Setting up some triple integrals

[STEMucator]

Homework Statement

I want to know if I've gone about setting up these integrals in these questions properly before I evaluate them.

(i). Find the mass of the cylinder $S: 0 ≤ z ≤ h, x^2 + y^2 ≤ a^2$ if the density at the point (x,y,z) is $δ = 5z^4 + 6(x^2 - y^2)^2$.

(ii). Evaluate the integral of $f(x,y,z) = (x^2 + y^2 + z^2)^{1/2}$ over the region D which is above the cone $z^2 = 3(x^2 + y^2)$ and between the spheres $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 9$.

Homework Equations

Cylindrical polars and spherical polars.

The Attempt at a Solution

(i). Switching to cylindrical polars, x=rcosθ, y=rsinθ and z=z.

So $S → S^{'}: 0 ≤ z ≤ h, -a ≤ r ≤ a, 0 ≤ \theta ≤ 2 \pi$
and the density becomes $δ = 5z^4 + 6r^4cos^2(2θ)$.
The Jacobian of the cylindrical polars is just r so our integral over S' becomes :

$\int_{0}^{2π} \int_{-a}^{a} \int_{0}^{h} 5rz^4 + 6r^5cos^2(2θ) dzdθdr$

(ii). Switching to spherical polars, x=ρcosθsinφ, y=ρsinθsinφ and z=ρcosφ and we also know that $x^2+y^2+z^2 = ρ^2$.

So f(x,y,z) now becomes $(ρ^2)^{1/2} = ρ$
The two spheres yield $1 ≤ ρ ≤ 3$
θ is as is in cylindrical, so $0 ≤ θ ≤ 2π$
Also $0 ≤ φ ≤ π$

The purpose of the cone is not yet clear to me.

 

[Mark44]

Homework Statement

I want to know if I've gone about setting up these integrals in these questions properly before I evaluate them.

(i). Find the mass of the cylinder $S: 0 ≤ z ≤ h, x^2 + y^2 ≤ a^2$ if the density at the point (x,y,z) is $δ = 5z^4 + 6(x^2 - y^2)^2$.

(ii). Evaluate the integral of $f(x,y,z) = (x^2 + y^2 + z^2)^{1/2}$ over the region D which is above the cone $z^2 = 3(x^2 + y^2)$ and between the spheres $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 9$.

Homework Equations

Cylindrical polars and spherical polars.

The Attempt at a Solution

(i). Switching to cylindrical polars, x=rcosθ, y=rsinθ and z=z.

So $S → S^{'}: 0 ≤ z ≤ h, -a ≤ r ≤ a, 0 ≤ \theta ≤ 2 \pi$
and the density becomes $δ = 5z^4 + 6r^4cos^2(2θ)$.
The Jacobian of the cylindrical polars is just r so our integral over S' becomes :

$\int_{0}^{2π} \int_{-a}^{a} \int_{0}^{h} 5rz^4 + 6r^5cos^2(2θ) dzdθdr$

(ii). Switching to spherical polars, x=ρcosθsinφ, y=ρsinθsinφ and z=ρcosφ and we also know that $x^2+y^2+z^2 = ρ^2$.

So f(x,y,z) now becomes $(ρ^2)^{1/2} = ρ$
The two spheres yield $1 ≤ ρ ≤ 3$
θ is as is in cylindrical, so $0 ≤ θ ≤ 2π$
Also $0 ≤ φ ≤ π$

The purpose of the cone is not yet clear to me.

The cone is part of the definition of the region, which looks sort of like a tapered cork or stopper.

 

[STEMucator]

The cone is part of the definition of the region, which looks sort of like a tapered cork or stopper.

I'll take it (i) was good to go then.

For (ii) does the cone not impact some of the information for the integral?

Also, a tapered cork? Looks more like an ice cream cone with a bite in it I find.

EDIT : If anyone could help me verify the second one it would be greatly appreciated. I don't set this up often so it's hard to tell if I'm doing it correctly.

 

[Dick]

I'll take it (i) was good to go then.

For (ii) does the cone not impact some of the information for the integral?

Also, a tapered cork? Looks more like an ice cream cone with a bite in it I find.

EDIT : If anyone could help me verify the second one it would be greatly appreciated. I don't set this up often so it's hard to tell if I'm doing it correctly.

Did you sketch a graph of the region? If you know it's an ice cream cone with a bite, you probably did. Doesn't being above the cone imply some restriction on $\phi$? What's the angle of the cone?

 

[STEMucator]

Did you sketch a graph of the region? If you know it's an ice cream cone with a bite, you probably did. Doesn't being above the cone imply some restriction on $\phi$? What's the angle of the cone?

Well when I drew it out, I had a bigger sphere with a smaller sphere inside of it. Then the cones would come in and cut the region created by the boundaries of the spheres; both in the positive and negative portions, but we only care about the positive portion since we want 'above' the cone ( I'm guessing this is what is intended since in the question it said 'which is above the cone' ).

