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Simple surface integration blues {again}

  1. Jan 3, 2005 #1
    Hi there,

    I'm still reading on about surface integrals and stuff, and I've arrived at a formula that goes like this {based on plane projection methods} :

    [tex] \iint_S G(x,y,z) dS = \iint_R G(x,y,z) \sqrt{1 + \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}}[/tex]

    So this is the question at hand :

    Evaluate the surface integral : [itex] \iint_S G(x,y,z) dS [/itex]

    where S is the portion of the plane [itex] x + y + z = 1 [/itex] in the first octant, and [itex] G(x,y,z) = z [/itex]

    So, finding the partial derivatives etc.. i arrive at :

    [tex]\sqrt{3}\int_0^1 \int_0^1 z dx dy[/tex] which is equivalent to : [tex]\sqrt{3}\int_0^1 \int_0^1 (1 - x - y) dx dy[/tex]...I chose the limits for dx and dy as so.....

    x + y + z = 1.... in the x plane... y+z = 0 thus x = 1, y = 0, and the same reasoning for y...

    Computing this i get :

    [tex]\sqrt{3}\int_0^1 (1 - \frac{1}{2} - y) dy = \sqrt{3}\left[ y - \frac{y}{2} - \frac{y^2}{2} \right]_{0}^{1} = 0[/tex] if im not mistaken..

    The official solution however is [tex] \frac{\sqrt{3}}{6}[/tex]

    Help :( :(
  2. jcsd
  3. Jan 3, 2005 #2


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    Well, actually, that should be
    [tex] \iint_S G(x,y,z) dS = \iint_R G(x,y,z) \sqrt{1 + \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}}dx dy[/tex]

    Here's how I like to do problems like this: the surface is given by x+ y+ z= 1 and we can think of that as a "level surface" of f(x,y,z)= x+ y+ z. The gradient of f: i+ j+ k is normal to that and its length is the differential of area. I like to think of it as the "vector differential of area" . In this particular case, the differential of area, projected onto the xy-plane, is
    [itex]\sqrt{3}dxdy[/itex], just as you have.

    HOWEVER, you have calculated the limits of integration wrong. The plane x+ y+ z= 1, in the first quadrant, is a triangle with vertices at (1,0,0), (0,1,0), and (0,0,1). Projecting down to the xy-plane, we get a triangle with vertices at (0,0), (1,0), and (0,1). Of course, the sides of that triangle are the x-axis, the y-axis, and the line x+ y= 1.

    If we choose to order the integrals with the x-integral outside, then x ranges from 0 to 1, just as you say. BUT, FOR EACH X, y ranges from 0 up to y= 1-x, the slant line. Your integral should be
    [tex]\sqrt{3}\int_{x=0}^{1}\int_{y= 0}^{1-x}(1-x-y)dydx[/tex]

    Integrate that and see what you get.
    Last edited by a moderator: Jan 3, 2005
  4. Jan 3, 2005 #3
    Saviour...I suppose I should polish up on the theory of iterated integrals...The book I currently have contains alot of mathematical language, such of that a mere engineering student finds a bit much on first glance.. Could someone just brief me up on the theory of these iterated integrals in which a limit is a function.. I know its analogous to partial derivatives, but thats as far as i go :D

    Cheers guys, another triumph!! :D:D

    Edit : Just to get the ball rolling, the next question is of the same form; but :

    [tex] G(x,y,z) = \frac{1}{1 + 4(x^2 + y^2)}[/tex] and [tex] z = x^2 + y^2 [/tex]
    from Z = 0 to Z = 1; If someone could just push me in the right direction, regarding limits and whether or not i should convert to polar coordinates.

    Thanks again :biggrin: :biggrin:
    Last edited: Jan 3, 2005
  5. Jan 3, 2005 #4


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    Yes,the surface [itex] z=x^{2}+y^{2} [/tex] is the of a rotation paraboloid round the Oz axis,with the lowest point:the origin of the cartesian coordinate system.It's projection on the Oxy plane is a disk of radius 1.
    Use polar coordinates in the Oxy plane.The limits of integration will be:
    [tex] 0\leq \rho\leq 1;0\leq \phi\leq 2\pi [/tex]
    The integrals won't look very pretty,though...

  6. Jan 3, 2005 #5
    I wish there was a real time chat facility :biggrin:
  7. Jan 4, 2005 #6
    :( this integral is pretty tricky, any takers?! I've got the numerical solution but my answers just wont agree...sigh :frown:
  8. Jan 4, 2005 #7


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    It's this "baby":
    [tex] \int_{0}^{1} d\rho\int_{0}^{+\pi} d\phi \frac{\sqrt{1+2\rho(\sin\phi+\cos\phi)}}{1+4\rho^{2}} [/tex]
    ,which i guess it cannot be expressed through elementary functions.At least the integral of the angle seems to be put in connection with a complete elliptic integral of the second kind of Legendre.

  9. Jan 4, 2005 #8
    Man that's mighty unfair...I arrived at the same integral as you've stated but evaluating it was somewhat nasty and led no where really.. Ah well, onto the next I suppose.. Thanks to everyone though, the aid is much appreciated.

    All the best!!
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