<s>[INST] Find $\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}$ 
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:
$D_n=\left[\frac{1}{n},1\right]\times[0,2\pi]$.
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?
 [/INST] There is an error in the computation of the Jacobian: it is equal to $\frac r{2\sqrt3}$.
</s>