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Surface Integral of a Square Pyramid

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the following surface integral for the four slanted faces of a square pyramid with a square base in the xy plane with corners at (0,0) (0,2) (2,0) (2,2) and a top vertex at (1,1,2) is equal to 4 by evaluating it as it stands:

    [tex]\int\int (\nabla \times V)\cdot n dS[/tex]

    where V is:
    V = (x2z - 2)i + (x + y + z)j - (xyz)k


    2. Relevant equations
    [tex]\int\int (\nabla \times V)\cdot n dS[/tex]

    V = (x2z - 2)i + (x + y + z)j - (xyz)k

    3. The attempt at a solution
    The main aim of this exercise is to prove how much simpler it can be use use Stokes' theorem to find the surface integral of a certain object that is bound by a simple shape, in this case the square base of the pyramid. I can easily prove that, by Stokes' theorem, the surface integral is indeed supposed to be 4 but the trouble comes when i have to calculate the integral as it is. I calculated the curl to be:

    [tex](\nabla \times V) = (-xz + 1)\hat{i} + (yz + x^{2})\hat{j} + \hat{k} [/tex]

    Now, for the next step, from what i understand the surface integral should be calculated for each of the four faces separately, since each surface will have a different normal. I've calculated the normals for each surface by simply finding the cross product of two vectors in the surface, such that the normals are:

    [tex]n1 = \frac{4\hat{j} + 2\hat{k}}{\sqrt{20}}[/tex]

    [tex]n2 = \frac{-4\hat{i} + 2\hat{k}}{\sqrt{20}}[/tex]

    [tex]n3 = \frac{-4\hat{j} + 2\hat{k}}{\sqrt{20}}[/tex]

    [tex]n4 = \frac{4\hat{i} + 2\hat{k}}{\sqrt{20}}[/tex]

    The next step would then be to find an equation (which should be the equation of the plane in which the triangular face is found) to put one variable in terms of the other two and integrate over the remaining variables, such that, for example writing y in terms of x and z:

    [tex]\int\int dS = \int\int \sqrt{\left(\frac{y(x,z)}{\partial x}\right)^{2} + \left(\frac{y(x,z)}{\partial y}\right)^{2} + \left(\frac{y(x,z)}{\partial z}\right)^{2}} dx dz[/tex]

    I found the equations of the planes to be, where the numbers correspond to the same surfaces as the normals with the same numbers:

    [tex]p1: z = 4 - 2y[/tex]

    [tex]p2: z = 2x[/tex]

    [tex]p3: z = 2y[/tex]

    [tex]p4: z = 4 - 2x[/tex]

    The trouble i'm having is with the j + k normals ... since dotting the normal and the curl previously found will give an equation with a y, a z and a x2 term in the integral. Now, since the equations of the planes only involve two variables, there is no way i can think of relating the third variable to the other two (in this case the x to the y or z). It is possible to write y in terms of z thus leaving the x and z variables, but what i can't figure out in this case is the upper and lower integration limits over which to integrate the x and z variables (since x cannot be written in terms of z or vice versa). This is the equation i get for surface 1:

    [tex]\frac{1}{\sqrt{20}}\int\int 4yz + x^{2} + 2 dS[/tex]

    It's been a very long time since i last did any surface integrals, so it could just be that i'm forgetting some basic concepts. Any help would be much appreciated as i've been trying to solve this equation all week. Many thanks in advance!
     
    Last edited: Feb 21, 2009
  2. jcsd
  3. Feb 21, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    There is an error in the x-component of this result; check your calculation again.

    Your triangular faces are not the entire planes that you have above, just triangular sections of them. Along the boundary (perimeter), of those triangles, x and y will be related (along the base, either x or y will be constant, while the other varies).

    For example, the face which is a section of p1, has vertexes at (0,2,0), (2,2,0) and (1,1,2). You should be able to see that [itex]z=2x=4-2y[/itex] for [itex]0\leq x\leq 1[/itex] and [itex]z=4-2x=4-2y[/itex] for [itex]1\leq x\leq 2[/itex].
     
  4. Feb 22, 2009 #3
    thanks a lot for that! i'll give it a go and should hopefully get the right answer now!
     
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