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Homework Help: Gauss' Theorem

  1. Nov 30, 2006 #1
    I have a problem that involves computing the vector surface of z=e^(1-x^2-y^2) where z is greater then or equal to 1, for a Vector Field F=xi+yj+(2-2*z)k. This is to be done using Gauss' theorem. I got an answer of 0, which doesn't seem right I just want to ask if anyone can confirm or deny the result. Here is how I did it.

    So this is a region sort of like a hump. The normal on the hump points upwards, and the hump and the plane z=1 I believe forms a closed surface, cutting out a circle if you look at down on it. So I calculate the Divergence of the Field and find that it equals to 0. x dx + y dy +(2-2*z)dz = 1+1-2. So then since the triple integral over the solid region is 0 I should be able to say the surface integral over the hump is equal to the sum of the double integral over the intersection of the hump and the double integral of the plane circle, which I think isof x^2+y^2=1. Therefore I calculate simply the surface integral of the circular region with a downward pointing normal (0,0,-1). Therefore F dot product n = 2-2*z. Here is the part where things get confusing for me. I try and subsitute z=e^(1-x^2-y^2). Then converting to polar coordinates so that I get e^(1-cos^2(a)-sin^2(a))=e^-0=1. There that makes the inner term I am integrating 2*1-2. Which equals 0 and making the whole thing 0.

    I don't think this is correct, can someone tell me where I am going wrong?
    Last edited: Nov 30, 2006
  2. jcsd
  3. Dec 2, 2006 #2
    I'm not as knowledgeable on this as others on the forum, so don't take this as a proper answer. However, there are a few things that concern me here:

    You say, "computing the vector surface" and then say you should use Gauss's Law. What do you mean by this?

    The Divergence Theorem (Gauss's Law) relates a closed surface integral to a volume integral. You are calculating the flux through the surface. I don't really know what you mean by computing the vector surface.

    Also, you say that the normal of the hump points upwards. Well the normal vector will point outwards, but I don't know about upwards (assuming upwards means in the +z direction). The normal on the surface is orthogonal to the surface so it is a space function (you must relate it to x,y,z in general). On the plane (the circle that you talk about) then yes the normal is constant over space since you are on a plane and so happen to be on a plane parallel to the x-y plane then yup, the normal is just [itex] -\hat z [/itex].

    Now if you are to calculate the flux, I would imagine that calculating the volume integral would be much simpler. I have not done any calculation on this problem so I may be wrong.

    Recall the Divergence Theorem:
    [tex] \int_V \nabla \cdot \vec A \,\, dv = \oint_S \vec A \cdot d \vec s[/tex]
  4. Dec 3, 2006 #3


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    You are correct that the integral of [itex]\nabla \cdot F[/itex] over the volume is 0. That means, by Gauss' theorem, that the integral of F itself over the entire surface is 0. But you are only asked for the integral over the "hump". Since the surface consists of the "hump" and the circular base, you have that the integral over the "hump" is the negative of the integral over the circular base.

    Of course, the base is given by z= 0 so you need to look at F(x,y,0)= xi+ yj+ 2k. The outward normal to that base is -k and so ds= -k dxdy. The surface integral on the base is just [itex]\int -2 dxdy[/itex] and so the integral is -2 times the area of the base, -2 times the area of the unit circle. It follows from that that the integral over the surface of the "hump" is two times the area of the unit circle, [itex]2\pi[/itex].
    Last edited by a moderator: Dec 3, 2006
  5. Dec 3, 2006 #4


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    "Of course, the base is given by z= 0"
    ??? the original question said "where z >= 1".

    In which case, I think the integral over the base (a circle radius 1 on the plane z=1) is zero as Lancen found, so the integral over that part of the "hump" is also zero.

    If the base was at z=0, the base covers the whole XY plane so the integrals are infinite - that doesn't make much sense.
  6. Dec 3, 2006 #5


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    You're right. I knew there was some reason I hadn't replied to this before!

    Yes, with the base at z= 1, F(x,y,1)= xi+ yj+ 0k so the integrand on the base is also 0.

    Lancen, I can only conclude that you are completely correct: the integral over the region is 0!
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