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Surface area - Double integrals

  1. Jan 22, 2015 #1

    etf

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    Hi!
    Here is my task:
    Calculate surface area of sphere $$x^{2}+y^{2}+z^{2}=16$$ between $$z=2$$ and $$z=-2\sqrt{3}$$.
    Here are 3D graphs of our surfaces:

    1.jpg

    2.jpg

    Surface area of interest is P3. It would be P-(P1+P2), where P is surface area of whole sphere. Is it correct?
    Here is how I calculated P1:

    33.jpg

    Similarly for P2:

    22.jpg

    We calculate P using formula 4pi*r*r.
    Is it correct?
     
  2. jcsd
  3. Jan 22, 2015 #2

    SteamKing

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    Can't really decipher your answer. It looks like 2πscribble.
     
  4. Jan 22, 2015 #3

    etf

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    P1=8*pi*ln2, P2=2*pi*ln(16/9), P=4*pi*16
    P=4*pi*16-(8*pi*ln2+2*pi*ln(16/9))=201.0619-21.0358=180.0261
     
  5. Jan 22, 2015 #4

    SteamKing

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    You can check your result using the formula in this article:

    http://en.wikipedia.org/wiki/Spherical_segment

    It seems there is a disagreement somewhere in your calculations between the formula and your result.
     
  6. Jan 22, 2015 #5

    LCKurtz

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    etf, have you given any consideration to working the problem in spherical coordinates? You can calculate it directly in about two steps.
     
  7. Jan 22, 2015 #6

    etf

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    I'm not familiar with spherical coordinates :(
     
  8. Jan 22, 2015 #7

    SteamKing

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  9. Jan 23, 2015 #8

    Zondrina

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    Doing multiple integrals and not being familiar with integral transforms is not a good thing. Polar, spherical and cylindrical co-ordinates are important to be able to integrate properly, and without them, you wont get very far.

    An interesting way to do this problem is to describe each face of the surface and then perform a surface integral over each one:

    $$\iint_S \space dS = \iint_{S_1} \space dS_1 + \iint_{S_2} \space dS_2 + \iint_{S_3} \space dS_3$$

    So you have a flat top surface ##S_1## described by the plane ##z = 2## and bounded by the region ##D_1## given by ##x^2 + y^2 = 12##. You also have a flat bottom surface ##S_2## describe by the plane ##z = -2 \sqrt{3}## and bounded by the region ##D_2##. Finally, you have the spherical middle section bounded by the planes.
     
  10. Jan 23, 2015 #9

    SteamKing

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    I think the OP was supposed to calculate only the area of the spherical surface lying between the planes z = -2√3 and z = 2. I don't think the areas of S1 or S2 were ever intended to be included.
     
  11. Jan 23, 2015 #10

    Zondrina

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    Sorry, I saw the problem a little differently when I first read it this morning. The sphere with its ends cut off seems to be more appropriate, so the ##dS_3## integral would be the only thing required. I believe the conventional method of evaluation then proceeds as usual:

    $$\iint_S \space dS = \iint_D \sqrt{z_x^2 + z_y^2 + 1} \space dA$$

    The two planes and sphere give two circles in the x-y plane from which the limits can be deduced.
     
  12. Jan 23, 2015 #11

    LCKurtz

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    That is certainly not the "natural" method to use. The surface is not symmetric vertically so you would have to work the top and bottom portions separately. And rectangular coordinates also are not the natural choice.
     
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