Calculating Mass of Solid Using Spherical Coordinates

[hils0005]

Homework Statement

W is the solid bounded above by $$\rho$$=3 and below by$$\phi$$=$$\pi$$/3, calculate the mass W if the density at each point is directly proportional to the square of its distance above the xy plane.

The Attempt at a Solution

I am having a difficult time starting out this problem, I am assuming I need to use spherical coordinates considering the problem is given in terms of rho and phi, however it says bounded above by rho=3 and below by phi =pi/3, I thought phi determined the angle from the z axis to rho??

Any thoughts direction would be appreciated

 

[Dick]

They didn't describe that very well, but it looks to me like they want you to integrate rho from 0 to 3 and phi from 0 to pi/3 and over all theta. The region looks like a cone with a spherical cap on it. I think by above and below they mean relative to the z axis.

 

[hils0005]

I completely stuck
How do I determine what my density function will be?

$$\int$$$$\int$$$$\int$$ $$\delta$$(x,y,z)$$\rho$$^2sin$$\theta$$d$$\rho$$d$$\phi$$d$$\theta$$
0$$\leq$$$$\rho$$$$\leq$$3
0$$\leq$$$$\leq$$$$\pi$$/3
0$$\leq$$$$\theta$$$$\leq$$2pi

would I use $$\rho$$^2??

 

[Dick]

No. They tell you that the density is proportional to z^2. What's z in spherical coordinates?

 

[hils0005]

z=$$\rho$$cos$$\phi$$

z^2=($$\rho$$cos$$\phi$$)^2 ??

 

[Dick]

Why the question marks?? Sounds right to me. They only said 'proportional to', so you might want to call it C*z^2, just for generality.

 

[hils0005]

$$\int$$$$\int$$$$\int$$ $$\rho$$^2cos^2$$\theta$$$$\rho$$^2sin$$\theta$$d$$\rho$$d$$\phi$$d$$\theta$$
0$$\leq$$$$\rho$$$$\leq$$3
0$$\leq$$$$\leq$$$$\pi$$/3
0$$\leq$$$$\theta$$$$\leq$$2pi

 

[Dick]

Looks ok so far. But be careful about theta and phi. You said z=r*cos(phI) and then you put in cos(theta). theta and phi tend to get mixed up. There are two different conventions for this. Is theta the polar angle or phi?

 

[hils0005]

my mistake, should be p^2cos^2(phi) p^2sin(theta)
can this be broken apart:

(Integral rho^4 from 0 to 3)*(Integral Cos^2 phi from 0 to pi/3)*(Integral sin theta from 0 to 2pi)

(1/5rho^5) * (pi/6 + .5sin(pi/3)cos(pi/3)) * (-cos(2pi) - -cos(0))

(48.6)(.2165)(0)=0 ?

 

[Dick]

The angle in the z and the angle in the dV part of the integral should both be the same angle (the polar angle), shouldn't they? I'm just worried you are mixing formulas.