# Homework Help: Multivariable Calculus - Integration Assignment

1. Aug 11, 2014

### ConnorM

1. The problem statement, all variables and given/known data
Hello PF! I'm having some trouble on the last part of my assignment, it's question 4 part "c".
Here is a picture of the assignment [http://imgur.com/1edJ3g5] ! I'll post this instead of writing it out so we know that we're all looking at the same thing!

2. Relevant equations
The change of variables given at the beginning of question 4 are,
x=au, y=bv, z=cw

From part "a" I used the change of variables given in the question and found that the ellipsoid equation became u^2 + v^2 + w^2 = 1. I found the Jacobian to be equal to abc. Next I set up my integral to determine the volume over the region S, ∫∫∫abc dV, Since a sphere with the radius 1 will have a volume of 4pi/3 I found my volume to be abc*4pi/3.

I think what I need for part "c" is just the Jacobian. so the Jacobian = abc.

The equation for inertia that we were given in class was I=∫∫(x^2 + y^2)*ρ(x,y) dA

Changing from rectangular to spherical coordinates. (I think you need to use this)*

x = ρsin(β)cosΘ
y=ρsin(β)sinΘ
z=ρcos(β)

3. The attempt at a solution

So to start off since I'm working in 3 Dimensions would I have to change my formula for moment of inertia to,

I=∫∫∫(x^2 + y^2 + z^2)*ρ(x,y,z) dV,

Then from here since I am working with changed variables I changed the x, y, and z, also multiplied by the Jacobian,

I=∫∫∫((au)^2 + (bv)^2 + (cw)^2)*ρ(x,y,z)*abc dV

From here would I have to switch to spherical coordinates? I would obtain,

I=∫∫∫((ρsin(β)cosΘ)^2 + (ρsin(β)sinΘ)^2 + (ρcos(β))^2)*ρ(ρ,Θ,β)*abc dV

I=∫∫∫(ρ^2)*ρ(ρ,Θ,β)*abc*(ρ^2)sin(β) dρdΘdβ

Then my bounds of integration would be

0 ≤ ρ ≤ 1
0 ≤ Θ ≤ 2pi
0 ≤ β ≤ pi

Does this look right so far, Or am I off track? if it looks good just let me know and I will continue, I'll reply as soon as I have either finished or got stumped again!

2. Aug 11, 2014

### Zondrina

Hi there.

The question states the density is constant. Hence $\rho = K$ and it can be pulled outside the integral. Also, I believe you have to use:

$\int_V \rho r^2_{\perp} dV = \rho \int_V (x^2 +y^2) dx dy dz$

Make the substitution:

$u = x/a$
$v = y/b$
$w = z/c$

What happens to the integral?

3. Aug 11, 2014

### ConnorM

Since it says about the z-axis is that why you use (x^2 + y^2) and not (x^2 + y^2 + z^2)? Also how would I make the substitutions when there are no u's, v's, or w's?

4. Aug 12, 2014

### Zondrina

The substitution provided maps the ellipsoid $V → V'$, where $V'$ is the unit sphere.

Simplifying your integral, you should get:

$\rho \int_{V'} \space [(au)^2 + (bv)^2] \space \mid \frac{∂(x,y,z)}{∂(u,v,w)} \mid \space dudvdw$

5. Aug 12, 2014

### ConnorM

So then I would have ρ∫V′ [(au)2+(bv)2] *abc dudvdw ? Then I would use the bounds 0 ≤ u ≤ 1, 0 ≤ v ≤ 2pi, 0 ≤ w ≤ pi?

6. Aug 12, 2014

### Zondrina

Its probably easier to use spherical co-ordinates to map $V' → V''$. That was the purpose of the substitution to give you a nice sphere to work with.

7. Aug 12, 2014

### ConnorM

OK so then take ρ∫V′ [(au)2+(bv)2] *abc dudvdw , switch to spherical coordinates. Then would my bounds of integration be the same?

8. Aug 12, 2014

### Zondrina

What do you get for $\phi$, $\theta$ and $r$ when applying the transform?

I say $r$ since $\rho$ is already taken.

9. Aug 12, 2014

### ConnorM

Just so I'm clear since x=au I would take au and change it to (rsin(β)cosΘ), using r instead of ρ for the conversion to spherical.

I got,

abcρ∫V′ r^2 dϕdθdr

Last edited: Aug 12, 2014
10. Aug 12, 2014

### Zondrina

Right. Now what happens to the integral?

Just to be clear, I take $u = rcos(\theta)sin(\phi)$.

11. Aug 12, 2014

### ConnorM

Hey so I got,

abcρ∫V′ r^2 dϕdθdr

12. Aug 12, 2014

### Zondrina

Take $u$ as it was mentioned in the prior post. My apologies.

Take $v = rsin(\theta)sin(\phi)$.

13. Aug 12, 2014

### ConnorM

No worries, give me a second I will try that!

14. Aug 12, 2014

### ConnorM

Now I have,

abcρ∫V′ (r^2)(sin^2(θ))(a^2 + b^2) dϕdθdr

15. Aug 12, 2014

### Zondrina

You seem to be ignoring something. What about $|J|$?

16. Aug 12, 2014

### ConnorM

Isn't the |J| just abc? I moved it outside the integration.

17. Aug 12, 2014

### Zondrina

No, $|J|$ is different for the spherical transformation. Compute the jacobian of the $u, v, w$ transformation.

With $w = rcos(\phi)$.

18. Aug 12, 2014

### ConnorM

OK I will try that!

Last edited: Aug 12, 2014
19. Aug 12, 2014

### Zondrina

What are $u, v, w$ in terms of? Looks somethin like:

$\frac{∂(u,v,w)}{∂(, ,)}$

Last edited: Aug 12, 2014
20. Aug 12, 2014

### ConnorM

This is looking really really messy, the equations I'm using are

u=rsin(θ)cosϕ
v=rsin(θ)sinϕ
w=rcos(ϕ)

The |J| for this is the determinant of d(ϕ, θ, r) / d(u, v, w)?

21. Aug 12, 2014

### Zondrina

You have it upside down for some indeterminable reason.

EDIT: In the interest of saving a little time since its pretty early around where I am, I will make it clear that:

$\frac{\partial(u,v,w)}{\partial(r,\theta,\phi)}$

Is what you are trying to find the result of. That is the jacobian of the $(u,v,w) → (r, \theta, \phi)$, $V' → V''$ transformation you essentially "plug" in to the integral.

Last edited: Aug 12, 2014
22. Aug 12, 2014

### ConnorM

So I found the |J| = (r^2)(4sinθ).

23. Aug 12, 2014

### ConnorM

So I found the |J| = (r^2)(4sinθ).
Also since we converted to spherical coordinates I would have to add on,

(r^2)(sinϕ)

to my integral.

24. Aug 12, 2014

### ConnorM

After finding the |J| and multiplying by (r^2)(sinϕ) since I converted to polar I now have,

ρ∫V′ (r^2)(sin^2(θ))(a^2 + b^2)(r^2)(4sinθ)(r^2)(sinϕ) dϕdθdr

This looks pretty messy, do you think this is correct?

25. Aug 12, 2014

### Zondrina

I seem to be getting:

$abc \rho \int_{V''} (a^2cos^2(\theta) + b^2sin^2(\theta))r^4 sin^3(\phi) dV''$