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My book says that a vector normal to a tangent plane at point $(a,b)$ is $$<f_x(a,b), f_y(a,b), -1>$$ If I understand vectors correctly, wouldn't this mean that the vector always moves $1$ unit down in the $z$ coordinates? But this should be obviously untrue, as the normal vector to a point in a tangent plane can be going in any direction, depending on the plane. And my books gives an image where it seems to do just that as an example!

enter image description here

Obviously over here the normal vector is not moving $1$ unit down in the z-coordinates. So what does it mean that a vector normal to a tangent plane at point $(a,b)$ is of the form $<f_x(a,b), f_y(a,b), -1>$?

Also, my book gives the derivation of Stokes Theorem and says that the normal vector to the surface is $<-f_x(x,y), -f_y(x,y), 1>$. Why are the signs switched? Also , when my book discusses the flux of a vector field, it does the same thing and says "For the normal vector to point upward, we need a positive z-component." But in the image I pasted, it certainly looks like the normal vector is pointing upward, and yet it is of the form $<f_x(x,y), f_y(x,y), -1>$

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The issue here is $orientation.$ The normal vector shown in your graph is the one (presumably of unit norm) that points $away$ from the surface. You need to choose an orientation before doing any calculations, so that you can keep track of the "pieces" of the surface that you will want to integrate over.

So, once and for all, we will always choose the vector of unit norm that points away from the surface.

Maybe the best way to see this is to look at an example. Take $f(x,y)=4-x^2-y^2$ at $(1,1,4)$. These data fit your graph fairly well.

Now, $f_x(1,1)=f_y(1,1)=-2$ so one possible normal vector is $\vec n=(-2,-2,-1)$. However, this choice of $\vec n$ points $into$ the surface, which violates our original choice of orientation. So, we simply switch its sign, multilpying by $-1$, to obtain $\vec n_1=(2,2,1)$, which, after normalization, is the vector shown in your graph.

An interesting question is: is such a choice of orientation always possible? For example, consider the Mobius band and a point $P$ on it, attach a normal vector at $P$. How would you define the normal so that it points "away" from the surface?

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  • $\begingroup$ So basically, the image is not a great example of the the point the book is trying to make, that the form of the normal to tangent is $<f_x(a,b), f_y(a,b), -1>$. It should probably have given an image of a vector pointing inwards, so as to be more clear. $\endgroup$ – agblt May 7 '19 at 15:30
  • $\begingroup$ As for Mobius, I would split it up into two parts, one where the "away" direction is positive in z and one where the "away" direction is negative in z. Is this possible? $\endgroup$ – agblt May 7 '19 at 15:33
  • $\begingroup$ Right. The author should have pointed out that the assignment of the normal vector amounts to making a choice of orientation, and that usually, one picks the "outward" pointing vector. As for the Mobius, if you do what you suggest, which is a reasonable thing to do, when you get to integrating over the surface of the band, you will run into a problem. You will "overcount" the surface by a factor of two. In fact, the Mobius is a surface that is not orientable. Lu's book "Introduction to Manifolds" covers this nicely. $\endgroup$ – Matematleta May 7 '19 at 15:42
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There is not just one normal vector. Any non-zero multiple of a normal vector is another normal vector, and that includes multiplying by $-1$.

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  • $\begingroup$ So in the image, is z-component being multiplied by a negative number to give a positive direction? $\endgroup$ – agblt May 7 '19 at 15:05
  • $\begingroup$ Yes. When you apply Stokes' Theorem, you have to pay attention to the orientation. In your picture the orientation is "upwards." $\endgroup$ – B. Goddard May 7 '19 at 15:17
  • $\begingroup$ So to make a unit normal vector for the vector in the image, it would have to be of the form $<-f_x(a,b), -f_y(a,b), 1)>$? $\endgroup$ – agblt May 7 '19 at 15:21
  • $\begingroup$ No. For a unit normal, you need to divide the vector by its length. Then choose the sign to get the right orientation. $\endgroup$ – B. Goddard May 7 '19 at 15:26

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