Calculating double integral using variables substitution $\displaystyle
  D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \;  3 x^2 + 6 y^2 \leq 1  \right\rbrace$
Calculate $\displaystyle
   \iint_D  \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } }  \; dx dy{}$.
Attempt:
$x=\frac{r}{\sqrt3}cost,y=\frac{r}{\sqrt6}sint \implies |J|=\sqrt{\frac{2}{3}}r$
$3 x^2 + 6 y^2 \leq 1 \implies 0\leq r \leq 1$
$\iint_D  \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } }  \; dx dy{} =\int _0^1\:\int _0^{2\pi }\:\frac{cos^2t\sqrt{2}}{3\sqrt{3}}dtdr = \frac{\sqrt{2}\pi }{3\sqrt{3}}$
My answer isn't corect , can't find out what is wrong.
Appreciate any help.

With the change of variable $$(x,y)\to \left(\frac{1}{\sqrt{3}}r\cos t,\frac{1}{\sqrt{6}}r\sin t\right),$$ for $t\in [0,2\pi[$ and since $3x^2+6y^2\leqslant 1$ then $r\in [0,1]$.
The determinant of Jacobian is given by
$$\frac{\partial (x,y)}{\partial (r,t)}=\det\begin{pmatrix}\frac{\partial x}{\partial r}& \frac{\partial y}{\partial t}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial t}\end{pmatrix}=\frac{r}{3\sqrt{2}}>0.$$
Therefore, the integral is $$\frac{1}{3^2\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{1}\cos^{2}t\, drdt=\frac{\pi}{9\sqrt{2}}.$$
