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Surface Area

  1. Apr 7, 2005 #1
    Area/Volume Question

    Given the function on a 3D coordinate system:
    [tex] y = - z\sin \left( {xz} \right) [/tex] where [tex] \left| {x - \frac{\pi }
    {2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0 [/tex]

    How do find the surface area of the figure integrated from z=a to z=b (where 'a' and 'b' are constants) ?

    How would you also find the volume of this figure, from z=a to z=b (well, bounded by the top and by the trough, that is :blushing: ) ?
     
    Last edited: Apr 7, 2005
  2. jcsd
  3. Apr 7, 2005 #2

    dextercioby

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    That interval on "z" induces an interval on "x" (by the modulus inequality).So you have your limits to the Riemann version of the surface integral.
    U got the surface equation in explicit form

    [tex] y(z,x)=-z\sin xz [/tex]

    U must know the formula giving an area for a surface in [itex] \mathbb{R}^{3} [/itex] whose equation is given explicitely.

    I don't promiss the integration will be easy.

    Daniel.
     
  4. Apr 7, 2005 #3

    saltydog

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    Guys,

    I'm not sure, but it looks like this:

    [tex]
    \int_{z_a}^{z_b}\int_{\frac{\pi}{2}-\frac{\pi}{z}}^{\frac{\pi}{2}+\frac{\pi}{z}} \sqrt{z^4Cos^2(xz)+(zxCos(xz)+Sin(xz))^2+1}dxdz
    [/tex]

    Can someone check this please?

    Edit: The surface area I mean.
    Well, I guess I should explain to you how I arrived at this but I'll need to verify it before I do and perhaps others can provide additional input in the interim period . . .
     
    Last edited: Apr 7, 2005
  5. Apr 7, 2005 #4

    dextercioby

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    If it's really like that,then u can say good-bye from the integration.It's very elliptical...

    Daniel.
     
  6. Apr 7, 2005 #5

    saltydog

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    Ok, I think I'm wrong: The function f(x,z) oscillates above and below the x-z plane. The formula I'm using above works only for a surface above the x-z plane. Requires further study . . .
     
    Last edited: Apr 7, 2005
  7. Apr 7, 2005 #6

    Hurkyl

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    Why does that matter? The surface area of y = f(x, z) and of y = 10^10^100 + f(x, z) is the same, gives the same formula, and the latter does not go below the x-z plane.
     
  8. Apr 7, 2005 #7
    [tex] \left| {x - \frac{\pi }{2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0 [/tex] constricts [tex] y(x,z) [/tex] to a half-cycle, but I don't think it will oscillate. I mean, this given interval means half-cycle is either above the xz-plane or below the xz-plane (no "oscillation" between), depending on the sign of 'z' (did that to simplify to a half-cycle)--but not on both sides (unless you widen the interval, which might, i think, make it more difficult with oscillation)
     
    Last edited: Apr 7, 2005
  9. Apr 7, 2005 #8

    saltydog

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    Ok, thanks. I was looking at the whole plot without restricting it to the area between the lower and upper curve in z. Sooooo . . . maybe I do have it right. I'll spend some time on it and will ultimately just NIntegrate it in Mathematica and report the results. Can you report your results also?

    I've included a plot of the boundary of the integration limits. I'll numerically integrate it from say z=1 to z=5 between the lower and upper curve and report the results.
     

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    Last edited: Apr 7, 2005
  10. Apr 7, 2005 #9
    I think followed your approach to solve it--find the dL (arc length) for the y-curve with respect to x (using dx), next multiply by dz to get dA, and then use a double integral to solve it. However, I'm not quire sure if my approach is correct :bugeye: ...(and thus the thread-!)

    (My goal here for this problem is to find a general form (for the area And volume), given any bivariate function y(x,z) in 3D with x-interval [f(x,z), g(x,z)], integrated across from z=a to z=b. I started out here with a specific case :shy: )
     
    Last edited: Apr 7, 2005
  11. Apr 7, 2005 #10

    saltydog

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    Well, I'm just using the standard theorem for calculating surface area; it should be in your Calculus book. The one I'm looking at now is Leithold, p. 995. Anyway, I got a surface area between z=1 and z=5 of 59.135. I must tell you I'm not confident of it withoug further study of the matter however.
     
  12. Apr 7, 2005 #11
    Neither am I :shy:
    I got same answer (well, first I wrote it wrong, but then I fixed it and got 59.1)
    Check out the 3D graph--it looks kind of neat
     
    Last edited: Apr 8, 2005
  13. Apr 9, 2005 #12
    Hmm---it seems the domain of my equation was wrong from the start!:surprised In addition, (as it is a periodic function), the whole problem can actually be simplified as
    [tex] y\left( {x,z} \right) = z\cos \left( {zx} \right),\left| {zx} \right| \leqslant \frac{\pi }{2} [/tex]
    *And then one finds the surface area generated from z=a to z=b (across a much nicer interval)
     
    Last edited: Apr 9, 2005
  14. Apr 9, 2005 #13

    saltydog

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    Ok, in that case I get:

    [tex]\int_{z_a}^{z_b}\int_{-\frac{\pi}{2z}}^{\frac{\pi}{2z}} \sqrt{z^4Cos^2(xz)+(Cos(xz)-zxSin(xz))^2+1}dxdz[/tex]

    For the interval z=1 to z=5, I get 26.49.
     
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