Parametric Surfaces
Just as a curve can be described as a vector function \(\overline{r}\left(t\right)\) of a single parameter, \(t\), a surface can be described by a vector function \(\overline{r}\left(u,\nu\right)\) of two parameters \(u\) and \(v\), that is
\(\overline{r}\left(u,\nu\right)=x(u,\nu)\overline{i}+y(u,\nu)\overline{j}+z(u,\nu)\overline{k}\)
is a vector-valued function defined on a region \(D\) in the \(uv\)-plane. The component functions \(x\), \(y\), and \(z\) of \(\overline{r}\left(u,\nu\right)\) are functions of two variables \(u\) and \(v\) with domain \(D\). The set of all points \((x,y,z)\) in \(\mathbf{R}^{3}\) such that
\(x=x \begin{pmatrix} u,\nu \end{pmatrix},\quad y=y \begin{pmatrix} u,\nu \end{pmatrix},\quad z=z \begin{pmatrix} u,\nu \end{pmatrix}\quad \begin{pmatrix} 1 \end{pmatrix}\)
and \((u,v)\) varies throughout \(D\) is called a parametric surface \(S\). The parametric equations of \(S\) are (1).
Normal Vector to a Surface
Say that a vector \(\bar{\nu}\) is tangent to a surface \(S\) at the point \(P\) if \(\bar{\nu}\) is a tangent vector, at \(P\), to some curve that is contained in \(S\). This is analogous to the tangent line of a single variable function.
Assume that a surface \(S\) is represented by \(\overline{r}\left(u,\nu\right)=\left\langle x(u,\nu),y(u,\nu),z(u,\nu)\right\rangle\), where \(\left(u,\nu\right)\in D\subseteq\mathbb{R}^2\). At a point on the surface \(\left(u_0,\nu_0\right)\), there exists the following two tangent vectors, \(\bar{T}_u\left(u_0,\nu_0\right)\) and \(\bar{T}_\nu\left(u_0,\nu_0\right)\), given by
\(\bar{T}_u\left(u_0,\nu_0\right)=\left\langle\frac{\partial x}{\partial u}\right|_{(u_0,\nu_0)},\frac{\partial y}{\partial u}|_{(u_0,\nu_0)},\frac{\partial z}{\partial u}|_{(u_0,\nu_0)}\rangle\),
being the partial derivatives of the components \(x=x\left(u,\nu\right),y=y\left(u,\nu\right),\mathrm{and~}z=z\left(u,\nu\right)\) with respect to u, and
\(\bar{T}_\nu\left(u_0,\nu_0\right)=\left\langle\frac{\partial x}{\partial\nu}\right|_{(u_0,\nu_0)},\frac{\partial y}{\partial\nu}|_{(u_0,\nu_0)},\frac{\partial z}{\partial\nu}|_{(u_0,\nu_0)}\rangle\),
being the partial derivatives of the components \(x=x\left(u,\nu\right),y=y\left(u,\nu\right),\mathrm{~and~}z=z\left(u,\nu\right)\) with respect to \(\bar{\nu}\).
The vector \(\bar{N}\left(u_0,\nu_0\right)=\bar{T}_u\left(u_0,\nu_0\right){\times}\bar{T}_\nu\left(u_0,\nu_0\right)\), being perpendicular to both \(\bar{T}_u\left(u_0,\nu_0\right)\) and \(\bar{T}_{\nu}(u_{0},\nu_{0})\) is called a normal vector. The normal vector is computed by
\(\overline{N}=\overline{T}_u\times\overline{T}_\nu= \begin{vmatrix} \\ \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial\nu} & \frac{\partial y}{\partial\nu} & \frac{\partial z}{\partial\nu} \end{vmatrix}.\)
where we have dropped explicit reference to the point \((u_0,\nu_0).\)
Stokes' Theorem
Let \(S\) be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise boundary curve \(C\) with positive orientation. Let \(\bar{F}\) be a vector field whose components have continuous partial derivatives on an open region in \(\mathbf{R}^{3}\) that contains \(S\). Then
$$\int_\mathbb{C}\overline{F}\cdot d\overline{r}=\int\int_\mathbb{S}\mathrm{curl}\overline{F}\cdot d\overline{S}.$$
The positively oriented boundary curve of the oriented surface \(S\) is often written as \(\partial\mathbf{S}\). Then, Stokes' Theorem can be expressed as
$$\int\int_{\mathcal{S}}\mathrm{curl}\overline{F}\cdot d\overline{S}=\int_{\partial\mathcal{S}}\overline{F}\cdot d\overline{r}.$$
The left side of the equation involves an integral involved with derivatives because of \(\mathrm{curl}\bar{F}\)，and the right side has the values of \(\bar{F}\) at the boundary of \(S\).
