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Which Normal to Use?

  1. Jun 19, 2011 #1
    In evaluating a surface integral, I know that the surface integral of 1, dS over the surface S will give me surface area. That means that a regular double integral of the magnitude of the normal over the region R, in the xy-plane, will give the surface area.

    For a general sphere, x^2+y^2+z^2=a^2, can I use the normal obtained from the gradient? Or must I solve g(x,y)=sqrt(a^2-x^2-y^2) and then take the magnitude? The first way, I get that the normal is 2a and the second way I get that the normal is a/z.

    What is wrong?
     
  2. jcsd
  3. Jun 19, 2011 #2

    I like Serena

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    Hi schaefera! :smile:

    You're talking about the length of the gradient, but that's not what a normal unit vector is.

    Btw, you do not just need a normal vector, but you need a normal unit vector.

    A normal unit vector is perpendicular to the surface and has length 1.

    You can use the gradient to find the direction of a normal vector.
    But you need to divide it by its length to get a normal unit vector.
     
  4. Jun 19, 2011 #3

    HallsofIvy

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    A normal is a normal! It doesn't matter how you get it. However, the normal is a vector, not a number, so neither of those is correct. If you write [itex]f(x,y,z)= x^2+ y^2+ z^2= a^2[/itex], then the normal to that is the gradient, [itex]2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}[/itex]. The length of that normal is
    [tex]\sqrt{4x^2+ 4y^2+ 4z^2}= 2\sqrt{x^2+ y^2+ z^2}= 2a[/tex]

    I'm not sure what you mean by "solve g(x,y)= sqrt(a^2- x^2- y^2) and then take the magnitude". What do you mean by "solve g(x,y)"? Of course,
    [tex]z= \sqrt{a^2- x^2- y^2}[/tex]
    but then what are you going to do with it?
     
  5. Jun 19, 2011 #4
    Sorry, above I did mean to write the length of the normal.

    From that z that you solved, you can then make a parametric surface, with parameters x and y, where g(x,y)=z, correct? Then, cross the partial derivatives with respect to each parameter to get a normal.

    Also, in a surface integral isn't it actually dS=(magnitude of normal) dA?
     
  6. Jun 20, 2011 #5

    I like Serena

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    Sorry, but as HoI already stated, the magnitude of the normal is irrelevant.
    In your case dS = dA.

    The question is, how big is dS or dA in terms of the coordinates and their infinitesimals.

    Usually the surface of a sphere is integrated with spherical coordinates [itex](r, \theta, \phi)[/itex].
    In this case we have [itex]\text{d}S = a^2 \sin \theta \text{ d}\theta \text{ d}\phi[/itex].

    With your z = g(x, y) you could integrate as well, but that becomes way too complicated.

    Cheers! :smile:
     
  7. Jun 20, 2011 #6
    First off, thanks for your help- you guys are great!

    But here, for example, I see them showing that dS= |n| dA, where |n| is the magnitude of the normal. (The link: http://www-users.math.umd.edu/~jmr/241/surfint.html).

    Why do you say dS=dA in my original case?
     
  8. Jun 20, 2011 #7

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    My bad. I misinterpreted your questions. But then, you did not define dA.

    I take it, dA = dxdy in your case?

    Well, the length of the normal unit vector n is still 1.
    It is the surface dS that has a different size from dxdy.

    However, the gradient is not involved, nor its length.



    In your text a definition for a vector P is made.
    Based on your definition of g(x,y), you would have:
    [tex]\vec P(x,y) = \begin{pmatrix}x \\ y \\ g(x,y) \end{pmatrix}[/tex]

    The outer product mentioned, would be:
    [tex]\frac {\partial \vec P} {\partial x} \times \frac {\partial \vec P} {\partial y} = \begin{pmatrix}\frac x {g(x,y)} \\ \frac y {g(x,y)} \\ 1 \end{pmatrix}[/tex]

    This is indeed a normal vector, but one with the very special property that its length defines the ratio between the surface dS and dxdy.
    The normal unit vector [itex]\vec n[/itex] will have the same direction, but length 1.

    The result would be:
    [tex]dS = \sqrt {1 + \left(\frac x {g(x,y)} \right)^2 + \left(\frac y {g(x,y)} \right)^2} dxdy[/tex]

    To calculate the top half surface of the sphere with radius a, you would get:
    [tex]\int_{-a}^{+a} \int_{-\sqrt{a^2-x^2}}^{+\sqrt{a^2-x^2}} \sqrt {1 + \left(\frac x {g(x,y)} \right)^2 + \left(\frac y {g(x,y)} \right)^2} dy dx[/tex]

    Note that this is a difficult integral to integrate.
     
    Last edited: Jun 20, 2011
  9. Jun 20, 2011 #8
    I suppose that is the heart of my question! You obtain the normal there by crossing the partials of the parametrized surface. Why couldn't you, instead, use the gradient to find a normal that works to define the proportionality between dS and dxdy?
     
  10. Jun 20, 2011 #9

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    You need the ratio between dS and dxdy.
    This is given by the length of the outer product of the partial derivatives of the parametrized surface.
    As you can see the gradient simply does not give this ratio.

    A more detailed explanation is as follows.

    dS is an infinitesimal surface element that can be approximated by a parallelogram
    The 2 relevant sides as vectors are given by the partial derivatives.
    The length of the outer product gives the surface area, which is what we need.

    The gradient gives a measure how "fast" your sphere "expands".
    This will give you a vector that is perpendicular to the surface, but its length has nothing to do with the surface area.
     
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