If I have the surface of a hemisphere $S : x^2 + y^2 + z^2 = 4; z\geq 0$, then using the gradient to calculate the unit normal vector yields $\hat n = <\frac{x}{2}, \frac{y}{2}, \frac{z}{2}>$. But my textbook uses the explicit formula of the hemisphere $z=f(x,y)$ which yields $\hat n = <\frac{x}{z}, \frac{y}{z}, 1>$. Why is there a discrepancy between these two equations, and (in the case of the 2nd equation) how does it make sense physically for the z-component of the normal to always be 1? Also why is it that the normal coming from the explicit equation is undefined for $z=0$ when it shouldn't be? If I use the first normal vector in Stokes' theorem for some vector field $\vec F $, I get a different solution than if I were to use the 2nd one.
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2 Answers
Usually "normal vector" is understood to mean unit normal vector. In your answer, $$\hat{n}\cdot\hat{n} = \frac{x^2+y^2+z^2}{4} = 1$$ and you do indeed have a unit vector. On the other hand, if you check the textbook's answer, $$\hat{n}\cdot\hat{n} = 1+\frac{x^2+y^2}{z^2} = \frac{x^2+y^2+z^2}{z^2} = \frac{4}{z^2}$$ which is not usually 1 (and moreover there is a problem at $z=0$, as you noted). Normalizing the book answer gives $$\frac{z}{2}\left(\frac{x}{z},\frac{y}{z}, 1\right) = \frac{1}{2}(x,y,z)$$ which matches your answer.
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But as I understand Stokes' theorem which relates a surface integral to the line integral of its boundary, the normal vector of the surface to integrate must be a unit vector, but the textbook used the normal vector derived from the explicit formula which is not a unit vector. So is the textbook wrong? When I use the other normal vector to do the surface integral, I'm off by a factor from the textbook's answer. – hesson Dec 20 '13 at 04:32
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Maybe the book takes the magnitude of $n$ into account somewhere else? Without more information, the only thing I can say is that yes, you should use the unit normal in Stokes's theorem. – user7530 Dec 20 '13 at 04:38
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I posted a picture from the textbook so you can see how its used. The result that the textbook gets is $8\pi$ which agrees with the result from the line integral which can be calculated easily. Using the other form of the normal vector seems to yield a different result for me for the surface integral. – hesson Dec 20 '13 at 04:44
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Could you please post the full page, including the original problem statement? Are you sure you are using the correct area element? – user7530 Dec 20 '13 at 06:49
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I posted the original problem - they get $8\pi$ using the line integral, and then they get the same answer using the surface integral. It's still beyond me why it works for the normal vector that they used if it's not a unit vector. For the surface integral, the page is cut off, but they eliminate the term with the $sin$ function and the integrand becomes $2rdrd\theta$ - the integration also gives $8\pi$. Using the other normal vector, I get the dot product to be $y + z$ and I get the integral to be $16\pi / 3$ (http://wolfr.am/1bUdi6R). – hesson Dec 20 '13 at 07:08
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The surface integral must be done over the surface of the hemisphere. You can use change of variables to turn this into an integral over the planar disk, but when you do so, you get a determinant-of-Jacobian term in the integral that measures the distortion in the area element. If you still need help, please post a new question, with the full text of the original problem, and the steps you've done in your calculation. – user7530 Dec 20 '13 at 15:27
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Thanks, that makes sense, I thought the normal they used was weird because the z-component was always 1, and that's not true for a hemisphere, but if you warp it into a planar disk then it makes sense. I looked at the way they derived the formula and I understand it now because they changed the area element - thanks again! – hesson Dec 21 '13 at 19:32
If you use $n = \langle x/2, y/2, z/2\rangle$, then the term in Stokes' Theorem is $n\mathrm dS = \langle x/2, y/2,z/2\rangle\mathrm dS$.
You still have to figure out how to integrate w.r.t. $\mathrm dS$. If you use $n = \langle x/z, y/z, 1\rangle$, then $n\mathrm dS$ becomes $\langle x/z, y/z, 1\rangle \mathrm dA$ where $\mathrm dA$ is the projected area on the $xy$-plane.
