Reverse the order of integration of $ \int_{-1}^1 \int_{x^2}^1 \sqrt{y}\ dydx$.

Question

I am computing the following integral $$ \int_{-1}^1 \int_{x^2}^1 \sqrt{y}\ dydx$$ and I want to verify the exchanging of order of integration. The region is described as $-1\leq x \leq 1, x^2\leq y \leq 1,$ so we can express it as $0\leq y \leq 1, -\sqrt{y}\leq x \leq \sqrt{y}.$ The new integral is $$ \int_{0}^1 \int_{-\sqrt{y}}^{\sqrt{y}} \sqrt{y} \ dx dy.$$ If we compute this last integral, we get $ \int_{0}^1 2y dy = 1,$ but with the first integral we have $\int_{-1}^1 \frac{2}{3} (1-x^3)dx = \frac{4}{3},$ so they integrals do not coincide. I am pretty sure that the change of order of integration is allowed, and the integrals are correctly calculated, so maybe I am doing a mistake in the exchange of order... Anyone can help me?

Thank you very much!

Answer 1

The change of variables is allright, but the first integral is $$\int_{-1}^1 \frac{2}{3}(1-|x|^3)\,dx.$$ I think you made a mistake when evaluating the inner integral. $(x^2)^{3/2}=|x|^3$, not $x^3$.

Answer 2

you made a mistake in using absolute value. $2/3\int_{-1}^1(1-x^2|x|)dx=2/3(\int_{-1}^0(1+x^3)dx+\int_{0}^1(1-x^3)dx)=1$