I have the following double integral before me:
$$I=\int_{-1}^{1}\int_{0}^{x^2} \sqrt{x^2-y}\,dy\,dx$$
I got the answer of this integral as $0$ working in the following manner:
$$I=\int_{-1}^{1} \dfrac{-2}{3}(x^2-y)^\frac{3}{2}dx =\int_{-1}^{1}\dfrac{2}{3}x^3 dx$$
As $x^3$ is an odd function of $x$, $I$ must be zero.
But when I used Wolfram alpha to find the value of the integral, I instead got $\frac{1}{3}$ as the answer. I am not able to make out which one is correct and why. Please suggest.
We have, $$\begin{align} &\int_{-1}^1\int_0^{x^2}\sqrt{x^2- y}\ dy dx\\=&\int_{-1}^1\dfrac{-2(x^2 -y)^{3/2}}{3}\bigg|^{x^2}_0 dx\\=&\int_{-1}^1\dfrac{-2(x^2 -x^2)^{3/2}}{3} + \dfrac{2(x^2 - 0)^{3/2}}{3}dx\\=&\int_{-1}^10 + \dfrac{2(x^2)^{3/2}}{3}dx\\=&2\int_{0}^1 \dfrac{2(x^2)^{3/2}}{3}dx\qquad\rm{As \ it \ is \ an \ even\ function.}\\=&\dfrac43\int_{0}^1x^3dx\\=&\dfrac43\left(\dfrac{1}{4}\right)\\= &\dfrac13\end{align}$$