That portion of the spheres that the cone cuts into looks like someone took a bite out of the bottom of the cone.

Hmm so would I not need to find out when the cone intersects the first sphere and then similarly the second?

Just a thought. Not sure if it has merit.

 

[Dick]

Well when I drew it out, I had a bigger sphere with a smaller sphere inside of it. Then the cones would come in and cut the region created by the boundaries of the spheres; both in the positive and negative portions, but we only care about the positive portion since we want 'above' the cone ( I'm guessing this is what is intended since in the question it said 'which is above the cone' ).

That portion of the spheres that the cone cuts into looks like someone took a bite out of the bottom of the cone.

Hmm so would I not need to find out when the cone intersects the first sphere and then similarly the second?

Just a thought. Not sure if it has merit.

You aren't listening. Doesn't the angle of the cone have something to do with the range of the polar angle ϕ?

 

[STEMucator]

You aren't listening. Doesn't the angle of the cone have something to do with the range of the polar angle ϕ?

Yes it does, there are two rays of the cone which go off at an angle.

 

[Dick]

Yes it does, there are two rays of the cone which go off at an angle.

So ϕ doesn't go from 0 to pi, yes?

 

[STEMucator]

So ϕ doesn't go from 0 to pi, yes?

Indeed... by the looks of it by the way I'm picturing it, it would go from $-π/2$ along to $π/2$.

 

[Dick]

Indeed... by the looks of it by the way I'm picturing it, it would go from $-π/2$ along to $π/2$.

Does not. What's the angle of the cone from the z-axis? It's not pi/2.

 

[STEMucator]

Does not. What's the angle of the cone from the z-axis? It's not pi/2.

Ohhh wait so phi is the angle from the z axis to the ray of the cone... so I should...

Switch the cone to spherical and then calculate a value for phi I believe?

So in spherical the cone becomes :

$ρ^2cos^2 \phi = 3ρ^2sin^2φ$
$cos^2 \phi = 3sin^2φ$
$tan^2 \phi = 1/3$
$\phi = arctan( \sqrt{1/3} )$

So pi/6? So I would take my bounds from 0 to pi/6?

 

[Dick]

Ohhh wait so phi is the angle from the z axis to the ray of the cone... so I should...

Switch the cone to spherical and then calculate a value for phi I believe?

So in spherical the cone becomes :

$ρ^2cos^2 \phi = 3ρ^2sin^2φ$
$cos^2 \phi = 3sin^2φ$
$tan^2 \phi = 1/3$
$\phi = arctan( \sqrt{1/3} )$

So pi/6? So I would take my bounds from 0 to pi/6?

Now you've got it.

 

[STEMucator]

Now you've got it.

Ah I see now. So my final integral for (ii) would be :

$\int_{1}^{3} \int_{0}^{2π} \int_{0}^{π/6} ρ d \phi dθdρ$

and then I've got two integrals to evaluate.

Thank you for helping me understand what's going on. I should feel comfortable setting these up from now on.

 

[Coelum]

Zondrina: check the volume element in polar coordinates.

 

[STEMucator]

Zondrina: check the volume element in polar coordinates.

What do you mean check the volume element?

 

[Dick]

What do you mean check the volume element?

When you are integrating in polar coordinates, you don't just write dϕdθdρ after the function you are integrating. There's the jacobian part of the volume element as well.

 

[STEMucator]

When you are integrating in polar coordinates, you don't just write dϕdθdρ after the function you are integrating. There's the jacobian part of the volume element as well.

*facepalm.jpg*

Totally forgot about the jacobian there for a moment. The jacobian is ρ²sin(phi) which would change my integral a bit.

Thanks for reminding me guys :)

 

[Coelum]

Yes, the determinant of the Jacobian matrix can be seen as the scale factor of the unit volume when changing coordinates.

 

[STEMucator]

Sorry for bumping this, but I evaluated these integrals awhile ago and something has been bothering me.

For the cylindrical integral in (i), I get the mass to be 0 which does not make any sense at all?

Where might I have gone wrong? I'm guessing it has something to do with one of my limits. Also, I re-arranged my integral like this :

$\int_{-a}^{a} \int_{0}^{2 \pi} \int_{0}^{h} 5z^4r + 6r^5cos^2(2 \theta) dzd \theta dr$

 

[Dick]

Sorry for bumping this, but I evaluated these integrals awhile ago and something has been bothering me.

For the cylindrical integral in (i), I get the mass to be 0 which does not make any sense at all?

Where might I have gone wrong? I'm guessing it has something to do with one of my limits. Also, I re-arranged my integral like this :

$\int_{-a}^{a} \int_{0}^{2 \pi} \int_{0}^{h} 5z^4r + 6r^5cos^2(2 \theta) dzd \theta dr$

Why are you integrating over negative values of r?

 

[STEMucator]

Why are you integrating over negative values of r?

How careless of me. Thank you for pointing that out. So really r would run from 0 to a since we assume the radius is positive.

Thanks man :).