Find the volume of a cone using spherical coordinates

[clocksmith]

Find the volume of the portion of cone z^2 = x^2 + y^2 bounded by the planes z = 1 and z = 2 using spherical coordinates

I am having trouble coming up with the limits





Relevant equations
dV = r^2*sin(theta)*dr*d(theta)*d(phi)
r = sqrt(x^2+y^2+z^2)

the problem is actually 2 parts, the 2nd part asks to evaluate by cylindrical coordinates and I obtain 7pi/3 which i know is right, I just can't come up with the limits

Attempt
for r i have from sqrt2 to 2sqrt2
for theta i have from 0 to pi/4
and for phi i have from 0 to 2pi

 

[HallsofIvy]

In spherical coordinates, $x= \rho cos(\theta)sin(\phi)$, $y= \rho sin(\theta)sin(\phi)$, and $z= \rho cos(\phi)$
Putting those into $z^2= x^2+ y^2$ becomes $\rho^2 cos^2(\phi)= \rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 cos^2(\theta)sin^2(\phi)$$\rho^2 sin^2(\phi)$ which reduces to $cos^2(\phi)= sin^2(\phi)$. For $\phi$ between 0 and $\pi/2$, that is true only for $\phi= \pi/4$.

Now, for $z= \rho cos(\phi)= 1$ then $\rho= 1/cos(\phi)$ for $z= \rho cos(\phi)= 2$ then $\rho= 2/cos(\phi)$ the limits of integration are: $rho$ from $1/cos(\phi)$ to $2/cos(\phi)$, $\phi$ from 0 to $\pi/4$, and $\theta$ from 0 to $2\pi$.

It looks to me like cylindrical coordinates should be much easier! Again, $\theta$ runs from 0 to $2\pi$, z obviously runs from 1 to 2, and, since $z^2= x^2+ y^2= r^2$, z= r (z is positive) so r runs from 0 to z.