I have these particular exercise that i cannot solve. I know i have to change the variables, but i cannot figure out if i should use polar coords or any other change.

Let D be the region delimited by: $$ D = \{(x,y) \in \mathbb{R} ^{2} : (x-1)y \geq 0, \frac{(x-1)^2}{9} + \frac{y^2}{25} \leq 1 \} $$ Calculate:

$$ \iint\limits_D \sin((x-1)^2 + \frac{9y^2}{25}) \,dxdy $$

I've tried using $u = \frac{(x-1)}{3}$ and $v = \frac{y}{5}$ so that i can replace in the integral the following:

$$ \iint\limits_D \sin(9(u^2 + v^2)) \frac{1}{15} \,dudv $$

knowing the Jacobian is $J(x,y)=\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{15}$.

But i don't know where to follow, or if the variable changes i've made are correct. Can I use that $u^2 + v^2 = 1$, or that's just for polar coords?

Thanks a lot for your help!


You can indeed say now the the region (in $uv$-coordinates) is $D=\{(u,v)\in\mathbb{R}^2|u^2+v^2\leq 1\}$. So you've transformed your original elliptical region into a circular region using your $uv$-change of variable. Treating this as a problem in its own right, you are completely free to change now to polar coordinates.

If you think about it this makes perfect sense. Your first change of variable from $xy$ to $uv$ was a linear map. You can think about a linear map as stretching/compressing the $x$-axis and $y$-axis by some scalar factors and also changing the angle between them. Your transformation does this so that the elliptical region looks circular.


Integration domain $D = \{(x,y) \in \mathbb{R} ^{2}\colon \;\; (x-1)y \geqslant 0,\;\; \frac{(x-1)^2}{9} + \frac{y^2}{25} \leqslant 1 \}$ is two quarters of ellipse $\frac{(x-1)^2}{9} + \frac{y^2}{25} \leqslant 1$ which are located between pairs of rays $x\geqslant{1},\;y\geqslant{0}$ and $x\leqslant{1},\;y\leqslant{0}$ respectively. Mapping, given by \begin{gather} u = \frac{(x-1)}{3},\\ v = \frac{y}{5}, \end{gather} maps whole ellipse onto disc $K=\{(u,\, v) \in \mathbb{R} ^{2}\colon \;\; u^2+v^2 \leqslant {1} \}$. Polar coordinates \begin{gather} u=\rho\cos{\varphi} \\ v=\rho\sin{\varphi} \end{gather} are useful in this case with ${0}\leqslant{\rho}\leqslant{1}; \;\; {0}\leqslant{\varphi}\leqslant \frac{\pi}{2} \;\;\text{and}\;\;{\pi}\leqslant{\varphi}\leqslant \frac{3\pi}{2}.$ Therefore, \begin{gather} \iint\limits_K \sin(9(u^2 + v^2)) \frac{1}{15} \,dudv=\\ \frac{1}{15}\int\limits_{0}^{1}\left(\int\limits_{0}^{\frac{\pi}{2}}\sin(9\rho^2){\rho} \space d\varphi \right) d\rho+\frac{1}{15}\int\limits_{0}^{1}\left( \int\limits_{\pi}^{\frac{3\pi}{2}}\sin(9\rho^2)\rho \space d\varphi\right) d\rho \end{gather}

  • $\begingroup$ Well, i was missing the step of changing the ellipse onto the disc with polar coordinates! I've never seen an example like that (altough i imagine those were the logic steps). The p you add, is the jacobian of the polar coordinates, right? Thanks a lot for your help! $\endgroup$ – pmartelletti Dec 11 '12 at 23:02
  • $\begingroup$ Yes, $\rho$ is the jacobian of the polar coordinates. $\endgroup$ – M. Strochyk Dec 12 '12 at 20:41

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