# Ok another few problems

#### cronxeh

Gold Member
I really shouldnt be stuck on those problems but for whatever reason i cant solve them

Problem: A particular spherical cloud of gas of radius 3 km is more dense at
the center than towards the edge. The density, D, of the gas at a distance p km
from the center is given by $$D(p) = 9 - p^2$$. Write an integral representing the total mass of the cloud of gas, and evaluate it.

Solution: density = mass/area. The spherical cloud's area is 9pi/2
mass =9/2 pi (9-p^2).

So mass = $$\frac{9pi}{2} \int_{0}^{pi} \int_{0}^{3} (9-r^2) \ r \ dr \ dtheta$$

Is this correct?

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#### Lyuokdea

If you are looking for a spherical cloud of gas you should have an integral in 3 dimensions instead of an integral in two dimensions. It should be a triple integral. Also where did the term 9pi / 2 come from?

What you want to do is find a triple integral that will give you the total area of the sphere and count the density function over that volume.

~Lyuokdea

Last edited:

#### cronxeh

Gold Member
Volume of spherical cloud with r=3 ->> V=(4/3)*pi^3= 4pi/3

$$mass \ = \ \frac{4pi}{3} \ \int_{0}^{pi} \int_{0}^{2pi} \int_{0}^{3} (9-p^2) p^2 \ sin(phi) \ dp \ dtheta \ dphi$$

#### Lyuokdea

close, but you don't need the factor of 4*pi/3 there, the 4pi/3 factor will drop out of the integrals as such:

$$volume\ = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1}p^2 \ sin(\phi) \ dp \ d\theta \ d\phi$$

$$volume\ = 2\pi \int_{0}^{\pi} \int_{0}^{1}p^2 \ sin(\phi) \ dp d\phi$$

$$volume\ = 2\pi \frac{1}{3}p\right)^{3} \ \right]_{0}^{1} \int_{0}^{\pi} \ sin(\phi) \ dp d\phi$$

$$volume\ = \frac{2\pi}{3} \int_{0}^{\pi} \ sin(\phi) \ dp d\phi$$

$$volume\ = \frac{4\pi}{3}$$

Hope this helps

~Lyuokdea

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