# Change Variables in a Multiple Integral

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.

## 1 Answer

HINT

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?

UPDATE

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?

• 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. – gbgult Mar 5 at 19:43
• @gbgult not quite, but see the update – gt6989b Mar 5 at 21:32
• 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! – gbgult Mar 5 at 22:27