I'm preparing for an exam, and got stuck at this question when trying to do an exam from a few years ago.

Let $D:= \{(x,y) \in \mathbb{R}^2 \, \big|\, x \geq 0, y \geq 0, x^3 + y^3 \leq 1 \}$. Let $f :D \to \mathbb{R}$ be continuous so that $f(x,y)=-f(y,x)$ for all $(x,y) \in D$. Show that $$\iint_D f(x,y) dA = 0$$ using change of variables.

If we would integrate this over $D$, we could use for the boundaries that $0 \leq x \leq 1$, then $0 \leq y \leq \sqrt[3]{1-x^3}$.

I asked a fellow student for a hint regarding what transformation to use, and they suggested I use $\varphi(x,y) = (y,x)$. We know the following formula for change of variables, where $D\varphi$ is the Jacobian. $$\iint_{\varphi(E)} f = \iint_E (f\circ \varphi)|\det D\varphi|.$$ For $D\varphi$ we find $\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$, and $|\det D\varphi| = |-1| = 1$. Now, using the formula, we get $$\iint_{D=\varphi(E)} f(x,y) dA = \iint_E f(y,x)dA.$$

But I don't understand how this would make it easier, nor do I understand how to apply the new boundaries ($0 \leq y \leq 1$ and $\sqrt[3]{1-y^3}$?).

Any help would be appreciated, thanks in advance.

  • 2
    $\begingroup$ Is it unclear that $\phi(E)=E$? $\endgroup$ Commented Aug 4, 2016 at 11:34
  • $\begingroup$ @DanielLittlewood No, that's what I thought. Because this transformation flips $f$ over the line $x=y$, right? $\endgroup$
    – Abby
    Commented Aug 4, 2016 at 11:36
  • 4
    $\begingroup$ Exactly - so $\iint_{D}f(x,y)dA=\iint_{D}f(y,x)dA$, right? $\endgroup$ Commented Aug 4, 2016 at 11:41
  • $\begingroup$ @DanielLittlewood Ahh! I think I see it. Then, because $f(x,y)=-f(y,x)$, you can say $\iint_D f(x,y)dA=\iint_D -f(y,x)dA = \iint_D f(y,x)$. Then it follows that $\iint_D f(x,y)dA=0$. Agree? $\endgroup$
    – Abby
    Commented Aug 4, 2016 at 11:44
  • 2
    $\begingroup$ Precisely, yes :) $\endgroup$ Commented Aug 4, 2016 at 11:53

1 Answer 1


For the sake of closing this question, let me give a brief explanation of what we discussed in the comments. Defining $\varphi(x,y)=(y,x)$, $\varphi$ is a linear bijection, and so in particular it is differentiable, and the Jacobian matrix is given by the matrix of $\varphi$. This matrix has determinant $1$, and since the region $D$ is symmetric in the line $y=x$, we have $\phi(D)=D$. As you point out, $$\iint_{D}f(x,y)dx dy=\iint_{\varphi(D)}f(y,x)|\det \varphi|dydx=-\iint_{D}f(x,y)dx dy$$ and hence $\iint_{D}f(x,y)dx dy=0$.


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