Evaluating a volume integral $\iiint_ T (x+y-z)\,dV$

Question

Like the limits of my integration are which? $$\iiint_T \left(x+y-z\right)\,dV \quad\quad T=\left\{(x,y,z)\in \mathbb{R}^3 \,|\, x^2+2y^2\leq z\leq3-2x^2-y^2 \right\}$$

Answer 1

Here is the region.

We transform this into cylindrical coordinates. The $z$ bound would just be the top and bottom functions, which are $x^2+2y^2$ and $3-2x^2-y^2$.

Removing the zed axis, the "shadow" of the region of integration seems to be a circle. Hence, the $r$ bound would be from $0$ to $1$, and theta going from $0$ to $2\pi$.

Thus, the integral is $$\iiint_T x+y-z \text{ d}V = \int_0^{2\pi} \int_0^1 \int_{r^2(\sin^2(\theta)+1)}^{3-r^2(\cos^2(\theta)+1)} (r\cos(\theta)+r\sin(\theta) - z)\cdot r \text{ d}z\text{ d}r\text{ d}\theta$$ $$\implies \int_0^{2\pi} \int_0^1 -\frac32 r(r^2-1)(r^2\cos(2\theta)+2r\sin(\theta)+2r\cos(\theta)-3) \text{ d}r\text{ d}\theta$$ $$\implies \int_0^{2\pi} \frac1{40}(16\sin(\theta)+16\cos(\theta)+5\cos(2\theta)-45) \text{ d}\theta = -\frac{9\pi}4$$

This seems to be what @cpiegore got in the comments. Although I'm pretty confident in my answer, I'm not sure what Quanto did in the comments (and since he is pretty reputable, there's a chance my value is wrong), so verify my work here.