In a special case where the surface \(S\) is flat and lies in the \(xy\)-plane with upward orientation, the unit normal is \(\bar{N}\), the surface integral becomes a double integral, and Stokes' Theorem becomes
$$\int_\mathbb{C}\bar{F}\cdot d\overline{r}=\int\int_\mathbb{S}\mathrm{curl}\bar{F}\cdot d\overline{S}=\int\int_\mathbb{S}\mathrm{curl}\bar{F}\cdot\bar{N}dA.$$
Example 1.
Let C be the circle \(x^{2}+y^{2}=1\) and \(z=1\), oriented counterclockwise as seen from a point \((0,0,z)\) with \(z>1\) on the z-axis. For \(\vec{F}=2\vec{i}+x\vec{j}+y^{2}\vec{k}\), compute \(\int_{C}\vec{F}\cdot d\vec{r}\).
Solution. Calculating directly, we parametrize C by \(\overline{r}=\left\langle\cos t,\sin t,1\right\rangle\), where \(t\in[0,2\pi)\). Then \(\overline{r^{\prime}}(t)=\left\langle-\sin(t),\cos(t),0\right\rangle\) We then obtain
$$\begin{aligned}
\int_{C}{\bar{F}}\cdot d{\bar{r}}& =\int_{\mathrm{C}}\overline{F}\left(\overline{r}\left(t\right)\right)\cdot\overline{r}\left(t\right)dt \\
&=\int_{0}^{2\pi}\Big\langle2,\cos\bigl(t\bigr),\sin^{2}\bigl(t\bigr)\Big\rangle\cdot\Big\langle-\sin\bigl(t\bigr),\cos\bigl(t\bigr),0\Big\rangle dt \\
&=\int_{0}^{2\pi}\Bigl(-2\sin\bigl(t\bigr)+\cos^{2}\bigl(t\bigr)\Bigr)dt \\
&=\int_{0}^{2\pi}\biggl(-2\sin\bigl(t\bigr)+\frac{1}{2}+\frac{1}{2}\cos\bigl(2t\bigr)\biggr)dt \\
&=\left[2\cos\left(t\right)+\frac{1}{2}t+\frac{1}{4}\sin\left(2t\right)\right]_{0}^{2\pi}=\pi,
\end{aligned}$$
using the fact that \(\cos^2(t)=\frac{1}{2}+\frac{1}{2}\cos(2t).\)
Alternatively, we can solve this problem using Stokes' Theorem:
$$\int\limits_{C}\bar{F}\cdot d\bar{r}=\int\int_\mathbb{S}\mathrm{curl}\bar{F}\cdot d\overline{S}$$
We first calculate the curl:
$$\mathrm{curl}\overline{F}=\begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2&x&y^2\end{vmatrix}=2\:y\overline{i}\:+\overline{k}\:.$$
The surface bounded by C is the disk given by \(x^2+y^2\leq1\mathrm{~and~}z=1.\) We can parameterize this surface by \(\bar{r}\left(u,\nu\right)=\left\langle u,\nu,1\right\rangle,\mathrm{~for~}u^2+\nu^2\leq1.\) We can then calculate the tangent and normal vectors:
$$\overline{T}_{u}=\left\langle\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right\rangle=\left\langle1,0,0\right\rangle,\\\overline{T}_{y}=\left\langle\frac{\partial x}{\partial\nu},\frac{\partial y}{\partial\nu},\frac{\partial z}{\partial\nu}\right\rangle=\left\langle0,1,0\right\rangle,$$
$$\overline{N}=\overline{T}_u\times\overline{T}_\nu=\begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\1&0&0\\0&1&0\end{vmatrix}=\overline{k}\:.$$
Now, we have shown the surface is flat and lies in the \(xy\)-plane with upward orientation, so Stokes' Theorem becomes
$$\int_{C}\bar{F}\cdot d\bar{r}=\int\int_{S}\mathrm{curl}\bar{F}\cdot d\bar{S}=\int\int_{S}\mathrm{curl}\bar{F}\cdot\bar{N}dA.$$
Hence, we have
$$\int\limits_{\mathrm{C}}\overline{F}\cdot d\overline{r}=\int\int\limits_{\mathrm{S}}\mathrm{curl}\overline{F}\cdot d\overline{S}=\int\int\limits_{u^{2}+v^{2}\leq1}\left(2\:y\overline{i}+\overline{k}\right)\cdot\overline{k}dA=\int\int\limits_{u^{2}+v^{2}\leq1}1dA=\pi.$$
Example 2. Let S be the interior of the unit sphere, i.e. \(x^2+y^2+z^2\leq1.\) Compute \(\iiint_\mathrm{s}\mathrm{curl}\bar{F}\cdot d\bar{S},\mathrm{where~}\bar{F}(x,y,z)=z\overline{i}+y\overline{j}+x\overline{k}.\)
Solution. We use spherical coordinates to write \(\bar{F}\) in terms of two variables..
