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:

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