Let $D = \left \{ (x,y) : x^2 \leq y^2 \leq 2x^2, 1 \leq xy \leq 2\right \}$

It is given that for any $f \in C^0(\mathbb{R}^2, \mathbb{R}):$ $$ \int_Df(\frac{y}{x^2}, xy)|dxdy| = \int_{\Omega} f(u,v) J(u,v)|dudv|$$

For some $\Omega \subset \mathbb{R}^2$. Find $J(u,v)$.

So obviously $J$ is the Jacobian determinant from the change of variables theorem.

So what I tried is first expressing $x,y$ as functions of $u,v$ which gives us: $$\left\{\begin{matrix} u= \frac{y}{x^2}\\ v=xy \end{matrix}\right. \implies \left\{\begin{matrix} x=u^{-\frac{1}{3}}v^{\frac{1}{3}} \\ y=u^{\frac{1}{3}}v^{\frac{2}{3}} \end{matrix}\right.$$

And then by the version I read of the variable change theorem we should have: $$ J(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} -\frac{1}{3}u^{-\frac{4}{3}}v^{\frac{1}{3}} & \frac{1}{3}u^{-\frac{1}{3}}v^{-\frac{2}{3}}\\ \frac{1}{3}u^{-\frac{2}{3}}v^{\frac{2}{3}} & \frac{2}{3}u^{\frac{1}{3}}v^{-\frac{1}{3}} \end{vmatrix} = -\frac{1}{3u} $$

But apparently this answer is wrong, and the correct answer is $\frac{1}{3u}$. I noticed that if I'd change the order of the columns, i.e thinking of $v$ as the first variable, the answer would be correct. But I don't see why should I do that?

Can anyone point my mistake?


1 Answer 1


The change-of-variables formula has the absolute value $|J(u,v)|$ (or perhaps your text uses $J$ to denote the absolute value — I can't guess).

Here's a shortcut to compute the Jacobian. Do the Jacobian $\mathscr J$ of $u,v$ in terms of $x,y$: $$\mathscr J = \left|\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{matrix}\right| = \left|\begin{matrix} -\frac{2y}{x^3} & \frac1{x^2} \\ y & x \end{matrix}\right| = -3\frac y{x^2} = -3u.$$ Then $J = \dfrac1{\mathscr J} = -\dfrac1{3u}$. Regardless, you need the absolute value to proceed to the double integration.

  • $\begingroup$ Oh god... how could I miss that. Thank you, as usual, Ted. $\endgroup$ Commented Feb 11, 2021 at 22:36

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