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!

Surface integral - find mass

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the mass of [itex]z= \sqrt{x^{2}+y^{2}}[/itex] when 1 ≤ z ≤ 4.
    The density function is ρ(x,y,z) = 10 - z


    2. Relevant equations



    3. The attempt at a solution

    [itex]\int\int_{s} ρ dS[/itex]

    [itex]S = <x, y, \sqrt{x^{2}+y^{2}} > [/itex]


    therefore dS = [itex]< \frac{-x}{\sqrt{x^{2}+y^{2}}} , \frac{-y}{\sqrt{x^{2}+y^{2}}}, 1 >[/itex]

    This is where I get sort of lost. Skipping a few steps I paramterized the function into polar coordinates and ended up with this integral

    [itex]\int^{2pi}_{0}\int^{2}_{1} 10 \sqrt{1+r^2} - r\sqrt{1+r^2}drd\theta[/itex]


    This just doesn't seem right, what am I missing?
     
  2. jcsd
  3. Apr 13, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The mass of a surface would be 0 - unless the density function is a surface mass density. Is that the case?
    (surfaces have no volume)

    Did you use cylindrical-polar coordinates?
    I'd have worked out the surface element in those coordinates...
    Why not divide the surface into rings radius r, width dw (measured along the surface)

    ##dm = 2\pi r \rho(z) dw##

    Since you know how r and z are related, you should be able to express dw in terms of dz.
     
  4. Apr 13, 2014 #3
    I guess it is. The answer is supposed to be 108pi sqrt(2).
    Yes, I used cylindrical coords.

    That was exactly what I was attempting to do - however - my integral turned out to be unsolvable so I must have gone wrong somewhere. I can't figure it out.
     
  5. Apr 13, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    'Mass' implies volume, which is why using a surface integral is puzzling, unless you plan to use one of the vector integral theorems to reduce a volume integral to a surface integral.
     
  6. Apr 13, 2014 #5

    BruceW

    User Avatar
    Homework Helper

    err... OK, so we're talking about a problem in 2D. So really, the 'density function' is a mass per area. It is better to think of it as ##\rho(x,y) = 10 - \sqrt{x^{2}+y^{2}}## So that it is clearly a 2D problem, not a 3D problem.

    They have already given you one of the equations for a change of coordinates: ##z=\sqrt{x^{2}+y^{2}}## So now, you need to decide what you want your other coordinate to be, for your new coordinate system? (And you already mentioned polar coordinates, I think that is a good direction to head towards).

    really? I am puzzled where the sqrt(2) could have come from...
     
  7. Apr 13, 2014 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think I know:
    Going radially along the surface is to go along the hypotenuse of a 1-1-√2 triangle.
    You can use that to work out dS.
     
  8. Apr 13, 2014 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Lets see, your integral was:
    $$\int_0^{2\pi} d\theta\int_1^2 10\sqrt{1+r^2}+r\sqrt{1+r^2}\; dr$$... which looks doable to me: the theta integral is easy - it is the r one that is tricky...
    ... proceed by making the substitution ##r=\tan u##.
    ... I think it ends up as: $$\qquad = 2\pi\left[5x\sqrt{1+x^2}+5\sinh^{-1} x+\frac{1}{3}(1+x^2)^{3/2}\right]_1^2$$

    ... but that does not look like it will get you the expected answer.

    I suspected at the start that your surface element dS was not right, and tried to show you another way to compute it.
    You don't seem to want to do it that way. Oh well.
     
    Last edited: Apr 13, 2014
  9. Apr 13, 2014 #8

    BruceW

    User Avatar
    Homework Helper

    OK, I have no idea what the problem statement is. I'll leave this problem to you and the OP'er :)
     
  10. Apr 13, 2014 #9
    I'm open for any suggestions. I just thought you were suggesting what I was already doing. Hmm my dS is wrong?

    I just re-did it and can't seem to find anything wrong with it.

