How can I calculate the cumulative mass of a disk using disk mass density?

In summary: It looks like you're on the right track. The biggest mistake I see is that your equation starts off with ##\rho(r)## when it should be m(r), the cumulative mass. Again, I haven't worked through the problem, and am assuming that @pasmith's hint is correct.If so, all you need to do to finish is 1) evaluate ##\int h e^{-r/h}dr##, and then 2) evaluate the resulting large expression at r = 0 and r = R, subtracting the first expression (with r = 0) from the latter expression.
  • #1
independentphysics
26
2
Homework Statement
Find cumulative mass of a disk
Relevant Equations
Exponential mass function
I want to find the cumulative mass m(r) of a mass disk. I have the mass density in terms of r, it is an exponential function:
ρ(r)=ρ0*e^(-r/h)

A double integral in polar coordinates should do, but im not sure about the solution I get.
 
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  • #2
independentphysics said:
Homework Statement: Find cumulative mass of a disk
Relevant Equations: Exponential mass function

I want to find the cumulative mass m(r) of a mass disk. I have the mass density in terms of r, it is an exponential function:
ρ(r)=ρ0*e^(-r/h)

A double integral in polar coordinates should do, but im not sure about the solution I get.
What solution do you get?

Also, what is the significance of h in your equation?

You can write an iterated integral in LaTeX like this: ##\int_{\theta = 0}^b \int_{r = 0}^a f(r, \theta)~dr~d\theta##
 
  • #3
Mark44 said:
What solution do you get?

Also, what is the significance of h in your equation?

You can write an iterated integral in LaTeX like this: ##\int_{\theta = 0}^b \int_{r = 0}^a f(r, \theta)~dr~d\theta##
I am getting: ρ(r)=2*π*ρ0 * h^2 * [1 - e^(-r/h)]
h is a constant
 
  • #4
Is your intermediate result [itex]2\pi \rho_0 \int_0^R re^{-r/h}\,dr[/itex]?
 
  • #5
pasmith said:
Is your intermediate result [itex]2\pi \rho_0 \int_0^R re^{-r/h}\,dr[/itex]?
Yes, it is.
 
  • #6
pasmith said:
Is your intermediate result [itex]2\pi \rho_0 \int_0^R re^{-r/h}\,dr[/itex]?

independentphysics said:
Yes, it is.
Assuming @pasmith's work is correct (I haven't worked the problem), the integral shown can be evaluated using integration by parts.
 
  • #7
Mark44 said:
Assuming @pasmith's work is correct (I haven't worked the problem), the integral shown can be evaluated using integration by parts.
Yes, I am aware of it, but Im not sure about my final result.
 
  • #8
independentphysics said:
Yes, I am aware of it, but Im not sure about my final result.
Well, why don't you show us what you got, together with your work?
 
  • #9
Mark44 said:
Well, why don't you show us what you got, together with your work?

From [itex]2\pi \rho_0 \int_0^R re^{-r/h}\,dr[/itex]

I evaluate this integral using integration by substitution. Let u = r/h, then du/dr = 1/h and dr = h*du. Substituting, we get:

m(r) = 2π * ρ0 * h^2 * ∫ u * e^(-u) * du

Is this correct?
 
  • #10
independentphysics said:
From [itex]2\pi \rho_0 \int_0^R re^{-r/h}\,dr[/itex]

I evaluate this integral using integration by substitution. Let u = r/h, then du/dr = 1/h and dr = h*du. Substituting, we get:

m(r) = 2π * ρ0 * h^2 * ∫ u * e^(-u) * du

Is this correct?
It's technically correct, but it doesn't get you any closer to evaluating the integral. To integrate ##\int re^{-r/h}dr## you need to use integration by parts, something I mentioned before.
A substitution won't be useful here.
 
  • #11
Mark44 said:
It's technically correct, but it doesn't get you any closer to evaluating the integral. To integrate ##\int re^{-r/h}dr## you need to use integration by parts, something I mentioned before.
A substitution won't be useful here.
How do I integrate by parts the function?
 
  • #12
independentphysics said:
How do I integrate by parts the function?
Your calculus textbook should have a section on integration by parts, along with several examples.
 
  • #13
Mark44 said:
Your calculus textbook should have a section on integration by parts, along with several examples.
If I choose u=r and v=e^(-r/h) so that du=1, will this do?
 
