# Double integral over the region enclosed by 4 curves.

I'm a bit stuck on a problem that involves trying to integrate the area between 4 curves.

Problem:

$\int \int_D x^2+y^2 dxdy$

Where D is the region enclosed by the curves $xy =2, xy=7, y= 2x^2$ and $y=5x^2$.

For this, I set $u=y/x^2$, $v=xy$ and then work through the problem. In the end, I find that I can't fully rewrite the end integral in terms of $u$ and $v$, so am likely making a mistake somewhere.

Any help in solving this would be much appreciated.

Many thanks.

With $u=y/x^2$, $v=xy$, we first need to write $x^2+y^2$. Solving for $y$ on both equalities we obtain $x^2u=v/x$, so $x^3=v/u$. Then $x^2=(v/u)^{2/3}$. Squaring the second equality, $v^2=x^2y^2$, so $y/u=v^2/y^2$, or $y^3=v^2/u$. Then $$x^2+y^2=\left(\frac{v}{u}\right)^{2/3}+\left(\frac{v^2}{u}\right)^{2/3}.$$ The Jacobian is $$\begin{vmatrix} x_u&y_u\\ x_v&y_v \end{vmatrix} =\begin{vmatrix} -\dfrac13\,\dfrac{v^{1/3}}{u^{4/3}}&-\dfrac13\,\dfrac{v^{2/3}}{u^{4/3}}\\ \dfrac13\,\dfrac{v^{-2/3}}{u^{1/3}}&\dfrac23\,\dfrac{v^{-1/3}}{u^{1/3}} \end{vmatrix} =-\dfrac19\left(\frac{2}{u^{5/3}} -\frac{v^{4/3}}{u^{5/3}}\right)$$ since $x=(v/u)^{1/3}$, $y=(v^2/u)^{1/3}$. Now you can form the integral $$-\frac19\int_2^7\int_2^5\left(\left(\frac{v}{u}\right)^{2/3}+\left(\frac{v^2}{u}\right)^{2/3}\right)\,\left(\frac{2}{u^{5/3}} -\frac{v^{4/3}}{u^{5/3}}\right)\,dv\,du.$$ This is trivial after distributing.
• Thanks, I didn't expect the solution to be that complicated. Would you recommend changing what $u$ and $v$ are set to (to achieve a cleaner solution)? – user3199301 Nov 19 '15 at 14:46
• It is not surprising that the solution looks nasty. In the original integral the roles of $x$ and $y$ are symmetric, while $u$ and $v$ are not symmetric (because the two pairs of curves that limit your region are of a different kind). I'm not sure what you mean by "changing what $u$ and $v$ are". – Martin Argerami Nov 19 '15 at 15:01