Solving Integral for Mass of D in Spherical Coordinates

[Benny]

Hi, I'm having trouble setting up an integral for the following problem.

Q. Let D be the region inside the sphere x^2 + y^2 + z^2 = 4 in common with the region below the cone $$z = \frac{1}{{\sqrt 3 }}\sqrt {x^2 + y^2 }$$.

Using spherical coordinates find the mass of D if the mass density is z^2.

I keep on getting an answer which doesn't correspond to the given answer. I just need help setting up the integral. I get:

$$z = \frac{1}{{\sqrt 3 }}\sqrt {x^2 + y^2 } \Rightarrow \cos \phi = \frac{1}{{\sqrt 3 }}\sin \phi \Rightarrow \phi = \frac{\pi }{3}$$

$$x^2 + y^2 + z^2 = 4 \Rightarrow \rho = 2$$

$$m = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_E^{} {z^2 dV} } }$$

$$= \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_E^{} {\rho ^2 \cos ^2 \left( \phi \right)dV} } }$$

$$= \int\limits_{\frac{\pi }{3}}^\pi {\int\limits_0^{2\pi } {\int\limits_0^2 {\rho ^2 \cos ^2 \left( \phi \right)} } } \rho ^2 \sin \left( \phi \right)d\rho d\theta d\phi$$

$$= \int\limits_{\frac{\pi }{3}}^\pi {\int\limits_0^{2\pi } {\int\limits_0^2 {\rho ^4 \cos ^2 \left( \phi \right)} } } \sin \left( \phi \right)d\rho d\theta d\phi$$

The evaluation of the integral is fairly straight forward. It's just setting up the correct integral which is giving me problems. Can someone go through my working and show me where I stuffed up?

 

[Tide]

How do you know your setup is wrong?

It looks okay to me except that I would label things differently. I would use $\theta$ for the polar angle, $\phi$ for the azimuthal angle and $r$ for the radial coordinate.

 

[Cyrus]

He used rho instead of r, because in some books r is reserved for the radial direction in cylinderical coordinates, and rho is for sphereical. Math and physics text typically switch the phi and theta labels.

(What happened to consistency?) ;-p

 

[Tide]

Consistency? I think that's a bit much to ask! ;)

 

[Benny]

I might have just evaluated the integral incorrectly or the answer which is given might be incorrect (I doubt it though). Thanks for the help.