Limits on a triple integral

Solution 1:

but the (0,0,1) point makes me wanna use z=[-1,1] as z's limits.

Why? What would point (0, 0.5, 1) make you wanna use?

I would use this expression: $$\int_{-1}^1\left ( \int_0^2 (x-y) \left(\int_{-x^2+1}^{x^2-1} dz\right) dy \right) dx $$

Solution 2:

You can integrate in any order and the result should be the same. But for the given region, it is easiest to integrate wrt $dz$ before $dx$ otherwise you have one integral for the region above $z = 0$ and another for the region below $z = 0$. The order of $dy$ does not matter as both $z$ and $x$ are independent of $y$.

Also note that the region is symmetric to yz-plane and as $f(x) = x$ is an odd function of $x$, its integral over $x \lt 0$ and over $x \gt 0$ would be same in absolute value but with opposite sign and hence the integral will be zero for the entire region.

Now to find the bounds of the integral, simply see the projection of the region in xy-plane which is a square of side $2$ with sides parallel to coordinate axes and center at $(0, 1)$. For all points in the projection in xy-plane, $z$ is bound below by the parabolic cylinder $z = x^2 - 1$ and above by the parabolic cylinder $z = 1- x^2$

So the integral translates to,

$ \displaystyle \int_{-1}^1 \int_0^2 \int_{x^2-1}^{1-x^2} -y ~dz ~dy ~dx$