Let $D$ be the region that's bound by $y=x^2, y=2x^2, x=y^2, x=3y^2$. $D$ corresponds to the region $E$ where $u=\frac{x^2}{y}$ and $v=\frac{y^2}{x}$.

  1. Sketch $D$ and $E$
  2. Find the solution to the integral $$\iint_D xy\ dxdy$$

Okay, I'm pretty lost here. Obviously I know how to sketch D but I'm pretty clueless on how to sketch E.

I found the determinant of the Jacobian I got from $\frac{\partial (u,v)}{\partial (x,y}$ to be 3 but I don't really know how to use it here (I just know that you're suppose to put in the integral when changing variables). Some help would be appreciated since I can't figure out myself how to tackle this problem by my own.



Translate the curves. If $u = x^2/y$ and $v = y^2/x$, then $y=x^2$ translates to $u=1$, and $x=y^2$ to $v=1$. Can you translate the others?


You are almost correct in your comments, so the boundaries translate to a box $1 \le u \le 2$ and $1/3 \le v \le 1$, and the integrated fund is $xy=uv$, so the integral becomes $$ \iint_D xy dx dy = \int_{u=1}^{u=2} \int_{v=1/3}^{v=1} uv J(u,v) dudv, $$ where $J(u,v)$ denotes the Jacobian of the transformation. Can you find it and complete the problem?

  • $\begingroup$ Ah alright, so $y=2x^2$ translates to $u=\frac{1}{2}$ and $x=3y^2$ translates to $v=\frac{1}{3}.$ I guess that means that we have the integral $\int_{\frac{1}{2}}^1 $ $\int_{\frac{1}{3}}^1 \partial u \partial v$? Where to take it from there, I don't really know though. $\endgroup$ – gbgult Mar 5 at 19:43
  • $\begingroup$ @gbgult not quite, but see the update $\endgroup$ – gt6989b Mar 5 at 21:32
  • $\begingroup$ Oh okay I see, I get $J(u,v)=3$ but when comparing this to the answer I understand that it's supposed to be $\frac{1}{3}$ so I'm assuming I'm doing it the opposite way around. How do I know that? Thanks for your help! $\endgroup$ – gbgult Mar 5 at 22:27

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