  • #14
independentphysics said:
If I choose u=r and v=e^(-r/h) so that du=1, will this do?
No.

If ##u=r## (that's fine), then ##du=dr##.

For ##v##, you need ##\displaystyle dv=e^{-r/h}\,dr##.

Integrate to get ##v##.
 
  • #15
SammyS said:
No.

If ##u=r## (that's fine), then ##du=dr##.

For ##v##, you need ##\displaystyle dv=e^{-r/h}\,dr##.

Integrate to get ##v##.
ρ(r)= ρ0*e^(-r/h)

2π results from integration of angle in polar coordinates

integrate (r*e^(-r/h) dr)
u=r
du=dr
dv=e^(-r/h)*dr
v=-h*exp(-r/h)

ρ(r)=2π * ρ0 * ( r*h*(-e^(-r/h) + int(h*e^(-r/h)dr )

Is it correct?
 
  • #16
independentphysics said:
ρ(r)= ρ0*e^(-r/h)

2π results from integration of angle in polar coordinates

integrate (r*e^(-r/h) dr)
u=r
du=dr
dv=e^(-r/h)*dr
v=-h*exp(-r/h)

ρ(r)=2π * ρ0 * ( r*h*(-e^(-r/h) + int(h*e^(-r/h)dr )

Is it correct?
It looks like you're on the right track. The biggest mistake I see is that your equation starts off with ##\rho(r)## when it should be m(r), the cumulative mass. Again, I haven't worked through the problem, and am assuming that @pasmith's hint is correct.

If so, all you need to do to finish is 1) evaluate ##\int h e^{-r/h}dr##, and then 2) evaluate the resulting large expression at r = 0 and r = R, subtracting the first expression (with r = 0) from the latter expression.

The integral I mentioned is relatively straightforward -- a simple substitution will work.
 
  • #17
Mark44 said:
It looks like you're on the right track. The biggest mistake I see is that your equation starts off with ##\rho(r)## when it should be m(r), the cumulative mass. Again, I haven't worked through the problem, and am assuming that @pasmith's hint is correct.

If so, all you need to do to finish is 1) evaluate ##\int h e^{-r/h}dr##, and then 2) evaluate the resulting large expression at r = 0 and r = R, subtracting the first expression (with r = 0) from the latter expression.

The integral I mentioned is relatively straightforward -- a simple substitution will work.

Equation:
ρ(r)=ρ0*e^(-r/h)
In polar coordinates, a 2π factor multiplies the integratal (r*e^(-r/h) dr)
u=r
du=dr
dv=e^(-r/h)*dr
v=-h*exp(-r/h)
2π * ρ0 * ( r*h*(-e^(-r/h) + int(h*e^(-r/h)dr )
M (r)=2π * ρ0 * h * (h + r) * (-e^(-r/h)) + c
I find c by M(0)=c
 
  • #18
independentphysics said:
Equation:
ρ(r)=ρ0*e^(-r/h)
In polar coordinates, a 2π factor multiplies the integratal (r*e^(-r/h) dr)
u=r
du=dr
dv=e^(-r/h)*dr
v=-h*exp(-r/h)
2π * ρ0 * ( r*h*(-e^(-r/h) + int(h*e^(-r/h)dr )
M (r)=2π * ρ0 * h * (h + r) * (-e^(-r/h)) + c
I find c by M(0)=c
Your final answer for M should involve h and R, but should not involve r.
You need to evaluate M(R) and M(0), and your final answer for M will be M(R) - M(0).
For M(0) I get ##2\pi \rho_0 h^2##.
 
  • #19
Mark44 said:
Your final answer for M should involve h and R, but should not involve r.
You need to evaluate M(R) and M(0), and your final answer for M will be M(R) - M(0).
For M(0) I get ##2\pi \rho_0 h^2##.
The final answer for cumulative mass M(r) and depends on r.

R is end of the disk, which doesn't even need to exist (I've never mentioned if the disk in infinite).

M(0) is M(r=0).

The final answer for cumulative mass M(r) involves +constant an the end of the discussed integral, which is M(0).
 
