Finding the volume of a solid region I'm trying to find the volume of the solid region inside the sphere $x^2+y^2+z^2=4$, and the upper nappe of the cone $z^2=3x^2+3y^2$ (I only have to set up the triple integral itself, not evaluate it).
So, I need to do this in cylindrical coordinates. Thus, for the $r$ value, since $z^2+y^2=4$, $r=2$. Thus, $0 \leq r \leq 2$.
Next, I need to determine the value of $z$. I've considered the fact that $ y=\sin(\theta)$, and $x=\cos(\theta)$. Thus, $z^2 = 3(\cos^2(\theta)) + 3(\sin^2(\theta))$. This is equivalent to $z^2=3$. Thus, $z = \sqrt3$. This would be the lower limit, because it came from the equation of the cone. For the upper boundary of $z$, I'm not completely sure. Any help with this would be greatly appreciated.
Finally, for $\theta$, I think it would be $0$ to $2\pi$, since it is the solid region inside a sphere.
I'm not extremely confident in my abilities yet, and I don't want to lead myself astray. Any help/corrections would be greatly appreciated!
Thank you.
 A: In the conversion between rectangular and cylindrical coordinates, $x=r\cos\theta$ and $y=r\sin\theta$ (in your attempt, you dropped the $r$).
Then the equation of the cone becomes $$z^2=3r^2\cos^2\theta+3r^2\sin^2\theta =3r^2,$$ and since you're only interested on the upper half, you can use $z=\sqrt 3r$.
It is important not to drop the $r$ here because in cylindrical coordinates, $r$ is the radius on the $xy$-plane, not necessarily the distance of the point from the origin, so in a cone it is not fixed; it varies with the height $z$. (The equation $z=\sqrt 3$ actually gives you a plane that is parallel to the $xy$-plane. An equation where $r$ is fixed, like $r=2$, corresponds to a right circular cylinder.)
Now, the equation for your sphere will be $$4=z^2+x^2+z^2= z^2+r^2\cos^2\theta+r^2\sin^2\theta=z^2+r^2$$
Here again you're only interested in the upper half, so solving for $z$ you get $$z=\sqrt{4-r^2}$$
The bounds for $z$ will be these equations (the lower one is the cone, the upper one is the sphere). The fact that they depend on $r$ tells you that you must integrate with respect to $z$ before integrating with respect to $r$.
The bounds you found for $\theta$ are correct.
To figure out the bounds for $r$, you need to find the $r$ at which the cone and the sphere intersect:
$$\sqrt 3 r = \sqrt{4-r^2} \implies 3r^2=4-r^2 \implies 4r^2=4 \implies r=\pm 1$$
We can take the positive value and get the limits for $r$: $\; 0\leq r \leq 1$.
