• Support PF! Buy your school textbooks, materials and every day products Here!

Triple Density Integration

  • Thread starter 1d20
  • Start date
  • #1
12
0
I’ve got two questions about two problems. First I just want to confirm that I’m setting up this density integral properly:

“Find the mass of the conical solid bounded by [itex]z = \sqrt{x^2 + y^2}[/itex] and [itex]x^2 + y^2 + z^2 = 4[/itex] if the density at any point is proportional to the distance to the origin.

I’m taking that last part to mean that Density = ρ, so when I convert to spherical coordinates and set up the integral I get:

[itex] \int{0_2\pi} \int{0_\pi/4} \int{0_2} ρ^3 dρ d∅ dθ[/itex]

Is that [itex] ρ^3 [/itex] right? Or should it be [itex] (ρ^2 + ρ) [/itex]?
 

Answers and Replies

  • #2
12
0
Okay, next question.

“For the solid bounded by [itex] z = 4 - y^2[/itex], [itex]y = x[/itex], [itex]x = 0[/itex], [itex]z = 0 [/itex] find the volume using a double integral.”

I did this using a triple, but I’m stuck trying to use a double. Converting to cylindrical coordinates, z goes from [itex]0[/itex] to [itex] 4 - r^2sin^2θ [/itex] while θ goes from ∏/4 to ∏/2. All well and good.

But when I set up the integral:

[itex] \int{\pi/2_\pi/4} \int{4 - r^2sin^2θ_0} r dr dθ [/itex]

It’s clear that I’m going to end up with several unintegrated r terms. Gah, what am I missing?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,808
933
I’ve got two questions about two problems. First I just want to confirm that I’m setting up this density integral properly:

“Find the mass of the conical solid bounded by [itex]z = \sqrt{x^2 + y^2}[/itex] and [itex]x^2 + y^2 + z^2 = 4[/itex] if the density at any point is proportional to the distance to the origin.

I’m taking that last part to mean that Density = ρ, so when I convert to spherical coordinates and set up the integral I get:

[ itex ] \int{0_2\pi} \int{0_\pi/4} \int{0_2} ρ^3 dρ d∅ dθ[ /itex ]
Use instead
[ itex ] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/ itex ]
to get
[itex] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/itex]
as far as the LaTeX goes.

Is that [itex] ρ^3 [/itex] right? Or should it be [itex] (ρ^2 + ρ) [/itex]?
Have you thought about what you are doing? The mass of any object with density function [itex]\delta(x,y,z)[/itex] is [itex]\int\int\int \delta(x,y,z)dV[/itex]. I cannot imagine why you would think of adding.

However, you have the differential of volume wrong. For spherical coordinates it is
[tex]\rho^2 sin(\phi) d\rho d\phi d\theta[/tex]

Also, saying that the density is proportional to the radius does not mean it is equal to the radius. It means it is equal to some constant times the radius.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,808
933
Okay, next question.

“For the solid bounded by [itex] z = 4 - y^2[/itex], [itex]y = x[/itex], [itex]x = 0[/itex], [itex]z = 0 [/itex] find the volume using a double integral.”

I did this using a triple, but I’m stuck trying to use a double. Converting to cylindrical coordinates, z goes from [itex]0[/itex] to [itex] 4 - r^2sin^2θ [/itex] while θ goes from ∏/4 to ∏/2. All well and good.

But when I set up the integral:

[itex] \int{\pi/2_\pi/4} \int{4 - r^2sin^2θ_0} r dr dθ [/itex]

It’s clear that I’m going to end up with several unintegrated r terms. Gah, what am I missing?
I cannot see any good reason to convert to cylindrical coordinates. There is no circular symmetry that woud make cylindrical coordinates simpler. In Cartesian coordinates, the integral for volume would be
[tex]\int_{y= -2}^2\int_{z= 0}^{4- y^2}\int_{x= 0}^y dxdzdy[/tex]

Can you convert that to a double integral?
 
  • #5
12
0
Use instead
[ itex ] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/ itex ]
Thanks; this forum's code is very frustrating.

Have you thought about what you are doing?
Of course I've thought about it; that's why I'm here. If I hadn't thought about it, I'd have integrated according to my first instinct. Which in this case happened to be mostly right, but obviously I didn't know that.

I cannot see any good reason to convert to cylindrical coordinates. There is no circular symmetry that woud make cylindrical coordinates simpler. In Cartesian coordinates, the integral for volume would be
[tex]\int_{y= -2}^2\int_{z= 0}^{4- y^2}\int_{x= 0}^y dxdzdy[/tex]

Can you convert that to a double integral?
I've had no practice with this so let's find out...

[itex] \int_0^{2} \int_0^{y} (4 - y^2) dx dy [/itex]
 

Related Threads on Triple Density Integration

Replies
9
Views
3K
  • Last Post
Replies
2
Views
983
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
0
Views
675
  • Last Post
Replies
13
Views
744
  • Last Post
Replies
6
Views
473
  • Last Post
Replies
15
Views
685
  • Last Post
Replies
2
Views
4K
Top