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For a nice curve $C$ which is a boundary of a smooth surface $D$, Stokes' theorem says that

$$\begin{align*} \oint_C \mathbf{F}\cdot d \mathbf{s} = \iint_D (\nabla \times \mathbf{F} )\cdot d\mathbf{A} \end{align*}$$

However, I don't think I understand exactly what is going on with the $d\mathbf{A}$ term. I would think that it is the vector of magnitude differential area $dA$ and direction $\hat{n}$, where $\hat{n}$ is the unit vector in the direction perpendicular to the orientation of $dA$. However, by reading practice problems online, I see that this is not the case. What exactly is the definition of this $d\mathbf{A}$?

In addition, many sources online show that after evaluating the dot product $(\nabla \times \mathbf{F} )\cdot \mathbf{A}$ (no $dA$ term), the surface integral is magically done only in the $xy$ plane, rather than any other randomly defined plane. Is this legal?

Arturo don Juan
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1 Answers1

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The $d\vec S$ in the surface integral of Stokes' Theorem can be written $$d\vec S=\hat n dS$$

where $\hat n$ is the unit normal to the surface and oriented to conform with the right-hand rule.

The scalar $dS$ is precisely a differential element of surface area. So, $$\text{Surface Area of S}=\int_S dS$$

To compute $$\int_S V\cdot \hat n dS$$

often one projects the surface onto on of the coordinate planes (e.g., the $x-y$ plane). However, $dS\ne dxdy$! Rather, we need to account for the surface orientation upon projecting. So, in this example $dS =\frac{dxdy}{\cos \gamma}=\frac{dxdy}{\hat z \cdot \hat n}$, where $\gamma$ is the angle between the surface normal and the axis of projection (here, the $z$-axis).

If $\nabla \times \vec F$ is constant on $S$ and the surface normal $\hat n$ is constant (i.,e., a planar surface), then we have

$$\int_S \nabla \times \vec F\cdot \hat n dS=(\nabla \times \vec F\cdot \hat n)\int_S dS=(\nabla \times \vec F\cdot \hat n)S=\nabla \times \vec F\cdot \vec S$$

Mark Viola
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