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Differential equation for changing mass of a sphere

  1. Sep 13, 2013 #1
    The mass of a sphere with density as a function of radius is

    [itex]M=4\pi \int_0^r\rho(r) r^2dr[/itex]

    Lets say the radius increases and decreases as a function of time t. So:

    [itex]M(t)=4\pi \int_{0}^{r(t)}\rho (r) r(t)^2dr[/itex]

    I want to know the basic equation describing the mass added or removed from the sphere (mass increases when radius increases, mass decreases when radius decreases) as a function of t, starting from any t. The problem is I think I must use a differential form but I'm not sure what it looks like. What then is the differential form of dM(t)/dt? I think I must use a chain rule and write:


    is this right? How do I proceed to solve this with the integral?
  2. jcsd
  3. Sep 13, 2013 #2
    Hi !
    I am afraid that there is something wrong in your writting. It should be :
    [itex]M=4\pi \int_0^r\rho(u) u^2du[/itex]
    Lets say the radius increases and decreases as a function of time t. So:
    [itex]M(t)=4\pi \int_{0}^{r(t)}\rho (u) u^2du[/itex]
    You may use any other symbol than u, but not r.
  4. Sep 13, 2013 #3
    Ah yes, sure, the upper limit is the 'full' radius (r) and u is a radius. This doesn't solve my problem though.
  5. Sep 13, 2013 #4


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    Yes, the chain rule: [itex]\dfrac{dM}{dt}= \dfrac{dM}{dr}\dfrac{dr}{dt}[/itex]
    JJaquelines point helps make sense of the dM/dr.

    To find [itex]\dfrac{dM}{dt}[/itex] use the "fundamental theorem of Calculus":
    [tex]\frac{d}{dr}\int_0^r \rho(u)u^2 du= \rho(r)r^2[/tex]
  6. Sep 13, 2013 #5
    I guess the solution then is

    [itex]\frac{dM}{dt}=\frac{4}{3}\pi \rho(r)r^3\frac{dr}{dt}[/itex]

    does the solution change if [itex]\rho(r)[/itex] becomes [itex]\rho(r,t)[/itex] or could I write

    [itex]\frac{dM}{dt}=\frac{4}{3}\pi \rho(r,t)r^3\frac{dr}{dt}[/itex]

    Last edited: Sep 13, 2013
  7. Sep 13, 2013 #6
    I don't think so. A term involving the partial derivatine of rho relatively to t is missing into your last equation,
  8. Sep 13, 2013 #7
    Is there a rule I can apply to get this additional term?
  9. Sep 13, 2013 #8


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    If you have
    F(t) = \int_0^{a(t)} g(r,t)\,\mathrm{d}r
    [tex]F'(t) = \int_0^{a(t)} \left.\frac{\partial g}{\partial t}\right|_{(r,t)}\,\mathrm{d}r + a'(t)g(a(t),t)[/tex]
    assuming [itex]g[/itex] is sufficiently smooth.
  10. Sep 13, 2013 #9


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    Dearly Missed

    Please do not mix together dummy variables in the integrand with integral limits.
    Properly speaking, you have the following the relation:
    Thus, you have:
    which has as interpretation that only the outermost spherical shell at r(t) determines the total change of mass.

    Every compact ball strictly contained within the outermost shell (radii less than r(t)) remains constant in mass.

    Suppose you have a ball where at different times, the density at some fixed radius "x" may change as a function of time. Then, you have:
    In this case, the total mass of the ball will be due to two distinct effects:
    1. The ball shrinks or expands. This gives the contribution given above.
    2. The interior of the ball may change its mass. This effect is new.

    In sum, you'll then get:
    Last edited: Sep 13, 2013
  11. Sep 14, 2013 #10
    The general formula below shows the rule for derivation :

    Attached Files:

  12. Sep 14, 2013 #11
    Of course, the chain rule continues to applies !!!
    The formula given above is the application of the chain rule in case of an integral with the integrand and limits which are functions of the variable considered for derivation.
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