    [itex]

    S_{x} = < 1 , 0 , \frac{x}{\sqrt{x^{2}+y^{2}}} >

    S_{y} = < 0 , 1 , \frac{y}{\sqrt{x^{2}+y^{2}}} >

    [/itex]


    Taking the cross product:

    [itex]S = < - \frac{x}{\sqrt{x^{2}+y^{2}}}, - \frac{y}{\sqrt{x^{2}+y^{2}}}, 1 > [/itex]

    ps. the reason why I'm doing it this way is because I have test on surface integrals coming up next week and my professor specifically let us know that there will be a mass problem on the test. This problem is out of my book (7th edition Stewart, Calculus, California edition) in the chapter for surface integrals. For the test, I need to figure out how to do this as a surface integral.
     
  11. Apr 13, 2014 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember that the scalar surface element dS is not a vector. What you want is$$
    dS = |S_x \times S_y|dxdy$$You will find that magnitude is your ##\sqrt 2##. So set up your integral$$
    \iint_R \rho dS$$and change it to polar coordinates. It's easy.
     
  12. Apr 13, 2014 #11
    Thanks! I'm getting closer but still not quite there.

    So I end up with
    [itex]\sqrt{2}\int^{2\pi}_{0} \int^{2}_{1} 10r - r^{2} dr d\theta = 76\pi\sqrt{2}[/itex]



    Which is off by a little bit.
     
  13. Apr 13, 2014 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##z=r## on the surface and ##z## goes from ##1## to ##4##.
     
  14. Apr 13, 2014 #13
    Argh I took thought the radius varied from 1 to 2, but you're right. It varies from 1 to 4. Thank you very much!
     
  15. Apr 13, 2014 #14

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I am puzzled as to where you got the idea that this is a surface integral.

    The problem, you say, was to "Find the mass of [itex]z=\sqrt{x^2+y^2}[/itex] when 1 ≤ z ≤ 4.
    The density function is ρ(x,y,z) = 10 - z".

    There is nothing said about a "surface" there.

    That's a solid, a truncated cone. In cylindrical coordinates, [itex]z= \sqrt{r^2}= r[/itex] so the (3D) mass would be given by
    [tex]\int_{z= 1}^4\int_{r= 0}^z\int_{\theta= 0}^{2\pi} (10- z) rd\theta dr dz[/tex]
     
  16. Apr 13, 2014 #15

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [itex]z=\sqrt{x^2+y^2}[/itex] looks like the equation of a surface to me.
     
  17. Apr 13, 2014 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I thought it was supposed to be a volume integral too, particularly because of the use of ##\rho##, which usually denoted a volume density, as opposed to ##\sigma##. It would have helped to see the original problem statement, which is:

    From this, you can see the calculation is intended to be a surface integral (not to mention the fact that, as the OP noted, the problem came out of the section on surface integrals).
     
  18. Apr 13, 2014 #17

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    "mass" is sometimes used by mathematicians to mean something different from what physicists do.
    Think of it as charge instead - and you are given the surface charge density.
    Or maybe a uniform thickness sheet of something.
    [edit] AH - as in the original problem statement...

    Whatever: treating the whole thing as a surface integral gives the model answer.
    OP, indeed, has that part correct, even if we cringe somewhat at the wording of the problem.


    I think we've come far enough to do a bit of a walk-through.

    The surface is, indeed, a cone - with a right-angle at the apex.
    (treating z as the height above the x-y plane.)

    Since the density function is constant for constant radius, and so is z, we can take rings of area dS at radius r for our surface element. dm=ρ(r).dS would be the mass of the ring at radius r; and we can add up the dm's to find the overall mass.

    There is very little substitute for just using geometry instead of trying to remember which formula to use. This is where sketching the surface really pays off.

    The area of the ring on the surface between r and r+dr is 2πr.dw where dw is the length along the surface from r to r+dr. i.e. it is the distance between points (r,θ,z(r)) and (r+dr,θ,z(r+dr))

    Since z=r, dw=dr√2=dz√2.

    (I like to put everything in terms of z.)

    So ##dS=2\pi\sqrt{2}z\; dz##

    So ##dm=2\pi\sqrt{2}(10-z)z\; dz##

    Remains only to add up all the individual dm's: $$m=2\pi\sqrt{2}\int_1^4 (10-z)z\; dz$$ ... and you can take it from there.

    (Caveat: i have been known do mess up: so check - do not copy blindly.)
     
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: Surface integral - find mass
  1. Mass of a surface (Replies: 2)

Loading...