  • #20
The mass of the disc [itex]0 \leq r < R[/itex] is [tex]M(R) = 2\pi\rho_0 \int_0^R re^{-r/h}\,dr.[/tex] It follows immediately that [itex]M(0) = 0[/itex]. Since [tex]
\frac{d}{dr} re^{-r/h} = (1 - rh^{-1})e^{-r/h}[/tex] we have [tex]
re^{-r/h} = he^{-r/h} - h\frac{d}{dr} re^{-r/h}[/tex] and thus [tex]
\int_0^R re^{-r/h}\,dr = h^2(1 - e^{-R/h}) - hRe^{-R/h}.[/tex]
 
  • #21
pasmith said:
The mass of the disc [itex]0 \leq r < R[/itex] is [tex]M(R) = 2\pi\rho_0 \int_0^R re^{-r/h}\,dr.[/tex] It follows immediately that [itex]M(0) = 0[/itex]. Since [tex]
\frac{d}{dr} re^{-r/h} = (1 - rh^{-1})e^{-r/h}[/tex] we have [tex]
re^{-r/h} = he^{-r/h} - h\frac{d}{dr} re^{-r/h}[/tex] and thus [tex]
\int_0^R re^{-r/h}\,dr = h^2(1 - e^{-R/h}) - hRe^{-R/h}.[/tex]
There is a constant (+C) after integrating the density function to obtain cumulative mass M(r), and M(0) is not 0.
 
  • #22
independentphysics said:
There is a constant (+C) after integrating the density function to obtain cumulative mass M(r), and M(0) is not 0.

You seem confused. There is no arbitrary constant in a definite integral, which is what [itex]M(R) = 2\pi\int_0^R r\rho(r)\,dr[/itex] is, and a basic result of integration theory (Riemann or Lebesgue) is that for any function [itex]f[/itex], [itex]\int_a^a f(x)\,dx = 0[/itex].
 
  • #23
pasmith said:
You seem confused. There is no arbitrary constant in a definite integral, which is what [itex]M(R) = 2\pi\int_0^R r\rho(r)\,dr[/itex] is, and a basic result of integration theory (Riemann or Lebesgue) is that for any function [itex]f[/itex], [itex]\int_a^a f(x)\,dx = 0[/itex].
Cumulative mass in function of r is not a definite integral.
 
  • #24
independentphysics said:
Cumulative mass in function of r is not a definite integral.

The limits are specified: the lower limit is fixed at zero, and the upper limit is the radius we are considering. That makes it a definite integral.
 
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  • #25
independentphysics said:
There is a constant (+C) after integrating the density function to obtain cumulative mass M(r), and M(0) is not 0.
Yes:
You can use the indefinite integral to obtain the correct expression for the cumulative mass, M(r), of the disk. However, it makes sense that M(0) = 0 . That does not give you a value of 0 for the integration constant, C.

In Post#17, you had:
independentphysics said:
M (r)=2π * ρ0 * h * (h + r) * (-e^(-r/h)) + c
I find c by M(0)=c
Writing the LaTeX version of that, we have: ##\displaystyle \quad M(r)=2\pi \rho_0 (h^2 + rh) (-e^{-r/h}) + C##

Giving that ##\displaystyle \quad M(0)=C-2\pi \rho_0 h^2 ##.

Setting this to zero gives ##\displaystyle \quad 2\pi \rho_0 h^2 =C ##

Substitute that back into the expression for ##M(r)##, and after a little rearranging, you get:

##\displaystyle \quad M(r)=2\pi \rho_0 \left(h^2(1-e^{-r/h}) - rh\,e^{-r/h} \right)##
 
  • #26
SammyS said:
Yes:
You can use the indefinite integral to obtain the correct expression for the cumulative mass, M(r), of the disk. However, it makes sense that M(0) = 0 . That does not give you a value of 0 for the integration constant, C.

In Post#17, you had:

Writing the LaTeX version of that, we have: ##\displaystyle \quad M(r)=2\pi \rho_0 (h^2 + rh) (-e^{-r/h}) + C##

Giving that ##\displaystyle \quad M(0)=C-2\pi \rho_0 h^2 ##.

Setting this to zero gives ##\displaystyle \quad 2\pi \rho_0 h^2 =C ##

Substitute that back into the expression for ##M(r)##, and after a little rearranging, you get:

##\displaystyle \quad M(r)=2\pi \rho_0 \left(h^2(1-e^{-r/h}) - rh\,e^{-r/h} \right)##
I was wrong, I find this correct. Thank you very much
 
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