The Question :
Calculate
$$\int_A z \ dx \ dy \ dz$$
Where $A$ is the cone : $\ A=\{0\leq z \leq 1,\ z^2\geq x^2+y^2 \}$
My try :
I first try to establish the bounds.
$z$ is from $0$ to $1$.
Once $z$ is fixed, I changed for polar coordinates and so: $z\geq r$.
We also know that a radius is positive so I deduced $r$ is from $0$ to $z$.
I finally set $\theta$ from $0$ to $2\pi$ because my cone is a full one I guess. It gives me :
$$\int_0^1\left(\int_0^{2\pi} \left( \int_0^z zr\ dr \right) d\theta \right)dz$$
$$=\int_0^1\left(\int_0^{2\pi} \frac{z^3}{2} d\theta \right)dz$$
$$=2\pi \int_0^1 \frac{z^3}{2} dz$$
$$=\frac{2\pi}{8}$$
I have the feeling something is wrong, mostly when I find the bounds (the deductions about the bounds of $\theta$ and $r$). The difficulty for me in this case is that it's very difficult to have a visual interpretation.
My question: Is my bounds searching process good or am I in the wrong direction ?
Yes, your bounds are correct. Another way: since $0\leq r\leq z\leq 1$, it follows that $$\iiint_A z dx dy dz=\int_{r=0}^1\left(\int_{\theta=0}^{2\pi} \left( \int_{z=r}^1 z\ dz \right) d\theta \right)rdr=\pi\int_{0}^1(1-r^2)r dr=\frac{\pi}{4}.$$