# Math2111: Chapter 4: Surface integrals. Section 1: Parameterisations of surfaces

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

Previously we have seen parameterised curves. We now generalise this idea to surfaces.

Definition
Let $D\subseteq \mathbb{R}^2$ be Jordan measurable. By a parameterised surface we mean a continuously differentiable function $\boldsymbol{\Phi}:D\to\mathbb{R}^3$. The image $\boldsymbol{\Phi}(D)$ is called surface.

Notice that we included the condition that $\boldsymbol{\Phi}$ is continuously differentiable in the definition above since we only consider such parameterisations. One could of course consider more general surfaces.

As a first example, consider a continuously differentiable function $f:D\subset\mathbb{R}^2\to\mathbb{R}$. By setting

$\boldsymbol{\Phi}(x,y)=x \widehat{\boldsymbol{i}} + y \widehat{\boldsymbol{j}} + f(x,y) \widehat{\boldsymbol{k}} \hspace{2cm} (1)$

we obtain a parameterisation of the graph of ${}f$.

Not all surfaces can be parameterised as in (1) since not all surfaces can be represented by a function. For instance a sphere cannot be represented by a function since there are points $(x,y)$ for which one would have to assign two values (for a sphere of radius ${}1$ at $(0,0)$ this would be $-1$ and ${}1$).

Example
A sphere of radius $r$ centered at $(0,0)$ given by $x^2+y^2+z^2=r^2$ can be parameterised by

$\displaystyle \boldsymbol{\Phi}(\theta, \phi) = (r \cos \theta \sin \phi, r \sin\theta\sin\phi, r \cos\phi),$

with $0 \le\theta\textless 2\pi$ and $0\le \phi \le \pi.$ $\Box$

Example
Consider the cone $z^2 = x^2+y^2$ with $0 \le z\le 1.$ We can parameterise this surface by $\boldsymbol{\Phi}(u,v)= (u \cos v, u \sin v, u)$ with $0 \le v \textless 2\pi$ and $0 \le u \le 1.$ $\Box$

Hence, in the following we deal with functions of the form

$\displaystyle \boldsymbol{\Phi}(u,v)= X(u,v)\widehat{\boldsymbol{i}} + Y(u,v) \widehat{\boldsymbol{j}} + Z(u,v) \widehat{\boldsymbol{k}},$

where $X, Y, Z: D\to\mathbb{R}$ are continuously differentiable functions. The derivatives of $\boldsymbol{\Phi}$ are given by

$\displaystyle \begin{array}{rcl} \frac{\partial \boldsymbol{\Phi}}{\partial u} & = & \frac{\partial X}{\partial u} \widehat{\boldsymbol{i}} + \frac{\partial Y}{\partial u} \widehat{\boldsymbol{j}} + \frac{\partial Z}{\partial u} \widehat{\boldsymbol{k}} \\ && \\ \frac{\partial \boldsymbol{\Phi}}{\partial v} & = & \frac{\partial X}{\partial v} \widehat{\boldsymbol{i}} + \frac{\partial Y}{\partial v} \widehat{\boldsymbol{j}} + \frac{\partial Z}{\partial v} \widehat{\boldsymbol{k}}. \end{array}$

Notice that the vectors $\frac{\partial \boldsymbol{\Phi}}{\partial u}$ and $\frac{\partial \boldsymbol{\Phi}}{\partial v}$ are tangent to the surface.

The following notation is convenient to find normal vectors to the surface:

$\displaystyle \begin{array}{rcl} \frac{\partial (Y,Z)}{\partial (u,v)} & := & \left|\begin{array}{rr} \frac{\partial Y}{\partial u} & \frac{\partial Y}{\partial v} \\ & \\ \frac{\partial Z}{\partial u} & \frac{\partial Z}{\partial v} \end{array} \right| = \frac{\partial Y}{\partial u} \frac{\partial Z}{\partial v} - \frac{\partial Y}{\partial v}\frac{\partial Z}{\partial u} \\ && \\ \frac{\partial (Z,X)}{\partial (u,v)} & := & \left|\begin{array}{rr} \frac{\partial Z}{\partial u} & \frac{\partial Z}{\partial v} \\ & \\ \frac{\partial X}{\partial u} & \frac{\partial X}{\partial v} \end{array} \right| = \frac{\partial Z}{\partial u} \frac{\partial X}{\partial v} - \frac{\partial Z}{\partial v}\frac{\partial X}{\partial u} \\ && \\ \frac{\partial (X,Y)}{\partial (u,v)} & := & \left|\begin{array}{rr} \frac{\partial X}{\partial u} & \frac{\partial X}{\partial v} \\ & \\ \frac{\partial Y}{\partial u} & \frac{\partial Y}{\partial v} \end{array} \right| = \frac{\partial X}{\partial u} \frac{\partial Y}{\partial v} - \frac{\partial X}{\partial v}\frac{\partial Y}{\partial u} \end{array}$

Indeed, it can be verified that the normal vector $\boldsymbol{n}=\boldsymbol{n}(u,v)$ is given by

$\displaystyle \boldsymbol{n}= \frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v} = \frac{\partial (Y,Z)}{\partial (u,v)} \widehat{\boldsymbol{i}} + \frac{\partial (Z,X)}{\partial (u,v)} \widehat{\boldsymbol{j}} + \frac{\partial (X,Y)}{\partial (u,v)} \widehat{\boldsymbol{k}}.$

Example
Let $f:D\subset\mathbb{R}^2 \to\mathbb{R}$ where $f=f(u,v).$ Then the normal vector is given by

$\displaystyle \boldsymbol{n}= \frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v} = -\frac{\partial f}{\partial u} \widehat{\boldsymbol{i}} - \frac{\partial f}{\partial v} \widehat{\boldsymbol{j}} + \widehat{\boldsymbol{k}}.$

$\Box$

The surfaces which we consider in the following sections are assumed to be piecewise smooth, that is, they are assumed to be images of parameterised surfaces $\boldsymbol{\Phi}_i:D_i\to\mathbb{R}^3$ for which:

• $D_i$ is a Jordan-measurable subset of $\mathbb{R}^2$;
• the mappings $\boldsymbol{\Phi}_i$ are one-to-one;
• the normal vector $\boldsymbol{n}_i=\frac{\partial \boldsymbol{\Phi}_i}{\partial u} \times \frac{\partial \boldsymbol{\Phi}_i}{\partial v} \neq \boldsymbol{0}$ except possibly at a finite number of points;

A geometric explanation of parameterised surfaces is given in the following video.