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Areas, surfaces, volumes

  1. Aug 7, 2007 #1
    hi how can the following be proved using integral methods:

    a) prove surface area of sphere, radius a, is 4 [tex]\pi a^2[/tex]
    b) prove area of a disk, radius a, is [tex]\pi a^2[/tex]
    c) prove volume of ball, radius a, is [tex]\frac{4}{3} \pi a^3[/tex]
    d) prove volume of axisymmetric cone of height h and base with radius a, is [tex]\frac{1}{3}\pi a^2 h[/tex]

    .................................................. ...........
    i think my working of (a) is correct:

    working of (a):
    use spherical co-ordinates:
    |S| = [tex]\int\int_{D} ||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| dA[/tex]

    [tex]||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi[/tex]

    [tex]|S| = \int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta = \int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta = \int^{2\pi}_{0} 2a^2 d\theta = 4\pi a^2 [/tex]

    how can i do the rest please. and what integration methods should I be using for each? thnx xxxx
  2. jcsd
  3. Aug 7, 2007 #2


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    You have done the hardest one (a surface area) first in a way that doesn't make it clear how to do the others. Let's start with the second one. The area of a disk is the integral of the (circumference of a circle of radius r)*dr over r ranging from 0 to a. Can you draw a geometric picture to show you why this is so? For the last two divide the ball and the cone into disks.
  4. Aug 8, 2007 #3


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    And when you've done all that, you may try to divide the area of the sphere into disks and do it similarly to the other three :)
  5. Aug 8, 2007 #4
    are we talkinga bout using washer's method here?

    just wanted to ask. would washer's method actually be considered a "proof" though? or is it simply just a formula?
  6. Aug 8, 2007 #5
    i wanted to acutally know. is there a way to solve them using integral methods such as divergence, or greens, or stoke's??

    or simply using spherical/polar/cartesian co-ordinates??

    for example, i worked out (A) using spherical co-ordinates
    where r = a

    so then i used the parameter [tex]r(\theta,\phi) = (asin\phi cos\theta)i + (asin\phi sin\theta)j + (2cos\phi)k[/tex]

    now cross product of [tex]\frac{dr}{d\theta}X \frac {dr}{d\phi} = -(a^2sin^2\phi cos\theta)i - (a^2sin^2\phi sin\theta)j - (a^2sin\phi cos\phi)k[/tex]

    the magnitude is therefore:

    [tex]||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi[/tex]

    [tex]|S| = \int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta = \int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta = \int^{2\pi}_{0} 2a^2 d\theta = 4\pi a^2 [/tex]

    so that is how i managed to work out the surface area of the sphere.

    do you have a similiar simple way of solving the others instead of washer's method please?
  7. Aug 8, 2007 #6


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    That is NOT what is normally meant by "integral methods" for finding area and volume! But you certainly can use Stokes' or Green's or the divergence area: A volume is normally a triple integral. Which of those theorems equates a triple integral to on the boundary? What do you get if you apply that theorem to [itex]\int\int\int dxdydz[/itex]?
  8. Aug 8, 2007 #7

    are you referring to the divergence theorem..

    if so, could you show me an example from b,c, or d on how to solve it, so that i can solve the rest :)

  9. Aug 8, 2007 #8
    hi i managed to get the one for (b):
    using Green's theorem : [itex]\frac{1}{2}\oint_{C}[xdy-ydx][/itex]

    [itex]c(t)=(rcos(t), rsin(t))[/itex]



    so we can deduce:



    but i don't know how to start on with (C) and (D).. any help would be greatly appreciated.
  10. Aug 8, 2007 #9


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    Persistant, aren't you? :) Ok, you can do (C) by realizing that the exterior derivative of the form w=xdydz is dw=dxdydz which is the volume form. So integrating dw over the interior of the sphere (volume) is the same as integrating w over the boundary. Do it in spherical coordinates. Same basic idea for (D). I think I'd really prefer using washers...
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