Can you explain the absolute value sign when integrating $\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$?

Question

I'm doing homework on the following integral: $$\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$$ And here is the answer:

I try to think about it, but I couldn't answer why there is an absolute value sign in the third expression at $x^3$. Can somebody explain it for me why? Thank you! :D

Answer 1

Let's compute $\int_0^{x^2} \sqrt{x^2 - y} \, dy$. Applying the substitution $u = x^2 - y$, we have $du = -dy$ and hence

$$ \int_0^{x^2} \sqrt{x^2 - y} \, dy = -\int_{x^2}^0 \sqrt{u} \, du = \int_0^{x^2} \sqrt{u} \, du = \left[ \frac{2}{3} u^{\frac{3}{2}} \right]^{u=x^2}_{u=0} = \frac{2}{3} \left( (x^2)^{\frac{1}{2}} \right)^3 = \frac{2}{3} |x|^3. $$

Answer 2

To elaborate on the existing answers, notice what happens when you forget that $\sqrt{u^2} = |u|$ and substitute $u$ instead: $$ I_2 = \dots = \int_{-1}^1 \int_0^{x^2} \sqrt{x^2-y}\,dy\,dx = \color{red}{\frac{2}{3}\int_{-1}^1 x^3\,dx} = 0 $$ At this point, you should notice there must be some mistake. The integrand is nonnegative so the integral should be positive.