$$x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta,\quad0\leq\theta\leq\pi,\quad0\leq\phi<2\pi,$$
where \(r=1.\) The parametric representation of the surface \(x^2+y^2+z^2=1\) is given by.
$$\overline{r}\begin{pmatrix}\theta,\phi\end{pmatrix}=\sin\theta\cos\phi\overline{i}+\sin\theta\sin\phi\overline{j}+\cos\theta\overline{k}$$
and the tangent and normal vectors are given by
$$\begin{aligned}
&\bar{T}_{\theta}= {\frac{\partial x}{\partial\theta}}\:{\overline{i}}+{\frac{\partial y}{\partial\theta}}\:{\overline{j}}+{\frac{\partial z}{\partial\theta}}\:{\overline{k}}=\cos\theta\cos\phi{\overline{i}}+\cos\theta\sin\phi{\overline{j}}-\sin\theta{\overline{k}}\:, \\
&\bar{T}_{\phi}= \frac{\partial x}{\partial\phi}\overline{i}+\frac{\partial y}{\partial\phi}\overline{j}+\frac{\partial z}{\partial\phi}\overline{k}=-\sin\theta\sin\phi\overline{i}+\sin\theta\cos\phi\overline{j}, \\
&\bar{N}= \overline{T}_{\theta}\times\overline{T}_{\phi}=\sin^{2}\theta\cos\phi\overline{i}+\sin^{2}\theta\sin\phi\overline{j}+\sin\theta\cos\theta\overline{k}.
\end{aligned}$$
Finally, applying Stokes' Theorem, we obtain
$$\begin{aligned}
&\iiint_{\mathrm{S}}\mathrm{curl}\overline{F}\cdot d\overline{S}=\int\int_{\partial S}\overline{F}\cdot d\overline{r}=\int\int_{\partial S}\overline{F}\cdot\overline{N}dA \\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\Bigl(\cos\theta\overline{i}+\sin\theta\sin\phi\overline{j}+\sin\theta\cos\phi\overline{k}\Bigr)\cdot\Bigl(\sin^{2}\theta\cos\phi\overline{i}+\sin^{2}\theta\sin\phi\overline{j}+\sin\theta\cos\theta\overline{k}\Bigr)d\theta d\phi \\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\Bigl(2\sin^{2}\theta\cos\theta\cos\phi+\sin^{3}\theta\sin^{2}\phi\Bigr)d\theta d\phi \\
&=2\int_{0}^{\pi}\sin^{2}\theta\cos\theta d\theta\int_{0}^{2\pi}\cos\phi d\phi+\int_{0}^{\pi}\sin^{3}\theta d\theta\int_{0}^{2\pi}\sin^{2}\phi d\phi \\
&=2\int_{0}^{\pi}\sin^{2}\theta\cos\theta d\theta\Big[\sin\phi\Big]_{0}^{2\pi}+\int_{0}^{\pi}\Big(1-\cos^{2}\theta\Big)\sin\theta d\theta\Big]_{0}^{2\pi}\frac{1}{2}\Big(1-\cos\big(2\phi\big)\Big)d\phi \\
&=0-\int_{1}^{-1}\Bigl(1-u^{2}\Bigr)du\times\biggl[\frac{1}{2}t-\frac{1}{4}\sin\bigl(2\phi\bigr)\biggr]_{0}^{2\pi} \\
&=\left[\begin{matrix}{u-u^{3}}\\\end{matrix}\right]_{-1}^{1}\times\pi=\frac{4}{3}\pi,
\end{aligned}$$
where we use the \(\text{u -substitution }u=\cos\theta\) and the fact that \(\sin^2(t)=\frac{1+\cos(2t)}{2}.\)