Confusion with normalizing normals and gradients in surface integrals

Question

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?

Answer 1

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;

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

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!