Find volume within sphere outside of Cylinder

[mknut389]

Homework Statement

Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 16, above the xy plane, and outside the cone z = 3 \sqrt{{x^2+y^2}}.

Homework Equations

spherical system:
x=\rhocos\thetasin\phi
y=\rhosin\thetasin\phi
z=\rhocos\phi

cylindrical system
x^2+y^2=r^2
z=z

The Attempt at a Solution

I have tried this using both the spherical and cylindrical systems and arrived at the same answer, cylindrical is easiest here, so Ill use it to demonstrate what I have done

x^2+y^2+z^2=16
r^2+z^2=16
z=\sqrt{{16-r^2}}

z = 3 \sqrt{x^2+y^2}
z=3\sqrt{r^2}
z=3r

when z=0
0=\sqrt{16+r^2}
r=4
0=3r
r=0

Therefore the bounds are
z: \left[3r,\sqrt{16-r^2}\right]
r:\left[0,4\right]
\theta: \left[0,2\pi\right]

Which gives me
\int_{0}^{2\pi}\int_{0}^{4}\int_{3r}^{\sqrt{16-r^2}}r dzdrd\theta

After going through the steps i get
-(256\pi)/3

First off it is negative, so that tells me I am completely off base here, but when doing the same problem with spherical coordinates, I get (256\pi)/3. Since this answer is wrong, I am at a loss of what I should be doing. Please help.

 

[mknut389]

I solved it:

First the bounds had to be reworked for r in terms of Z
r: \left[z/3,\sqrt{16-z^2}\right]

Then by solving the two equations in terms of Z and setting them equal to each other, I determined the maximum value for r
\sqrt{16-r^2}=3r
r=(2*\sqrt{10})/5
after plugging that in and solving for z, z's maximum is 3.83863112788

Then using that as the bounds for Z and using the same bounds for \theta
The answer came out to be 127.137105725

Thanks to anyone that attempted to figure it out for me...