Multivariable Calculus - Integration Assignment

[ConnorM]

Homework Statement

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!

Homework 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)

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!

 

[STEMucator]

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?

 

[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?

 

[STEMucator]

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?

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

Simplifying your integral, you should get:

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

 

[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?

 

[STEMucator]

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?

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.

 

[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?

 

[STEMucator]

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?

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

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

 

[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

 

[STEMucator]

Just so I'm clear since x=au I would take au and change it to (ρsin(β)cosΘ)

Right. Now what happens to the integral?

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

 

[ConnorM]

Hey so I got,

abcρ∫V′ r^2 dϕdθdr

 

[STEMucator]

Hey so I got,

abcρ∫V′ r^2 dϕdθdr

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

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

 

[ConnorM]

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

 

[ConnorM]

Now I have,

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

 

[STEMucator]

Now I have,

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

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

 

[ConnorM]

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

 

[STEMucator]

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

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

With $w = rcos(\phi)$.

 

[ConnorM]

OK I will try that!

 

[STEMucator]

Do I need the abc then? So my equations would be u=rsin(θ)cos(ϕ), v=rsin(θ)sin(ϕ), w= rcos(ϕ), what would I differentiate in terms of?

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

$\frac{∂(u,v,w)}{∂(, ,)}$

 

[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)?

 

[STEMucator]

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)?

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" into the integral.

 

[ConnorM]

So I found the |J| = (r^2)(4sinθ).

 

[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.

 

[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?

 

[STEMucator]

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?

I seem to be getting:

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

Check your answer again.

 

[ConnorM]

Sorry do you mine telling me what your answer for the Jacobian was? Also did you multiply by the volume element that you get from converting to polar coordinates r^2 (sinϕ)?

 

[STEMucator]

Sorry do you mine telling me what your answer for the Jacobian was? Also did you multiply by the volume element that you get from converting to polar coordinates r^2 (sinϕ)?

When we map from$V' → V''$, the Jacobian of the spherical transformation is given by:

$|J| = r^2 sin(\phi)$

You correctly determined this already I believe.

The Jacobian IS the volume element.

 

[ConnorM]

After I found my determinant I had
ϕθ
r^2 ( sin^3θsin^2ϕ + sinθcos^2θcos^2θ + sinθcos^2θsin^2ϕ + sin^3θcos^2ϕ )

 

[ConnorM]

Oh I didn't know that thanks! So the Jacobian is r^2 (sinϕ). So now I have,

abcρ∫V′′(a2^+b2^)r4sin^3(ϕ)dV′′

Also the abc outside the integral do we need that? All that is is the jacobian before converting to spherical coordiantes.

The abc came from when x=au, y=bv, z=cw,

determinant of the matrix d(u, v, w) /d(x, y ,z) == abc

 

[STEMucator]

After I found my determinant I had
ϕθ
r^2 ( sin^3θsin^2ϕ + sinθcos^2θcos^2θ + sinθcos^2θsin^2ϕ + sin^3θcos^2ϕ )

I'm getting the feeling the main concept of the integration is what seems to be eluding you here.

When you originally had an ellipsoid in $x,y,z$ space, the integral had the form:

$\int_{V} f(x,y,z) \space dV$

We mapped $x,y,z$ to $u,v,w$ using an invertible transformation $V → V'$ that took the ellipsoid to the unit sphere. To do this, we had to multiply by the Jacobian of the $x,y,z$ transform:

$\int_{V} f(x,y,z) \space dV = \int_{V'} g(u,v,w) |J_{x,y,z}| \space dV'$

Similarly, when we mapped using spherical co-ordinates from $V' → V''$, we have to multiply by the Jacobian of the $u,v,w$ transformation:

$\int_{V} f(x,y,z) \space dV = \int_{V'} g(u,v,w) |J_{x,y,z}| \space dV' = \int_{V''} h(r, \theta, \phi) |J_{x,y,z}| |J_{u,v,w}| \space dV''$

 

[ConnorM]

Ok I see how it works now thank you. So then my integral would be,

abc(a^2 +b^2)ρ∫V′′r4sin^3(ϕ)dV′′

This has both of the Jacobians we found, from here I would integrate to find my answer then? My bounds of integration would have to be 0 ≤ θ ≤ 2pi , 0 ≤ ϕ ≤ pi , 0 ≤ r ≤ 1 ?

 

[STEMucator]

Ok I see how it works now thank you. So then my integral would be,

abc(a^2 +b^2)ρ∫V′′r4sin^3(ϕ)dV′′

This has both of the Jacobians we found, from here I would integrate to find my answer then? My bounds of integration would have to be 0 ≤ θ ≤ 2pi , 0 ≤ ϕ ≤ pi , 0 ≤ r ≤ 1 ?

Somehow the limits look good, but the integral looks wrong. I already showed you what you should wind up integrating a few posts ago. What do you get when integrating this?

Some advice: Work on determinants/trig/algebra a little bit to make this sort of thing easier in the future.

 

[ConnorM]

Ok I think I messed up and thought that I could factor out the sin^2θ + cos^2θ to simplify to 1. I will try again.

