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Homework Help: Surface Integral: dot product of two unit vectors

  1. Sep 17, 2011 #1
    Hello. :smile:

    I understand most of the work involved with these types of questions, but there is one point in an example I'm following that I don't understand.

    1. The problem statement, all variables and given/known data


    [tex]I = \int{(z^2)}dS[/tex] over the positive quadrant of a sphere, where (x,y > 0).

    2. Relevant equations

    [tex]x^2 + y^2 + z^2 = 1[/tex]
    [tex]\underline{\hat{n}} = \frac{\nabla f}{|\underline{\nabla}f|}[/tex]

    3. The attempt at a solution

    Project onto the xy plane:

    [tex]dS = \frac{dxdy}{\underline{\hat{n}} \cdot \underline{\hat{k}}}[/tex]

    At any point on the surface:

    [tex]z^2 = 1 - x^2 - y^2[/tex]


    [tex]\int{(z^2)}dS = \int{\int{(1-x^2-y^2)}}\frac{dxdy}{\underline{\hat{n}} \cdot \underline{\hat{k}}}[/tex]

    [tex]\underline{\hat{n}} = \frac{\nabla f}{|\underline{\nabla}f|}[/tex]

    [tex]= x \underline{\hat{i}}+y\underline{\hat{j}}+\left (\sqrt{1-x^2-y^2} \right)\underline{\hat{k}}[/tex]

    I don't understand how to get to the next step:

    [tex]\underline{\hat{n}} \cdot \underline{\hat{k}} = \sqrt{1-x^2-y^2}[/tex]

    Its probably something really simple, knowing me. :redface:

    I've tried the dot product, but couldn't get the answer from that.

    Would anyone be gracious enought to end my torment?

    I hope my Tex is okay. Its changed since I was last here, though it could be my reinstallation of ProText. Everything is in bold and the gaps between equations are much bigger.

  2. jcsd
  3. Sep 17, 2011 #2


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    Personally, I wouldn't do it that way. I particularly dislike the notation
    [tex]\vec{n}= \frac{\nabla f}{|\nabla f}[/tex]
    because you also have a factor of [itex]|\nabla f|[/itex] in dS itself- they always cancel and so there is no need to calculate it.

    Instead, I think of a surface integral this way: parametric coordinates for the surface of a sphere, of radius R can be got from the formulas for spherical coordinates with [itex]\rho[/itex] fixed at R. That is, [itex]x= Rcos(\theta)sin(\phi)[/itex], [itex]y= Rsin(\theta)sin(\phi)[/itex], [itex]z= Rcos(\phi)[/itex]. That is, we can write the "position vector" of any point on the surface of the sphere as
    [tex]\vec{r}(\theta, \phi)= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}[/tex]

    The derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex],
    [tex]\vec{r}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
    [tex]\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}[/tex]
    are vectors lying in the tangent plane to the sphere. Their cross product,
    [tex]R^2cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2 sin(\phi)cos(\phi)\vec{k}[/tex]
    where the order of multiplicatio/n has been chosen to give the outward pointing normal, is normal to the sphere and, with [itex]d\theta d\phi[/itex], gives the "vector differential of surface area".

    dS itself, the "scalar differential of surface area" is given by the magnitude of that vector:
    [itex]dS= R^2 sin(\phi)d\theta d\phi[/itex].

    Your integrand is [itex]z^2= R^2cos^2(\phi)[/itex] so your integral is
    [tex]R^4 \int\int cos^2(\phi)sin(\phi)d\phi d\theta[/tex]
    where the integral is to include only the first octant.

    (Of course, in this problem, R= 1.)
  4. Sep 17, 2011 #3


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    Having got that off my chest, you can, of course, to answer your actual question, yes, it is "something really simple"!

    You arrived at the fact that [itex]\vec{n}= x\vec{i}+ y\vec{j}+ \sqrt{1- x^2- y^2}\vec{k}[/itex].

    It's dot product with [itex]\vec{k}= 0\vec{i}+ 0\vec{j}+ 1\vec{k}[/itex] is
    [tex]0(x)+ 0(y)+ 1(\sqrt{1- x^2- y^2})= \sqrt{1- x^2- y^2}[/tex]
    Last edited by a moderator: Sep 17, 2011
  5. Sep 17, 2011 #4
    Wow. Thanks for the replies. :smile:

    I knew it would be something straightforwards like that. I keep overthinking.

    I completely agree about the notation. The method I've been using looks very messy, so I'll give the method you prefer a go.

    Just past the bit where I was stuck, the example has this:

    [tex]\int{(z^2)dS} = \int{\int{(1-x^2-y^2)\frac{dxdy}{\sqrt{1-x^2-y^2}}}}[/tex]

    [tex]= \int{\int{(1-x^2-y^2)}dxdy}[/tex]

    Is that what you meant about the cancelling terms? I don't understand it, though it looks to be saying z = 1, or the dot product of the two unit vectors is 1.
  6. Sep 17, 2011 #5


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    What? No,
    [tex]\frac{1- x^2- y^2}{\sqrt{1- x^2- y^2}}= \sqrt{1- x^2- y^2}[/tex]
    not just [itex]1- x^2- y^2[/itex]. You will need a couple of trig substitutions to integrat that.

    [tex]Is that what you meant about the cancelling terms? I don't understand it, though it looks to be saying z = 1, or the dot product of the two unit vectors is 1.[/QUOTE]
  7. Sep 18, 2011 #6
    Yup, my example is incorrect. It switches to polars coordinates on the next line and has the root sign there.

    I'm so rusty with this. I'm stuggling to do the division. :redface:


    Its coming back now.
    Last edited: Sep 18, 2011
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