Integrate mass density, mass density in terms of r: rho(r) = (rho 0)*e^(-r/h)

[manudaren]

I want to find the cumulative mass m(r) of a mass disk. I have the mass density in terms of r, its 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.

 

[stapel]

I want to find the cumulative mass m(r) of a mass disk. I have the mass density in terms of r, its 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.

Please reply showing your steps and the solution that you have gotten. Thank you!

Eliz.

 

[manudaren]

Please reply showing your steps and the solution that you have gotten. Thank you!

Eliz.

I change to polar coordinates, so:

m(r) = 2π * ρ0 * ∫(r * e^(-r/h))dr

and then substitute u = r/h so that du/dr = 1/h and dr = h*du. I get:

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

 

[blamocur]

I change to polar coordinates, so:

m(r) = 2π * ρ0 * ∫(r * e^(-r/h))dr

and then substitute u = r/h so that du/dr = 1/h and dr = h*du. I get:

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

You have "m(r)" in the left-hand side, but the right-hand side does not depend on "r". Hint: figure out your integration limits.
Also, do you know how to integrate [imath]\int ue^{-u} du[/imath] ?

 

[manudaren]

You have "m(r)" in the left-hand side, but the right-hand side does not depend on "r". Hint: figure out your integration limits.
Also, do you know how to integrate [imath]\int ue^{-u} du[/imath] ?

Is it -e^(-u)*(1 + u) ?

 

[blamocur]

Is it -e^(-u)*(1 + u) ?

Do you know how to verify results of integration?

 

[manudaren]

Do you know how to verify results of integration?

No, I dont

 

[blamocur]

No, I dont

You differentiate the result.
I believe your result is correct, but it is a good idea to verify the result by yourself first.
What about the integration limits?

 

[manudaren]

You differentiate the result.
I believe your result is correct, but it is a good idea to verify the result by yourself first.
What about the integration limits?

I finally got

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 )

 

[blamocur]

I finally got

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 )

This is looks very confusing to me. Can you add some explanations for your steps? For example, why do you need "u=r". Where do you use "v" ?

Also, I'd strongly recommend using LaTeX to make your formulas more readable, both for your readers and for yourself.

 

[khansaheb]

why do you need "u=r". Where do you use "v"

I believe s/he is trying to use - integration by parts, in its standard format, where

\(\displaystyle \int {u \cdot dv} \ = u \cdot v - \int {v \cdot du} \ \)

 

[manudaren]

This is looks very confusing to me. Can you add some explanations for your steps? For example, why do you need "u=r". Where do you use "v" ?

Also, I'd strongly recommend using LaTeX to make your formulas more readable, both for your readers and for yourself.

I believe s/he is trying to use - integration by parts, in its standard format, where

\(\displaystyle \int {u \cdot dv} \ = u \cdot v - \int {v \cdot du} \ \)

Yes, I am