I need to integrate $f(x,y):=x^2y^2$ over an area $B\subset\mathbb R^2$that is restricted by the following 4 functions: $$y=\frac x9;\;y=\frac x4;\;y=\frac 1x;\;y=\frac4x;$$ Of course due to the symmetry we can integrate just over the area in the positive $x$-$y$ region and then multiply the result by two.

I am confused by choosing the boundaries for integration. On the one hand it could be $$\frac x9 \leq y\leq \frac x4,\;2\leq x\leq6$$ or alternatively $$\frac 1x \leq y\leq \frac 4x,\;2\leq x\leq6$$

So how do I choose the boundaries to calculate $\int_Bx^2y^2d\mu(x,y)?$ B

  • $\begingroup$ If the two boundaries lead to different answers, then there is something wrong in your derivation. They are supposed to give you the same answer in the end, and as of such, you are free to choose which ever boundaries you please, preferably the ones that lead to the easiest calculations. $\endgroup$
    – fromGiants
    Jun 20, 2014 at 13:59
  • $\begingroup$ In our case $x$ goes obviously from $2$ to $6$, but $y$ is restricted by $$\frac1x, \; \frac4x,\;\frac x9,\; \frac x4$$ Say we choose $\frac x9 \leq y\leq \frac x4,\;2\leq x\leq6$. We restrict the area by two hyperbolas and two vertical lines $x=2$ and $x=6$, but not lines $y=x/9$ and $y=x/4$ $\endgroup$ Jun 20, 2014 at 14:07
  • $\begingroup$ Do you have the final answer? $\endgroup$ Jun 20, 2014 at 14:54

3 Answers 3


You need to use a nonstandard change of coordinates to solve this. Adopt $u = xy$ and $v = x/y$.

The motivation for this is as follows: you can rewrite the equations $y=1/x$ and $y=4/x$ as $xy =1$ and $xy=4$, therefore we obtain the change $u=xy$.

You can rewrite the other two equations $y=x/4$ and $y=x/9$ as $x/y=4$ and $x/y=9$, leading to $v=x/y$. Isolating $x$ and $y$ in terms of $u$ and $v$ as defined you get $$x = \sqrt{uv}, \quad y = \sqrt{\frac{u}{v}}.$$ Computing the Jacobian and taking the absolute value gives $$|J(u,v)| = \frac{1}{2v}.$$ The new limits are $1 \leq u \leq 4 $ and $4 \leq v \leq 9$. The function $f(x,y) = x^2y^2$ is now $g(u,v) = u^2$ and the integral is $$I = \frac{1}{2} \int_1^4 \int_4^9 \frac{u^2}{v} \, dv \, du.$$ My computation yields $$I = 21 \ln \left( \frac{3}{2} \right).$$

  • $\begingroup$ That's absolutely perfect. The one question is: do I always need to consider parametrization when coming across boundaries, that are not defined by vertical or horizontal lines? $\endgroup$ Jun 20, 2014 at 14:45
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    $\begingroup$ They don't even need to be lines. The problem was possibly set up to have a nice Jacobian to facilitate computations. If you were considering the region between hyperboles and ellipses the computations might have been utterly complicated. =) $\endgroup$ Jun 20, 2014 at 14:50
  • $\begingroup$ But you've miscalculated the integral :) Your integral yields $21\log\left(\frac32\right)$ $\endgroup$ Jun 20, 2014 at 15:14
  • $\begingroup$ You're right. Sorry! I've edited. $\endgroup$ Jun 20, 2014 at 18:22

Switch to polar coordinates. Then you have the following integral: $$\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}} \int_{\sqrt{2\csc(2\theta)}}^{2\sqrt{2\csc(2\theta)}}r^5\sin^2\theta\cos^2\theta\,dr\,d\theta=\frac{1}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\sin^2\theta\cos^2\theta (504\csc^3(2\theta))\,d\theta $$ $$=\frac{504}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\frac{\sin^2\theta\cos^2\theta}{8\sin^3\theta \cos^3\theta}\,d\theta=\frac{504}{24}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\csc(2\theta)\,d\theta$$ $$=\boxed{21\ln\left(\frac{3}{2}\right)}$$

  • $\begingroup$ Fantini's sub seems easier to be calculated afterwards. I am doing the calculations, so will check now, if one of the answers is right. $\endgroup$ Jun 20, 2014 at 15:06
  • $\begingroup$ @DmitryKazakov: I don't know much about what Fantini did because of my zero knowledge about Jacobians. Can you please confirm the final answer from the answer key if possible? Thanks! $\endgroup$ Jun 20, 2014 at 15:07
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    $\begingroup$ My calculations yielded $21\log\left(\frac32\right)$ $\endgroup$ Jun 20, 2014 at 15:12
  • $\begingroup$ @DmitryKazakov: And that is what I get. :D Thanks! :) $\endgroup$ Jun 20, 2014 at 15:12
  • 1
    $\begingroup$ Actually Fatini has just miscalculated his integral, because his integral yields the same numeric result as yours. $\endgroup$ Jun 20, 2014 at 15:15

EDIT: The first part is the calculation of the area B, the second part is the integration of function $f(x,y)$ in the area B.

The intersection of $y_1=\frac x4$ with $y_2=\frac4x$ is at $x=4,y=1$.

The intersection of $y_3=\frac x9$ with $y_4=\frac 1x$ is at $x=3,y=1/3$.

So the area is given by:

$$B=\int_{2}^{4}y_1(x) dx+\int_{4}^{6} y_2(x) dx-\int_{2}^{3}y_4(x) dx-\int_{3}^{6} y_3(x) dx$$ $$=\int_{2}^{4}\frac x4 dx+\int_{4}^{6} \frac4x dx-\int_{2}^{3}\frac 1x dx-\int_{3}^{6} \frac x9 dx=\ln\left(\frac{27}{8}\right)$$

EDIT: Now I add the part of the integration of function $f(x,y)$ in the area B.

$$\int\int_B f(x,y)dxdy=\int_{2}^3x^2 \left(\int_{y_4(x)}^{y_1(x)} y^2 dy\right)dx+\int_{3}^4x^2 \left(\int_{y_3(x)}^{y_1(x)} y^2 dy\right)dx+\int_{4}^6 x^2 \left(\int_{y_3(x)}^{y_2(x)} y^2 dy\right)dx=21\ln(3/2)$$

  • $\begingroup$ I am afraid you are missing something, have a look at the answer above yours. $\endgroup$ Jun 20, 2014 at 14:58
  • $\begingroup$ And I know. What, the problem asks not to compute the area, that is restricted by 4 functions, but to integrate a function $f(x,y)=x^2y^2$ over this area. $\endgroup$ Jun 20, 2014 at 15:02
  • $\begingroup$ I added the integration of f(x,y) in B and the answer is also $21\ln(3/2)$. $\endgroup$
    – mike
    Jun 20, 2014 at 15:31
  • $\begingroup$ Ok I see, that's correct. Thanks for your time, man! $\endgroup$ Jun 20, 2014 at 15:36

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