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If the surface over which we integrate is a level set of $f$, then the gradient of $f$ gives a normal $\nabla f(p)$ at the point. However, this normal may not be of unit magnitude.

By definition $\iint_\Omega \overrightarrow F\cdot \overrightarrow{dS} $ where $\overrightarrow{dS}=\hat n dS$, so the integrand is $\overrightarrow F\cdot \hat n$ with $\hat n=\widehat{\nabla f}$. So I understand we need to normalize the gradient to find the integrand.

However, I think $dS=\|r_u\times r_v\|dudv$. But $r_u\times r_v=\nabla f$ so we end up with $$\iint_\Omega \overrightarrow F\cdot \overrightarrow{dS}=\iint_\Omega (\overrightarrow F\cdot \hat n)\circ r\|\nabla f\|dudv.$$ To me this looks equal to $$\iint_\Omega (\overrightarrow F\cdot \nabla f)\circ rdudv.$$

In the question Using Stoke's theorem evaluate the line integral $\int_L (y i + zj + xk) \cdot dr$ where $L$ is the intersection of the unit sphere and x+y = 0, the gradient is normalized to obtain $$\iint_\Omega(\overrightarrow F \cdot \hat n)\circ r \ dS=\frac 1{\sqrt 2}\iint_\Omega dS$$ but it is said equal to $\frac 1{\sqrt 2}\pi$, which is the result of saying $\iint_\Omega dS=\pi$. But I think $dS=\sqrt dudv$ and the answer should be $\pi$.

Who is correct?

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This was a big mistake that I made as well the first time I was understanding surface integrals. The key here is that your surface normal is dependent on your parametrization. Hence, although $\nabla f$ is another normal vector for a patch on your surface, it may have a different magnitude than the normal you get as a result of using the parametrization.

For instance, let $G: D \subset \mathbb{R}^2 \to S \subset \mathbb{R}^3$ be a parametrization with $(u,v)$ be the coordinates on $\mathbb{R}^2$ and $(x,y,z)$ the coordinates on $\mathbb{R}^3$. Then given a point $(u,v)$, we have $G(u,v) \in S$ and $G_u\Bigr|_{u,v}, G_v\Bigr|_{(u,v)}$ are tangent vectors at $G(u,v)$ which span $T_{G(u,v)}S$ (notation for tangent plane to the surface at $G(u,v)$). Thus $G_u \times G_v := \textbf{n}(u,v)$ is the normal vector for the tangent plane at $G(u,v)$. If $F$ is a vector field on $S$ then $F(G(u,v)) \cdot \textbf{n(u,v)}$ is calculating the amount of $F$ which is in the direction of $n$ i.e how much substance has flowed through the patch at $G(u,v)$. Hence, the flux or vector surface integral is given by;

$$\int_s F = \iint_D F(G(u,v)) \cdot \textbf{n}(u,v) \ du \ dv = \iint_D \underbrace{F(G(u,v)) \cdot \frac{\textbf{n}(u,v)}{\|\textbf{n}(u,v)\|}}_{\textrm{projection on $F(G(u,v))$ onto $e_{\textbf{n}}$}} \ \|\textbf{n}(u,v)\| \ du \ dv$$

enter image description here

In the above $e_{\textbf{n}}$ is just notation for the unit vector in the direction of $\textbf{n}(u,v)$. So, now you see that although the normalize gradient vector at $G(u,v)$ points in the same direction as $\textbf{n}(u,v)$, it is not necessarily the case that $\|\nabla f(u,v)\| = \|\textbf{n}(u,v)\|$ i.e the volume of the parallelepipes spanned by;

$$\left\{G_u,G_v,\textbf{n}\right\} \ \ \textbf{and}\ \left\{G_u,G_v,\nabla f\right\}$$

are different. Recall that this volume calculates the fluid which has flowed through a very small patch at $G(u,v)$. Hopefully this helps!

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  • $\begingroup$ How does this fit in with the paraboloid example here tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx? $\endgroup$ – eleventeen Feb 9 '17 at 11:27
  • $\begingroup$ They chose to parametrize the surface $S$ by using its graph. Well, if you do that, then the surface normal for that parametrization is the gradient. Does that make sense? All I'm saying is that in the example you showed me, they used the parametrization associated to the gradient which is available if your surface arises as the graph of some function $f$ i.e all points are of the form $(x,y,f(x,y)$. $\endgroup$ – Faraad Armwood Feb 9 '17 at 11:29
  • $\begingroup$ I think I understand but I have some more question. Is there any chance you can open a chat room chat.stackexchange.com/… for me to ask you stuff? $\endgroup$ – eleventeen Feb 9 '17 at 11:32
  • $\begingroup$ Okay. Also, if this answers your question, you can accept the answer, to close the question. $\endgroup$ – Faraad Armwood Feb 9 '17 at 11:33
  • $\begingroup$ Go into the chat room that says Geometry and Topology. $\endgroup$ – Faraad Armwood Feb 9 '17 at 11:34

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