I have to compute $ \int \int_D dxdy$ over the region bounded by $x^2=ay$, $x^2=by$, $y^2=cx$, $y^2=dx$, where $0<a<b$ and $0<c<d$.

When I look at the problem from one-dimensional view and try to compute the integral as the area between 2 functions, clearly I get answer as sum/difference of $a$,$b$,$c$,$d$ in some rational powers. But I want to use another approach by converting the problem to another system of coordinates, to be more precise on $u=x^2$, $v=y^2$. The aim is to simplify the limits of integration, and I assume that I have the next equivalent definite integral: $\int_a^b \int_c^d \dfrac{du dv}{4uv}$. The question is that I get the answer in the terms of logarithm functions.

I suppose that the mistake lies in the wrong change of variables, but I'm not sure how to fix it.

  • 1
    $\begingroup$ How about variables $u=x^2/y$ ($u:a \to b$) and $v=y^2/x$? ($v:c \to d$)? $\endgroup$ – StackTD Mar 14 '16 at 15:12
  • $\begingroup$ I elaborated in an answer; to me this seems more the way to go for this exercise. $\endgroup$ – StackTD Mar 14 '16 at 16:27

With those substitutions, you would have the bounding functions $u^2=a^2v$, $u^2=b^2v$, $v^2=c^2u$, and $v^2=d^2u$. You are just changing the constants, not simplifying any integrals.

Instead, I would handle this by breaking down the domain of integration into $3$ regions

enter image description here

where the red region is between $y^2=cx$ and $x^2=ay$ and $a^2c\le x^3\le a^2d$, the green region is between $y^2=cx$ and $y^2=dx$ and $a^2d\le x^3\le b^2c$, and the blue region is between $x^2=by$ and $y^2=dx$ and $b^2c\le x^3\le b^2d$.


Although there is already an answer, I would suggest you do try this with new variables. You just need to pick them more carefully and I'll try to show you how.

For two of the given curves, $x^2=ay$ and $x^2=by$, the expression $x^2/y$ is a constant. Let $u=x^2/y$ and $u$ will run from $a$ to $b$.

Similarly, for the curves corresponding to $y^2=cx$ and $y^2=dx$, the expression $y^2/x$ is a constant. Let $v=y^2/x$ and $v$ will run from $c$ to $d$.

With these new variables $u$ and $v$, you will have constant limits for both variables! And even the Jacobian turns out to be very simple ($1/3$, you can check); so the integral becomes:

$$\iint_D \; dxdy = \int_a^b \int_c^d \frac{1}{3} \; dvdu = \frac{(b-a)(d-c)}{3}$$ Once you get the idea, you put a bit of time in choosing handy new variables and computing the new limits and Jacobian, but the calculations can become a lot easier.

  • $\begingroup$ Thank you. I got it from your comment. But the answer is surprisingly simple. Why is the Jacobian just a number? It must be because we have the functions and theirs inverses(multipled by some constants) as boundary functions. $\endgroup$ – Mihail Mar 14 '16 at 16:31
  • $\begingroup$ Well the Jacobian is what it is... :-). The fact that the equations in this problem and our choice of variables are so symmetric, 'causes' this. If you have computed the integral the 'straightforward way' (without new variables, as in the other answer): you should find the same answer - at least after simplifying. Although, probably, also after more lengthy calculations. $\endgroup$ – StackTD Mar 14 '16 at 16:35
  • $\begingroup$ If you compare both ways of calculating the integral, I guess this problem is a good example to show how much it can be simplified with the right choice of variables! $\endgroup$ – StackTD Mar 14 '16 at 16:38
  • $\begingroup$ Although your answer is more helpful, I accepted answer of @robjohn, because I was looking for the error in my approach and he indicated to my mistake in his second sentence. Thank you both for help. $\endgroup$ – Mihail Mar 14 '16 at 16:44
  • $\begingroup$ My answer is there to help, glad it did. You're welcome! $\endgroup$ – StackTD Mar 14 '16 at 16:45

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