# Find Surface Area Via a Line Integral (Stokes' Theorem)

I am trying to use Stokes' Theorem to calculate the surface area of a parametrized surface via a line integral. The surface is the part of $z= x^2+y^2$ below the plane $z=5$. I know this can be done the usual way, without Stokes' Theorem, but there is another surface whose area I'm trying to calculate for which the usual way doesn't work well due to tough limits of integration.

As stated in the question Un-Curl Operator, we can find the surface area via $\oint\vec{F}\cdot \vec{ds}$ over the curve of intersection of the paraboloid and plane, provided we can find a field $\vec{F}$ such that $\text{curl}(\vec{F})\cdot \vec{n}=1.$ This would mean that

$$\text{curl}(\vec{F}) = \frac{\vec{n}}{{\Vert \vec{n} \Vert}^2},$$

so that,

$$\text{curl}(\vec{F})\cdot \vec{n}=\left\Vert\frac{\vec{n}}{{\Vert \vec{n} \Vert}^2}\right\Vert\cdot\Vert \vec{n} \Vert=\frac{{\Vert \vec{n} \Vert}^2}{{\Vert \vec{n} \Vert}^2}=1,$$as required. But finding such a vector field has proven difficult. I first tried using the method described in the question Anti-Curl Operator, but realized that this wouldn't work because $\text{curl}(\vec{F})$ has a nonzero divergence. I then turned to the method described in an answer to the Un-Curl Operator question, known as the Helmholtz decomposition. The idea is that given our vector field, $\text{curl}(\vec{F})$ in this case, we can split it into a curl-free part and a divergence-free part:

$$\text{curl}(\mathbf{F})=-\nabla\Phi+\nabla\times \mathbf{A}$$

I believe (I could be wrong though) that it remains only to find $\mathbf{A}$, since the first term will not contribute to the line integral around a closed path. However, this is where my confusion begins. According to Wikipedia, $\mathbf{A}$ is found as follows:

$$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_V\frac{\nabla'\times\mathbf{F}(\mathbf{r}')}{\left\vert\mathbf{r}-\mathbf{r}'\right\vert}\text{d}V' - \frac{1}{4\pi}\oint_S\mathbf{\hat{n}}'\times\frac{\mathbf{F}(\mathbf{r}')}{\left\vert\mathbf{r}-\mathbf{r}'\right\vert}\text{d}S'$$

I do not know how to use this formula. Specifically, I do not know what most of the symbols represent in the context of this problem. What is the difference between $\mathbf{r}$ and $\mathbf{r}'$? What is the region of integration? Also, the Wikipedia page mentions that the second term can be dropped if the field satisfies certain conditions. Does that apply here?

I am also wondering if there is anything else I need to do to specify that the dot product of $\text{curl}(\mathbf{F})$ and $\mathbf{n}$ is $1$. The answer on the Un-Curl Operator question used the following notation:

$$\mathbf{G}\vert_{\partial\Omega}\cdot\mathbf{n}=1,$$

where $\mathbf{G}$ = $\text{curl}(\mathbf{F})$. I'm pretty sure that $\partial\Omega$ represents the boundary of a domain $\Omega$, but what is $\Omega$ in the context of this problem? Am I missing something?

How would we use the Helmholtz decomposition to find the vector field $\mathbf{A}$ in this problem?

## 1 Answer

I'm not sure whether this helps you or not. Suppose $\mathbf{B}$ is a constant vector. Define $$\mathbf{A}(\mathbf{r})=-\frac{1}{2}\mathbf{r}\times \mathbf{B}\Longrightarrow \nabla\times \mathbf{A}=\mathbf{B}$$ Then by Stokes theorem $$\mathbf{B}\cdot \mathbf{S}=\int_{\Omega} \mathbf{B}\cdot d\mathbf{a}=\oint_{\partial\Omega} \mathbf{A}\cdot d\mathbf{r}= -\frac{1}{2}\oint_{\partial\Omega} (\mathbf{r}\times\mathbf{B})\cdot d\mathbf{r}= \mathbf{B}\cdot \left(\frac{1}{2}\oint_{\partial\Omega} \mathbf{r}\times d\mathbf{r}\right)$$ where $\mathbf{S}$ is the total (vector) surface area of $\Omega$. Now choose $\mathbf{B}=\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ to obtain $$\mathbf{S}_\Omega=\int_{\Omega} d\mathbf{a}=\frac{1}{2}\oint_{\partial \Omega} \mathbf{r}\times d\mathbf{r}$$ This is a formula for calculating the surface area via line integrals.

• Thank you, I was looking for that kind of formula. I just have a couple questions: Does $\mathbf{r}$ represent $\langle x,y,z \rangle$ and $d\mathbf{r}$ represent $\langle dx,dy,dz \rangle$? – eg001 May 26 '16 at 18:56
• It might not be very useful to you though since it is the vector surface area which is very different from surface area. But maybe you can play around with it and find what you want. – Hamed May 26 '16 at 18:58
• Oh I see, that could explain why the final line integral is of a vector and not a scalar. I was wondering about the interpretation of that. I am also not familiar with vector surface area, but maybe there is a way to get regular surface area out of this. – eg001 May 26 '16 at 19:02