# Problem confirmation

1. Dec 16, 2004

### T@P

i had to do this some time ago:

assume you have a large cloud or radius 2 km, and its density is defined as D(ro) = 3 - ro (btw i cant find the letter 'ro' anywhere... )

what i the mass of the cloud?

i did it this way: mass = density * volume, so in this case it equals D(ro) * dV, so integrating (triple integral) over V yeilds (triple integral) (3 - ro) * (ro)^2 sin (phi) d ro d theta d phi. is that right?

i hope you can understand my *terrible* notation

2. Dec 16, 2004

### arildno

Is this what you meant?
$$\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{2}(3-\rho)\rho^{2}\sin\phi{d}\rho{d}\theta{d}\phi$$
If that's what you meant, I agree with you

Last edited: Dec 16, 2004
3. Dec 16, 2004

### dextercioby

For me it doesn't seem a problem in three dimentions,but rather in 2.I mean the cloud cound have a shape of a circle,and in this case,there should be integrations only after 2 coordinates:$$\rho$$ and $$\phi$$.

It looks that way to me,since u're given the radius (of the circle).I've never heard of cilindric clouds,neither of circular ones.But since you aren't given the height,then it should be a circle.
Try to make calcuations for this case and cf.to the result.

Daniel.

EDIT:On the other hand,it might be a sphere.Though it's weird.Anyway,Arildno may be right and disregard what i've written above.

Last edited: Dec 16, 2004
4. Dec 16, 2004

### arildno

On second thought, I guess they assumed a "disk-like" cloud.
So I would go with dexter's first suggestion.

5. Dec 16, 2004

### T@P

actually the problem was supposed to come out as a triple integral, and yes that was what i meant arildno thanks. ( i think they might have specified a spherical cloud too)

6. Dec 16, 2004

### HallsofIvy

Staff Emeritus
You said "assume you have a large cloud or radius 2 km"

I would have assume a spherical cloud. In that case, the simplest thing to do is set up a coordinate system with (0,0,0) at the center of the cloud. The mass then is exactly what arildno said.