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Double & Surface Integrals

  1. Jul 3, 2006 #1
    Hi eveybody,
    I'm having trouble in surface integrals. I know already what the double integrals measure; a multivariable function ( drawing surface) over "a region of domain"..
    Now, the surface integrals are for surfaces given by 2-dependent parametrizations over " the surface".
    my questions are:
    1. What is the difference between the integrations of the surfaces in the 2 cases? in other words, what difference would it make if it was over the domain or the surface? aren't they the same and isn't the domain just the projection of the surface on the xy-plane?
    2. Do both measure the "height" of the surface from the xy-plane, but over different veiws of space?

    thanks alot in advance:smile:
  2. jcsd
  3. Jul 3, 2006 #2


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    Are you asking what the difference is between integrating over some parametrization of the surface vs projecting down to the xy plane and integrating over its "shadow"? Well the latter is just a special kind of parametrization, so there isn't really a difference. And, yes, in a way you are measuring the height of a surface above the xy plane, where this surface corresponds to the function you actually integrate, after finding the surface area element dA in terms of dxdy and multiplying the original function for the surface by this transformation.
    Last edited: Jul 3, 2006
  4. Jul 3, 2006 #3


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    A surface is two dimensional and so can be written in terms of two parameters. If I understand what you are saying, when you talk about an integration over a "region of domain". You are really just talking about the situation where the parameters are x and y themselves: z= f(x,y) as opposed to x= h(u,v), y= g(u,v), z= f(u,v). The first is just a special case of the second: x= x, y= y, z= f(x,y) is the same as x= u, y= v, z= f(u,v).
    When you ask, "Do both measure the "height" of the surface from the xy-plane" it's not clear what you mean by "both". z measures the height above the x,y plane but parametric equations give x, y, and z.

    One advantage of the parametric form is with surfaces where z is NOT a function of x and y. For example, to integrate a function over the surface of the unit sphere, [itex]x^2+ y^2+ z^2= 1[/itex] to write z= f(x,y), I would have to do top and bottom separately:
    [itex]z= \sqrt{1- x^2- y^2}[/itex] and [itex]z= -\sqrt{1- x^2- y^2}[/itex]

    However, I can also write [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], [itex]z= cos(\phi)[/itex] (in other words, use spherical coordinates with [itex]\rho[/itex] set to 1). That will cover the entire sphere with [itex]0\le \theta \le 2\pi[/itex] and [itex]0\le \phi\le \pi[/itex]
  5. Jul 3, 2006 #4
    thanks a lot..it really helped me..
    just to make sure I got it, when the parametrization gives a plane not curved surface, then ,in this special case, the surface integral will equal its double integral since the "shadow" will be the same as the surface, and dS will have " the length of cross product of r u and r v" equal to the arc length in front of dA. Therefore, in this case, will both integrals give the same answer? ; whether you parametrize or write it as
    z=f(x,y) (x, y parameters), it is different ways to the same solution.

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