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Area of surface

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the area of the surface ##x^2+y^2+z^2 = R^2 , z \ge h , 0 \le h \le R##

    2. Relevant equations
    ##A(S_D) = \iint_D |\mathbf r'_s \times \mathbf r'_t|dsdt##
    where ##S_D## is the surface over ##D##.

    3. The attempt at a solution
    We write the surface in parametric form using spherical coordinates
    ##\mathbf r(\theta ,\phi) = R(\sin \theta \cos \phi , \sin \theta \sin \phi , \cos \theta)##
    which gives us
    ##\mathbf r'_\theta = R(\cos \theta \cos \phi , \cos \theta \sin \phi , -\sin \theta)##
    and
    ##\mathbf r'_\phi = R(-\sin \theta \sin \phi , \sin \theta \cos \phi , 0 )##
    so we end up with
    ##|\mathbf r'_\theta \times \mathbf r'_\phi | = R^2|\sin \theta | |(\sin \theta \cos \phi , \sin \theta , \sin \theta \sin \phi , \cos \theta ) | = R^2|\sin \theta |##
    and the area
    ##A(S_D) = \iint _D R^2|\sin \theta | = \int _?^? \int _0^{2\pi } R^2|\sin \theta | d\phi d? = 2\pi \int _?^? R^2|\sin \theta | d? ##
    So my problem is i don't know what im supposed to integrate over. I suppose it should be ##\theta## but I'm not sure between what limits. Or even if the switch to spherical coordinates was a good idea at all.

    The answer according to the book should be ##A(S_D) = 2\pi R(R-h)##
    which makes me think i made another error somewhere else as well since I got an excess of ##R##.

    Appreciate any help. Cheers!
     
  2. jcsd
  3. Feb 24, 2015 #2

    LCKurtz

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    OK, you are using ##\theta## as the angle from the ##z## axis, and yes, spherical coordinates is what you want. Draw a cross section of your sphere of radius ##R## with a horizontal line at ##z=h##. The triangle that forms should give you the limits for ##\theta##.
     
  4. Feb 25, 2015 #3

    HallsofIvy

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    The only two variables you have are "[itex]\phi[/itex]" and "[itex]\theta[/itex]" and you have already integrated with respect to [itex]\phi[/itex] so the only possible remaining variable is [itex]\theta[/itex]. [itex]\theta[/itex] is the angle a line from (0, 0, R) to (x, y, z) makes with the z-axis. When (x, y, z)= (0, 0, R), [itex]\theta= 0[/itex]. When [itex]z= R cos(\theta)= h[/itex], [itex]cos(\theta)= \frac{h}{R}[/itex] so [itex]\theta= cos^{-1}\left(\frac{h}{R}\right)[/itex].

     
  5. Feb 25, 2015 #4
    Thanks for the help both of you! I got it from Kurtz advice earlier but couldn't respond before now. The confusion with if i should integrate over ##\theta ## was more that i couldn't find the limits and thought that perhaps spherical was a bad idea and i should use the radius somehow.

    Gonna add the rest of the solution in case anyone else come upon the thread.
    From the attachment pdf we can see that ##cos(\theta)= \frac{h}{R}## so therefore ##0 \le \theta \le \arccos (\frac{h}{R} ) ##

    ##A = 2R^2\pi \int_0^{\arccos \frac{h}{R}} \sin \theta d\theta = 2R^2\pi ( -\frac{h}{R} + 1) = 2R\pi(R-h)##
     

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