1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculate volume of body T between planes

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the volume of a body given by plane ##z=0## and ##z=1## and ##x^2+y^2+z^2=4##.



    2. Relevant equations
    ##detJ=r^2sin \theta## for spherical coordinates


    3. The attempt at a solution
    ##V=\iiint_{T}^{}dV=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dz\int_{\theta _{0}}^{\pi /2}r^2sin\theta d\theta ##
    ##V=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dzr^2\frac{1}{2}##
    ##V=2\pi r^2\frac{1}{2}=\pi r^2=4\pi ##

    Why is this wroooong? -.-

    one more thing: trying to integrate by dx,dy and dz (having some troubles with integral borders):
    ##V=\int_{0}^{1}dz\int_{-2}^{2}dy\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}}dx## or not?

    Thanks for all the help!
     
  2. jcsd
  3. Oct 24, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You seem to have a mix spherical and cylindrical coordinates. How are you defining r here? θ should not be going all way to π/2.
     
  4. Oct 24, 2013 #3
    am... the way I imagined this is that r=2 is constant and ##\theta ## angle between R and axis z like on this picture: http://phobos.ramapo.edu/~aferry/wk1principles/principle_images/form.png [Broken]

    Hmm, I guess what I need is r as function of z, where r would be radius of a circle at given z.
    So...if I am not wrong, this should than work perfectly?

    ##V=\iiint_{T}^{}dV=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dz\int_{0}^{\sqrt{R_0^{2}+z^2}}rdr ##

    Where ##R_0=2## is sphere radius.

    And if I a not wrong these are cylindrical coordinates?
     
    Last edited by a moderator: May 6, 2017
  5. Oct 24, 2013 #4
    Is it even possible to write this integral with spherical coordinates in one piece or only using the sum of two integrals?
     
  6. Oct 24, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, cylindrical. But you have a sign wrong in the upper bound for r.
    You could write it in one piece, but the limits would include functions like min{}, so it's effectively a sum or difference of two.
     
    Last edited by a moderator: May 6, 2017
  7. Oct 24, 2013 #6
    You could just use volume by washers up the z-axis:

    [tex]V=\int_0^1 \pi r^2 dz[/tex]

    and the radius, [itex]r=f(z)[/itex] since it's spherically-symmetric, is only a function of z.
     
  8. Oct 25, 2013 #7
    I have a similar problem again so I won't start new topic...
    Calculate the gravitational force between body S, which is defined as space between ##x^2+y^2\leq R## and ##0\leq z \leq R##, and point ##T(0,0,h)## where ##h>R##.


    2. Relevant equations
    ##F=G\frac{mM}{r^2}## where G is constant and r is the distance between tho bodies.


    3. The attempt at a solution
    Well, my idea was to calculate the gravitational force for a disc with radius R and for distance d above the disk and than sum (integrate) the discs from 0 to z.
    Let's try:
    ##dF=G\frac{mdM}{r^2}## where m is mass the point and M mass of the cylinder below the point, now here comes the very first problem because ##dM=\rho dV## but ##V(r,z)## so ##dM=\rho 2\pi rdrz + \rho \pi r^2dz##

    Now what? :/
     
  9. Oct 25, 2013 #8
    Or I could try

    ##dF=G\frac{mdM}{d^2}=G\frac{m\rho dV}{d^2}=Gm\rho\frac{dV}{d^2}=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{r}{d^2}dv## Where dv is thickness of the cylinder.
    I also have to take only the vertical component of the force since horizontal component is 0 due to symmetry, thereby
    ##F=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{rv}{d^3}dv## where ##d=\sqrt{v^2+r^2}##
    so

    ##F=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{rv}{(v^2+r^2)^{3/2}}dv## which now (If I calculated everything right) leads me to some suspicious logarithms:
    ##F=Gm\rho \pi \begin{bmatrix}
    ln(1+\frac{R^2}{(h-z)^2})-ln(1+\frac{R^2}{h^2})
    \end{bmatrix}##

    hmm, if ##z=0## the logarithms are the same therefore the force would be zero which is not what I expected... I more or less expected the force between a disk and point T for z=0...
     
  10. Oct 25, 2013 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You've lost me. I thought this was for a disc, so how does the mass of the cylinder come into it? If you meant to write 'mass of the disc', you can't reduce the disc to be equivalent to a point mass at its centre.
     
  11. Oct 25, 2013 #10
    Well m would be the be the mass of a body in ##T## while M is the mass of the entire cylinder. My idea was to sum up the gravitational force ##dF## in point ##T## which is ##v## above the disc, for all the discs with height ##dz##. All the discs together are than whole cylinder from ##0## to ##z##.

    So yes, what I meant was mass of the disc.
     
