Find $\displaystyle\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}.$

I got an answer, however the computer marks it as wrong and I can't find my mistake (if there is one, because sometimes it deems correct answers incorrect).

First of all I will use the substitution:

\begin{align*} x&=\frac{1}{2}\,r\cos(\theta)\\ y&=\frac{1}{\sqrt{3}}\,r\sin(\theta). \end{align*}

The Jacobian I get is $\dfrac{r}{\sqrt{3}}$.

Now, I will use this sequence of domains:


Now I tried to calculate the integral:

\begin{align*} \iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy &=\iint_{D_n} \frac{\frac{1}{4}r^2\cos^2(\theta)}{(r^2)^{3/2}}\frac{1}{\sqrt{3}}\,r \,dr\,d\theta \\ &=\int_{0}^{2\pi}\left(\int_{\frac{1}{n}}^{1} \frac{1}{4\sqrt{3}}\,\cos^2(\theta)\, dr \right)d\theta\\ &=\int_{0}^{2\pi}\left(1-\frac{1}{n}\right)\frac{\cos^2(\theta)}{4\sqrt{3}} \,d\theta. \end{align*}

From here this is a simple integral that gets the value $\dfrac{\pi}{4\sqrt{3}}$ when $n \rightarrow \infty$, however this is not the correct answer.

Where is my mistake?

  • 1
    $\begingroup$ Isn't the Jacobian equal to $\frac r{2\sqrt3}$? $\endgroup$ Jun 25 '19 at 22:29
  • $\begingroup$ Yup, found it after I made a similar mistake in another problem. Thank you. $\endgroup$ Jun 25 '19 at 22:52
  • $\begingroup$ Did this solve your problem? $\endgroup$ Jun 26 '19 at 3:50
  • $\begingroup$ It did solve it, that was my only mistake. $\endgroup$ Jun 26 '19 at 13:55
  • $\begingroup$ Then I will post my comment as an answer. $\endgroup$ Jun 26 '19 at 13:57

There is an error in the computation of the Jacobian: it is equal to $\frac r{2\sqrt3}$.


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