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**Failure** Ok, the limits +/-2 for the outermost integral with respect to x are right, now just imagine that an x from that range has been chosen. The y over which the next inner integral varies must still satisfy $\displaystyle \sqrt{x^2+y^2}\leq z\leq 2$.

Since z can still be chosen to be from a suitably small intervall, the relevant condition for the limits for y is $\displaystyle \sqrt{x^2+y^2}\leq 2$, from which it follows that $\displaystyle y^2\leq 4-x^2$, in other words $\displaystyle -\sqrt{4-x^2}\leq y\leq +\sqrt{4-x^2}$, agreed?