 

[ConnorM]

So now after integrating abcρ∫V′′(a2cos2(θ)+b2sin2(θ))r4sin3(ϕ)dV′′ , I have

(8pi*a^3 bcρ) / 15

 

[STEMucator]

So now after integrating abcρ∫V′′(a2cos2(θ)+b2sin2(θ))r4sin3(ϕ)dV′′ , I have

(8pi*a^3 bcρ) / 15

You have an iterated integral when you plug in the limits, right? So it doesn't matter too much what order of integration you decide to use.

Usually, you would want to pick the convenient one first. Either $\phi$ or $r$ in this case.

Check your answer again though.

 

[ConnorM]

This time I got, (3pi^2 r^4 (a^2 + b^2)abcρ)/8

 

[STEMucator]

This time I got, (3pi^2 r^4 (a^2 + b^2)abcρ)/8

Still not quite. I'm not sure where you are making the error, but it's in there. As a hint, what is the result of doing the first integration inside the big brackets:

$$abc \rho \int_0^1 \int_0^{2\pi} (a^2 cos^2(\theta) + b^2 sin^2(\theta)) r^4 \left[ \int_0^{\pi} sin^3(\phi) \space d \phi \right] d \theta d r$$

 

[ConnorM]

Doing the first integration I found that,

1/12 (cos(3ϕ) - 9cos(ϕ))

then subbing in ϕ = pi and ϕ = 0

1/12 (cos(3pi) - 9cos(pi)) - 1/12 (cos(0) - 9cos(0)) = 2/3 - (-2/3) = 4/3

 

[ConnorM]

After that I can integrate for dr,

(a2cos2(θ)+b2sin2(θ))r^5 /5

then subbing in r = 1 and r = 0

(a^2cos^2(θ)+b^2sin^2(θ))/5 - 0 = (a^2cos^2(θ)+b^2sin^2(θ))/5

 

[STEMucator]

Doing the first integration I found that,

1/12 (cos(3ϕ) - 9cos(ϕ))

then subbing in ϕ = pi and ϕ = 0

1/12 (cos(3pi) - 9cos(pi)) - 1/12 (cos(0) - 9cos(0)) = 2/3 - (-2/3) = 4/3

Okay, so now what happens here:

$$abc \rho \int_0^1 \frac{4}{3} r^4 \left[ \int_0^{2\pi} (a^2 cos^2(\theta) + b^2 sin^2(\theta)) d \theta \right] d r$$

Hint: You need to integrate $cos^2(\theta)$ and $sin^2(\theta)$ separately, treating $a^2$ and $b^2$ as constants.

 

[ConnorM]

Then I can integrate for dθ,

(a^2 θ)/2 + (a^2 sin(2θ)) /4 + (b^2 θ)/2 - (b^2 sin(2θ)) /4

Subbing in 2pi and 0,

(a^2 2pi)/2 + (a^2 sin(4pi)) /4 + (b^2 2pi)/2 - (b^2 sin(4pi)) /4 - ( (a^2 0)/2 + (a^2 sin(0)) /4 + (b^2 0)/2 - (b^2 sin(0)) /4 )

= a^2 pi + b^2 pi

So my final answer is pi(a^2 + b^2)

 

[STEMucator]

Then I can integrate for dθ,

(a^2 θ)/2 + (a^2 sin(2θ)) /4 + (b^2 θ)/2 - (b^2 sin(2θ)) /4

Subbing in 2pi and 0,

(a^2 2pi)/2 + (a^2 sin(4pi)) /4 + (b^2 2pi)/2 - (b^2 sin(4pi)) /4 - ( (a^2 0)/2 + (a^2 sin(0)) /4 + (b^2 0)/2 - (b^2 sin(0)) /4 )

= a^2 pi + b^2 pi

So my final answer is pi(a^2 + b^2)

Right, now just clean up the last integration, what do you get?

 

[ConnorM]

Oh by the way is there a tutorial or like a forum post somewhere on how to format the equations like you do? Makes it a lot easier to read, sorry for typing it out like I am!

 

[ConnorM]

Oops forgot about abcρ that was outside, so the answer is

abcρ * pi (a^2 + b^2)

 

[STEMucator]

Why not relax and carefully write the steps out as you did before? Your answer seems to be missing a factor of $\frac{4}{15}$. Anytime you rush math and forget to write something down, you're going to make a mistake like that.

Also, a useful LaTeX reference is located here:

https://www.physicsforums.com/showthread.php?p=3977517#post3977517

 

[ConnorM]

Ok I see now, yes I was rushing through things quickly and forgot that I had taken out the 4/3 from integrating dϕ and I had taken out 1/5 after integrating dθ. So that gives me the factor of 4/15.

final answer is now

(4/15) *abcρ * pi (a^2 + b^2)

 

[STEMucator]

Ok I see now, yes I was rushing through things quickly and forgot that I had taken out the 4/3 from integrating dϕ and I had taken out 1/5 after integrating dθ. So that gives me the factor of 4/15.

final answer is now

(4/15) *abcρ * pi (a^2 + b^2)

Good. That is indeed the answer and I'm glad you see why.

 

[ConnorM]

Thanks a lot, thank you for bearing with me all the way through!