(\theta)$. The radius is now expressed as t. Then, I traverse the full revolution of the circle with t rather than theta. The limits will be the same: from 0 to 2pi. $\endgroup$ Aug 22 '16 at 17:46

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