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Unit Normal Vector normal, but not normalized?

  1. Dec 3, 2008 #1
    I'm reading Matthew's Vector Calculus. In problem 2.8, I'm asked to find the surface integral for a surface which can be parametrized as [tex](x,y,x+y^2)[/tex]. I understand that to find a normal vector, I can cross product the x and y partial derivatives. When I do so, I get

    [tex](1,0,1)\times(0,1,2y) = (-1, -2y, 1)[/tex]

    The problem specifies that the normal has to have a negative z component. Not a complication -- reverse the signs: [tex](1,2y,-1)[/tex].

    At this point, I thought, it's in the bag. Just normalize and work the double integral. BUT, when I checked my work up to that point with the answers in back, I saw that the solution DID NOT involve normalization at all. He just ran with (1, 2y, -1) as the unit normal vector.

    Am I crazy in thinking this is wrong? Did I miss something? I just started, and I can't think that I'm already good enough at this to spot a major error by page 43. I keep thinking I've missed something, but the fact is, there's no way I can imagine anything of the form (1, whatever, -1) has length = 1. (Well, without throwing in a few imaginary numbers, but you know.)
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  3. Dec 4, 2008 #2


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    There are two different variations of "normal" involved here. A "normal" vector to a surface is simply a vector that is perpendicular to the surface. But to "normalize" a vector (I try never to use that term) means to make it of length 1.

    And in this case, you don't WANT a unit vector.

    If a surface is given by f(x,y)= constant, then with [itex]\vec{r}= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}[/itex], the [itex]\vec{r}_x[/itex] and [itex]\vec{r}_y[/itex], are vectors in the tangent plane to the surface and their cross product is "normal" (i.e. perpendicular) to the surface (which is true of the cross product of any two independent tangent vectors). More, the length of that cross product, often called the "fundamental vector product" for the surface, gives the "differential of area". That is, if dS is the differential of surface area for the surface, [itex]dS= |\vec{r}_x\times\vec{r}_y| dxdy[/itex]. [itex]d\vec{S}= \vec{r}_x\times\vec{r}_y| dxdy[/itex] is called the "vector differential of surface area".

    You don't want to "normalize" that vector- its length gives important information about the surface.
  4. Dec 4, 2008 #3
    I'll be darned. That makes perfect sense. I'm looking at n dS as two separate items, when in reality, it's no different than the parametric r and r' functions from a line integral. Those aren't normalized either. But the 'rate' at which they change is the key to making the whole integral work.

    Brilliant! Thank you so much.
  5. Dec 4, 2008 #4


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    Many textbooks, by the way, deal with the integral of a vector function on a surface using "[itex]\vec{f}\cdot\vec{n}dS[/itex] where [itex]\vec{n}[/itex] is the unit normal vector to the surface. But finding dS itself involves finding the length of the "normal vector" which you then divide to find the unit normal vector! Much better to just think "[itex]\vec{f}\cdot d\vec{S}[/itex]".
  6. Dec 4, 2008 #5
    After some investigation, I just figured that out this evening. My confusion came from a few steps that were elided in the Matthews book -- the ones you mentioned. I remembered a bit about surface integrals from years before, and I think the formula I'm recalling was used to find the actual area of the surface:

    \int f dS = \int \int f(u,v) \left\| n \right\| du dv

    where n is the normal vector (which we can get from the cross product, as above) and [tex]\left\| n \right\|[/tex] is its length. Then -- of course -- if you alter this integral to measure flow, not surface area, you have to integrate the dot product of f and the unit normal vector. But since

    [tex]\hat{n} = \frac{ n }{ \left\| n \right\|}[/tex]

    this reduces to

    \int f dS = \int \int f(u,v) \bullet n du dv

    just as you said. I think the sources of my confusion were the references to the "unit normal vector" throughout the chapter -- with the two former steps missing, the latter makes little sense in this context.
  7. Dec 5, 2008 #6


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    Be careful. If u and v are general parameters, dS may not be du dv but the length of the "fundamental vector product" times du dv. Since that length is exactly what you need to divide the fundamental vector product by to get n, rather than finding that length and both multiplying and dividing by it, it is better to think
    [tex]\int\int \vec{f}(u,v)\cdot d\vec{S}(u,v)[/tex]
    where [itex]d\vec{S}(u,v)[/itex] is just the fundamental vector product times du dv.

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