  12. Oct 25, 2013 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Then you need to integrate over the area of the disc. You don't need to consider a test mass at (0, 0, h), just ask what the field is at that point. What is the field at Cartesian (0, 0, h) due to a disc element of mass ##\rho rdr d\theta## at polar (r, θ) in the XY plane? What component of that field will be relevant to the integration?
     
  13. Oct 25, 2013 #12
    Am...
    ##F=Gm\int_{0}^{2\pi }d\theta \int_{0}^{R}\frac{r}{d^2}cos\varphi dr##
    ##F=Gmh\int_{0}^{2\pi }d\theta \int_{0}^{R}\frac{r}{(h^2+r^2)^{3/2}}##
    ##F=Gmh2\pi \int_{h^2}^{h^2+R^2}\frac{dt}{t^{3/2}}##
    ##F=-Gmh\pi \begin{bmatrix}
    \frac{1}{\sqrt{h^2+R^2}}-\frac{1}{h}\end{bmatrix}=Gmh\pi \begin{bmatrix}
    \frac{1}{h}-\frac{1}{\sqrt{h^2+R^2}}\end{bmatrix}##

    right?
     
  14. Oct 25, 2013 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think you have some factor of two errors. When substituting t, what is dt equal to?
    When you then integrated, looks like you multiplied by the -1/2 instead of dividing by it.
     
  15. Oct 25, 2013 #14
    Ah yes, that's right.

    ##dt=2rdr## and than when integrating multiplying by -2.
    All together now
    ##F=Gmh2\pi \begin{bmatrix}
    \frac{1}{h}-\frac{1}{\sqrt{h^2+R^2}}\end{bmatrix}## Sorry, I have exactly this on the paper yet...

    Now I have to integrate this again, for discs form 0 to z. How do I do that?
     
  16. Oct 25, 2013 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    First, you need to replace h with a function of z, right?
     
  17. Oct 26, 2013 #16
    THIS is the step I just couldn't remember!

    ##F=Gm2\pi \rho (1-\frac{h}{\sqrt{h^2+R^2}})## where now ##h=h_0-z## therfore ##dh=-dz##
    ##F=-Gm2\pi \rho \int_{0}^{z}(1-\frac{h_0-z}{\sqrt{(h_0-z)^2+R^2}})dz##
    ##F=-Gm2\pi \rho \begin{bmatrix}
    \int_{0}^{z}dz-\int_{0}^{z}\frac{h_0-z}{\sqrt{(h_0-z)^2+R^2}}dz
    \end{bmatrix}##
    ##F=-Gm2\pi \rho \begin{bmatrix}
    z+\frac{1}{2}\int_{h_0^2+R^2}^{(h_0-z)^2+R^2}t^{-1/2}dt
    \end{bmatrix}##
    ##F=-Gm2\pi \rho \begin{bmatrix}
    z+\sqrt{(h_0-z)^2+R^2}-\sqrt{h_0^2+R^2}
    \end{bmatrix}##
    or
    ##F=Gm2\pi \rho \begin{bmatrix}
    -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2}
    \end{bmatrix}##

    Now second part is ##\lim_{R->0}F(R)=\lim_{R->0}Gm2\pi \rho \begin{bmatrix}
    -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2}
    \end{bmatrix}##
    ##\lim_{R->0}F(R)=Gm2\pi \rho \begin{bmatrix}
    -z-\sqrt{(h_0-z)^2}+\sqrt{h_0^2}
    \end{bmatrix}=0##

    Again, not something I expected. What I would expect is the same force we get over a "stick" Since my english is not very good I tried to draw what I expected (in attachment).
    In that case F should be ##F=Gm\rho (\frac{1}{h-z}-\frac{1}{h})##


    or....not?
     

    Attached Files:

  18. Oct 26, 2013 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The ρ before you take the limit is a volume density. Taking the limit turns the total mass to 0. Replace ρ with M/πR2 first.
     
  19. Oct 26, 2013 #18
    You mean ##\rho = \frac{M}{\pi R^2z}## since we are talking about a cylinder here. Right?

    ##\lim_{R->0}Gm2\pi \rho \begin{bmatrix}
    -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2}
    \end{bmatrix}##
    ##\lim_{R->0}\frac{2GmM}{zR^2}\begin{bmatrix}
    -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2}
    \end{bmatrix}=\lim_{R->0}\frac{2GmM}{z}\begin{bmatrix}
    -\frac{z}{R^2}-\frac{\sqrt{(h_0-z)^2+R^2}}{R^2}+\frac{\sqrt{h_0^2+R^2}}{R^2}
    \end{bmatrix}##

    ammm... Now if R -> infinity, the limit is 0 which is perfect, yet if R -> 0... than... Am? I kind of forgot how to deal with limits such as this one (konst/0)... L hospital doesn't help much.
     
  20. Oct 26, 2013 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You need to expand the square roots using the Taylor expansion/binomial theorem for small R.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculate volume of body T between planes
Loading...