<s>[INST] 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.
 [/INST] $$J = \begin{bmatrix} x_r & x_t \\ y_r & y_t \end{bmatrix} = \begin{bmatrix} \frac1{\sqrt3}\cos(t) & -\frac r{\sqrt3} \sin(t) \\ \frac1{\sqrt6}\sin(t) & \frac r{\sqrt 6}\cos(t) \end{bmatrix}$$
$$\implies \det(J) = x_r y_t - x_t y_r = \frac r{\sqrt{18}} \cos^2(t) + \frac r{\sqrt{18}} \sin^2(t) = \frac r{\sqrt{18}}$$
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