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Surface Area of a Multivariable function

  1. Jun 1, 2012 #1
    I don't know how to calculate the surface area after setting everything up. I have tried both MAPLE 15 program and wolfram alpha, but I can't find the answer.

    I have the function:
    f(x,y)= x*(y^2)*e^-((x^2+y^2)/4)

    The form I found for surface are was the square root of (sum of squares of partials with respect to x and y).
    The work I have so far
    ∫∫√(e^(-(x^2+y^2)/4))*(y^4-(3x^2*y^4)+(4y^2*x^2)+((x^4*y^4)/4)+((x^2*y^6)/4)+1 dx dy
    I have to integrate from -3 to 3 for both x and y.
    I have tried putting it into MAPLE 15, but it won't solve the function for a value.
    What should I do? Is there any way to simplify this algebraically so that it is easier to solve? Is it even possible to solve it by hand?
     
  2. jcsd
  3. Jun 1, 2012 #2
    try spherical/cylindrical coordinates
     
  4. Jun 1, 2012 #3
    What would the limits of integration be though? It is not shaped anything like a cylinder or sphere.
     
  5. Jun 1, 2012 #4

    Ray Vickson

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    I doubt that there is a closed-form formula for the area, so try a numerical approach. If you call dA your integrand, the following works for me in Maple 11:
    Jx:=evalf(Int(dA,y=-3..3));
    Area:=evalf(Int(Jx,x=-3..3));
    Area := 44.43229369

    RGV
     
  6. Jun 1, 2012 #5
    [STRIKE]Isn't it an odd function in the appropriate sense? You do not have to do the integration, you can realize the value "by inspection".[/STRIKE]

    Ray Vickson has pointed out below that I made a (large) oversight.
     
    Last edited: Jun 2, 2012
  7. Jun 1, 2012 #6

    Ray Vickson

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    I think you have misunderstood the problem. For a graph of the form [itex] z = f(x,y)[/itex] the surface area S of the graph over a region [itex] A \subset R^2 [/itex] in (x,y)-space is
    [tex] S = \int\int_{A} \sqrt{1 + f_x^2 + f_y^2} \: dx \, dy,[/tex]
    where [itex] f_x = \partial f/\partial x,\; f_y = \partial f/\partial y.[/itex] For [tex]f = x y^2 \exp{\left(-\frac{x^2 + y^2}{4}\right)} [/tex] we have
    [tex] f_x = \frac{1}{2}y^2 (x^2-2) \exp{\left(-\frac{x^2 + y^2}{4}\right)}\\
    f_y = \frac{1}{2}x y (y^2-4) \exp{\left(-\frac{x^2 + y^2}{4}\right)}.[/tex]
    Do you really think the double integral S can be evaluated by inspection?

    RGV
     
  8. Jun 2, 2012 #7
    Yes, that was my mistake, thank you Ray!
     
  9. Jun 2, 2012 #8
    Try parametrising it in some other way.
     
  10. Jun 2, 2012 #9
    [STRIKE]So dimension10, are you thinking some thing like polar coords., r goes from 0 to 3 cos theta?[/STRIKE]

    Oh yeah sorry, I'm not reading very carefully.
     
    Last edited: Jun 2, 2012
  11. Jun 2, 2012 #10
    [Post cleared]
     
    Last edited: Jun 2, 2012
  12. Jun 2, 2012 #11
    never mind, deleted
     
  13. Jun 2, 2012 #12
    how would i parametrize this function?
     
  14. Jun 2, 2012 #13
    I'm not sure but I would suggest a way through which you can use some algebraic/trigonometric identities OR reverse chain rule/reverse product rule.
     
  15. Jun 3, 2012 #14

    Ray Vickson

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    No matter what you do the function will not have a closed-form double integral. The only reason to bother parametrizing is to make it easier to fit the shape of the region [itex] A \subset R^2[/itex] over which you want to take the surface area. So, if A is rectangular with sides parallel to the x and y axes, just use the original parametrization. If A is a circular disc, you could switch to polar coordinates. If A is an elliptical disc, re-scale x and y to make it a circle, then switch to polar coordinates. In every case you will be forced to get a numerical answer.

    RGV
     
  16. Jun 3, 2012 #15
  17. Jun 3, 2012 #16
    with your region, in wolfram alpha, I think you need to put the 1 outside the parenthesis in the square root
     
  18. Jun 3, 2012 #17

    Ray Vickson

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  19. Jun 3, 2012 #18
    observe some of the symmetries, even along y, odd along x. so your surface area would basically be 4x (integral over region of [0,3]^2)
     
  20. Jun 3, 2012 #19
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