<s>[INST] Finding the volume of a solid under a region (Triple integrals) Let S be the solid enclosed above by $x^2+y^2+z^2=2$, below by $x^2+y^2=z^2$ and $y=0$ compute the integral
$$ \iiint_S \frac{z}{\sqrt{x^2 +y^2 +z^2} }\text{d}x\,\text{d}y\,\text{d}z $$
What i tried
     The iterated integral becomes $x<z<\sqrt{2-x^2-y^2}$ for the $z$ component. Then setting $z=0$, it becomes $x^2+y^2<2$, hence it becomes obvious to use polar coordinates from here. Converting it to polar coordinates it becomes, $$2cos\theta<z<\sqrt{2-r^2}$$, $$0<r<{\sqrt2}$$ and $$0<\theta<2\pi$$ Is my iteriated integral correct. Could anyone explain.Im unsure whether my iteriated integral for $z$ is correct. Thanks
 [/INST] You need to sketch the region. It is bounded below by the cone $x^2+y^2=z^2$, above by the sphere of radius $\sqrt{2}$, $x^2+y^2+z^2=2$. The $y=0$ confines the region to one-side of the xz-plane. 
If you take z-slices of this volume, you'll find that at lower z, the boundary of the 2D section of the volume has the equation $x^2+y^2=z^2$, $y \geq 0$. This is a semi-circle of radius $z$. Above $z=1$, the slices have a boundary given by the latitude semi-circles of the sphere.
[To determine this, you can check where the cone and sphere intersect; it is in the circle $x^2+y^2=1, z=1$].
Again, the z-slices in the upper region are also semi-circles, but they are bounded by $x^2+y^2=2-z^2$, $y \geq 0$. The upper limit of $z$ is obviously $\sqrt{2}$.
So your triple integral is a sum of two integrals as follows 
$$
\int_{z=0}^{z=1} \int_{x=-z}^{x=+z} \int_{y=0}^{y=\sqrt{z^2-x^2}} \frac{z}{\sqrt{x^2+y^2+z^2}} dy dx dz + \\ \int_{z=1}^{z=\sqrt{2}} \int_{x=-\sqrt{2-z^2}}^{x=+\sqrt{2-z^2}} \int_{y=0}^{y=\sqrt{2-z^2-x^2}} \frac{z}{\sqrt{x^2+y^2+z^2}} dy dx dz   
$$
You can now convert the inner double integrals into polar. Recall $dx dy = r dr d\theta$.
$$
\int_{z=0}^{z=1} \int_{r=0}^{r=z} \int_{\theta=0}^{\theta=\pi} \frac{r z}{\sqrt{r^2+z^2}} d\theta dr dz + \\ \int_{z=1}^{z=\sqrt{2}} \int_{r=0}^{r=+\sqrt{2-z^2}} \int_{\theta=0}^{\theta=\pi} \frac{r z}{\sqrt{r^2+z^2}} d\theta dr dz   
$$
The innermost $\theta$ integrals are trivial;
$$
\pi \int_{z=0}^{z=1} \int_{r=0}^{r=z} \frac{r z}{\sqrt{r^2+z^2}} dr dz + \pi \int_{z=1}^{z=\sqrt{2}} \int_{r=0}^{r=+\sqrt{2-z^2}} \frac{r z}{\sqrt{r^2+z^2}} dr dz   
$$
These are two easy integrals and you should be able to proceed from this point.
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