Finding Mass of Non-Uniform Density

[Mr LoganC]

This is just a practice problem, not actual homework. I'm studying for my final but am having a bit of difficulty in understanding this concept.

Homework Statement

Consider a solid of non-uniform density ρ=x²+y+z, consisting of all points inside the sphere x²+y²+z²=1
a) Find the mass of the solid (use spherical coordinates.)
b) Find the moment of inertia of the solid with respect to the z-axis (use spherical coordinates.)

Homework Equations

$$M=\int \rho dV$$
$$dV= r^{2}sin\theta dr d\theta d\phi$$

The Attempt at a Solution

I am unsure if since the density equation is given, should I bring it out infront of the integral as if it's a constant and just integrate the spherical part of dV. Or do I also integrate the density?
My textbook has no examples of this, only uniform densities where rho is considered a constant and brought out infront of the integral

 

[tiny-tim]

Hi Mr LoganC!

(have an integral: ∫ and a theta: θ and a phi: φ )

I am unsure if since the density equation is given, should I bring it out infront of the integral as if it's a constant and just integrate the spherical part of dV. Or do I also integrate the density?
My textbook has no examples of this, only uniform densities where rho is considered a constant and brought out infront of the integral

You have to keep it inside the ∫ …

you can only take actual constants outside (including functions of variables other than the one belonging to that ∫)

 

[Mr LoganC]

Okay, well then I'm confused again, because I would have this: (Using Spherical coordinates)
$$<br /> M=\int_{\phi=0}^{2\pi} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} (x^{2}+y+z)r^{2}sin\theta dr d\theta d\phi<br />$$

But if I'm only integrating with respect to radius, theta and phi, then the x, y, and z would be the same, acting like a constant as if I were to bring it out front of the integral. Again, my textbook is not helping at all as there is no example with the density inside the integral

 

[tiny-tim]

ah, you need to convert x² + y + z into r θ and φ, and then integrate

(btw, only one of the integrals is to 2π)

 

[Mr LoganC]

Rightttt! It's been a while since doing spherical!
So I need to convert those.
Also, does it matter which I change from 2pi to pi? Forgot about that too, having both at 2pi is just like sweeping it out twice. So should I only let phi go from 0-pi, or does it not matter which one I choose?
Thanks again, you've been very helpful!

 

[Mr LoganC]

So I worked through the problem and got an answer. Took me a good 30-40mins. There's no way he would give us one question that takes 40mins to do on the final. So I must have either done something wrong or did it a very difficult way. Not only that, the integrals were very difficult!
I ended up getting an answer of
$$\frac{4\pi}{15} - \frac{\pi}{4}$$

Is there any easy way to check this answer to see if it's right? Unfortunately, the practice problems do no have solutions for them.

 

[tiny-tim]

Hi Mr LoganC!

(just got up :zzz: …)

So I worked through the problem and got an answer. Took me a good 30-40mins. There's no way he would give us one question that takes 40mins to do on the final. So I must have either done something wrong or did it a very difficult way. Not only that, the integrals were very difficult!

hmm … as soon as I saw ρ = x² + y + z, I thought "I wouldn't like to try to integrate that!"

I ended up getting an answer of
$$\frac{4\pi}{15} - \frac{\pi}{4}$$

Is there any easy way to check this answer to see if it's right?

(have a pi: π )

Nope.

 

[Mr LoganC]

So if this question were to show up on the exam, (This is a practice question from last years exam), How should I go about doing it? And I still have to use spherical coordinates!

 

[tiny-tim]

Do it the same way!

Apart from the tediousness, what's wrong with that?

 

[Mr LoganC]

Nothing I guess. I'm a terribly slow test writer, so perhaps I'll leave that one 'till the end, but at least I can show my work and show that I do know how to go about the problem!

Thanks again Tiny-Tim! A thumbs up to you, sir!
-LoganC