Strange double integral What is wrong with this computation of $\int_0^1\int_{-y}^y \sqrt[3]{x} \, dx \, dy$?
I'm considering real functions only.
Since $x^{4/3}$ is an antiderivative of the integrand, we will get $\frac{3}{4}[x^{4/3}]_{-y}^y =\frac{3}{4}(y^{4/3}-(-y)^{4/3})=\frac{3}{4}(y^{4/3}-y^{4/3})=0$. Thus $\int_0^1 \int_{-y}^y \sqrt[3]{x} \, dx \, dy=0$.
However, maple is giving  me a complex (nonzero) number as the answer. Why is that? Any hint?

As others have already stated, Maple is interpreting $\sqrt[3]{x}$ as a complex function. Here's how it's being calculated
$$ \int_0^1\int_{-y}^y\sqrt[3]{x}\ dxdy= \int_0^1\int_{-y}^y x^{\frac{1}{3}}\ dxdy $$
$$ = \int_0^1\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{-y}^y dy = \int_0^1\left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]_{-y}^y dy $$
$$
=\frac{3}{4}\int_0^1\left[x^{\frac{4}{3}}\right]_{-y}^y dy=\frac{3}{4}\int_0^1 y^{\frac{4}{3}}-(-y)^{\frac{4}{3}} dy
$$
$$
= \frac{3}{4}\left(\int_0^1 y^{\frac{4}{3}}dy-\int_0^1 (-y)^{\frac{4}{3}} dy\right)
$$
Let $u=-y$, then $-du=dy$. So now we have
$$
 \frac{3}{4}\left(\int_0^1 y^{\frac{4}{3}}dy+\int_0^{-1} u^{\frac{4}{3}} du\right)
$$
$$
= \frac{3}{4}\left(\left[\frac{y^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right]_0^1 +\left[\frac{u^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right] _0^{-1} \right)
$$
$$
= \frac{3}{4}\left(\left[\frac{y^{\frac{7}{3}}}{\frac{7}{3}}\right]_0^1 +\left[\frac{u^{\frac{7}{3}}}{\frac{7}{3}}\right] _0^{-1} \right)
$$
$$
= \frac{3^2}{4\cdot 7}\left(\left[y^{\frac{7}{3}}\right]_0^1 +\left[u^{\frac{7}{3}}\right] _0^{-1} \right)
$$
$$
= \frac{9}{28}\left(1-0 +(-1)^{\frac{7}{3}}-0\right)
$$
$$
= \frac{9}{28}\left(1+(-1)^{\frac{7}{3}}\right) = \frac{9}{28}\left(1+(-1)^{\frac{6}{3}+\frac{1}{3}}\right)
$$
$$
=\frac{9}{28}\left(1+(-1)^{\frac{6}{3}}(-1)^{\frac{1}{3}}\right)= \frac{9}{28}\left(1+(-1)^{2}\sqrt[3]{-1}\right)
$$
$$
=\frac{9}{28}\left(1+\sqrt[3]{-1}\right)
$$
