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$\displaystyle D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \; 3 x^2 + 6 y^2 \leq 1 \right\rbrace$

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.

A. P.
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Algo
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  • Well obviously you should go over the intermediate steps one by one looking for arithmetic mistakes. It happens that it's in your computation of the Jacobian. – JBL Jan 02 '23 at 19:01

2 Answers2

1

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 $$\frac{1}{3^2\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{1}\cos^{2}t\, drdt=\frac{\pi}{9\sqrt{2}}.$$

A. P.
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  • I think your last integral is missing some $r,$s :$$ \frac{1}{3^2\sqrt{2}}\int_0^{2\pi}\int_0^1 \frac{\color{red}{r^2}\cos^2t}{\color{red}{r^{3/2}}}\color{red}{r},dr,dt=\frac{\pi\sqrt{2}}{45} $$ – Kurt G. Jan 03 '23 at 07:45
  • @KurtG. Let me check: $$\int_{0}^{2\pi}\int_{0}^{1}\frac{(\frac{r}{\sqrt{3}}\cos(t))^2}{(3(\frac{r}{\sqrt{3}}\cos t)^2+6(\frac{r}{\sqrt{6}}\sin t)^2)^{3/2}}\cdot \frac{r}{3\sqrt{2}}, drdt$$ $$=\frac{1}{3^2\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{1}\frac{r^2\cos^2(t)}{(r^{2}(\cos^2(t)+\sin^{2}(t)))^{3/2}}r, drdt$$ $$=\frac{1}{9\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{1}\cos^2(t), drdt.$$ I think it is correct. – A. P. Jan 03 '23 at 12:24
  • You are right. In my denominator there should be an $r^3$ so they cancel indeed. – Kurt G. Jan 03 '23 at 13:54
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$$J = \begin{bmatrix} x_r & x_t \\ y_r & y_t \end{bmatrix} = \begin{bmatrix} \frac1{\sqrt3}\cos(t) & -\frac r{\sqrt3} \sin(t) \\ \frac1{\sqrt6}\sin(t) & \frac r{\sqrt 6}\cos(t) \end{bmatrix}$$

$$\implies \det(J) = x_r y_t - x_t y_r = \frac r{\sqrt{18}} \cos^2(t) + \frac r{\sqrt{18}} \sin^2(t) = \frac r{\sqrt{18}}$$

user170231
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