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What *is* a surface integral?

  1. Aug 5, 2015 #1

    B3NR4Y

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    I'm beyond multi-variable calculus, where this is taught, but I still don't know what the hell a surface integral is. I understand that [itex]d\sigma[/itex] is the surface element, and [itex] | \frac{\partial \vec{r}}{\partial u}du \times \frac{\partial \vec{r}}{\partial v}dv | = d\sigma = |\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} | \, dudv[/itex], and that cross product is equal to the area of the infinitesimally small parallelogram described by the two vectors, so summing across all of that (with double integrals yields a surface area). But when you have a function in there like this [itex]\iint_\Sigma f d\sigma[/itex] I don't know what the heck this is supposed to represent. My intuition tells me it's a volume, because it's a function (height) times an area, but I don't think that's right.

    I know (or think I know) a line integral of the form [itex] \int_C f ds [/itex] is equivalent to the area under the function traced by the curve C. So the natural extension of this should be a surface integral equaling a volume.

    Basically I know how to calculate surface integrals, but I don't know what in the world I'm calculating.
     
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  3. Aug 5, 2015 #2

    jedishrfu

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  4. Aug 5, 2015 #3

    B3NR4Y

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    I've read that before, but I have just read it again. So a regular integral is integrating a function over a region that you specify with the limits, a surface integral is integrating a function over a surface. So like a triple integral of a density function is the mass of the solid, a surface integral would yield mass if you had a density function per unit area. So flux being the dot product of the vector field and the normal vector of the surface, you sum over the entire surface which is a surface integral, right?

    I guess what confuses me is the common picture showing the projection of the surface onto the plane, where does this come in? The projection is the region you integrate over for the double integral? Also, how do I conceptualize something like [itex] \iint_{\Sigma} z d\sigma [/itex] where the surface is the upper half of a sphere of radius 2? I'm summing all the z's across the surface? This is confusing to me.
     
  5. Aug 5, 2015 #4

    HallsofIvy

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    I am not sure if you are asking about a "surface integral" itself or about applications of a surface integral. The surface integral itself is just the integral of some function that takes on a given value at each point on the surface, just as the "ordinary" Calculus I integral is a an integral of a function that takes on values at every point from x= a to x= b. The integral being an "area" or the line integral being "the area under the function traced by the curve C" are specific applications, not the meanings of the integrals themselves. For example, a line integral might well be the work done by a given force moving an object along that curve. One possible application of a surface integral is that the integral of a "pressure" function over a surface is the total force on that surface.
     
  6. Aug 5, 2015 #5

    B3NR4Y

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    I think I understand now, but they only thing I don't understand is what [itex] \iint_\Sigma f \, \, d\sigma [/itex] represents. I know that [itex] \int_C f \, \, ds [/itex] represents the area under the curve C on f, so what does the surface integral represent?
     
  7. Aug 6, 2015 #6

    HallsofIvy

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    And what I said was that [itex]\int_C f ds[/itex] does NOT 'represent' the area under the curve. That is one possible application of the integral, not what the integral is or "represents".
     
    Last edited: Aug 6, 2015
  8. Aug 6, 2015 #7

    jedishrfu

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    The basic idea behind these volume and surface integrals is to reduce it down to a single integral.

    So if you have a surface integral z=f(x,y) for a surface that projects a circle in the xy plane then you'd integrate over say y first and then evaluate using the y limits.

    Notice the y limits are dependent on x so the resultant expression become the integrand of an integral over x.

    Integrating that integral with respect to x and applying the x limits produces your answer.

    Sometimes you'll have some extra steps where you'll have to do extra steps or recognize some symmetry to get the complete answer.

    As a simple example, find the area of a circle in xy coordinates will get you half the circle as apply the x limits of -r to +r and then you have to do it for the other half or note the symmetry of the half circles.

    I hope that wasn't too confusing.
     
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