Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
Thank you very much! – pmartelletti Dec 11 '12 at 23:03

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}

share|cite|improve this answer
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! – pmartelletti Dec 11 '12 at 23:02
Yes, $\rho$ is the jacobian of the polar coordinates. – M. Strochyk Dec 12 '12 at 20:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.