# Swapping the limits of integration $\int_0^4 \int_\sqrt x^2 \sin(x^3) \,dy\,dx$

I want to solve this double integral.

$$\int_0^4 \int_\sqrt x^2 \sin(x^3) \,dy\,dx$$ I tried to swap the limits of integration so I got $$\int_0^2 \int_0^{y^2} \sin(x^3) \,dx\,dy$$

But when I looked on te answers (see the link at the bottom) I saw that they also change x to y and got $$\int_0^2 \int_0^{y^2} \sin(y^3) \,dx\,dy$$

My question is what is the justification for this?

• It will remain $\sin x^3$, in the last expression. – MAN-MADE Jul 10 '17 at 11:21