MATH1011 - Lecture 19

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MATH1011 - Lecture 18
Skeleton Slides

Parametric Form of a Surface

• See slides for full definition of a parametric form of a surface
  • A surface has many parametric equations
  • There is not a unique choice for $D$
    • However, often it is rectangular
      i.e. $a\le u\le b,c\le v\le d$
• A surface is specified by a vector function involving two parameters
  $\mathbf{S}(u,v)=(x(u,v),y(u,v),z(u,v)\text{position vector}$
• ${\mathbf{S}}_{\mathbf{u}}(u,v)$ and ${\mathbf{S}}_{\mathbf{u}}(u,v)$ lie in the tangent plane
  ${\mathbf{S}}_{\mathbf{u}}(u,v)=\frac{\partial \mathbf{S}}{\partial u}(u,v)\text{a tangent vector}$
  ${\mathbf{S}}_{\mathbf{v}}\mathbf{u}(u,v)=\frac{\partial \mathbf{S}}{\partial v}(u,v)\text{another tangent vector}$
• Thus the normal vector to the tangent plane at the point: $\mathbf{N}(u,v)={\mathbf{S}}_{\mathbf{u}}(u,v)\times {\mathbf{S}}_{\mathbf{v}}(u,v)$

Practice parametrising different surfaces, examples are on slides
Use your coordinate changes from previous lectures!
e.g. make $u=\theta$ and $v=\varphi$

Area of Surfaces

Proof in slides, utilising familiar concepts of Riemann sums and using the area of a parallelogram as approximation
$\Delta \mathbf{S}\approx |\mathbf{N}(u,v)|\Delta u\Delta v$
Hence total surface area of $S$: $\sum |\mathbf{N}(u_i,v_j)|\Delta u_i\Delta v_j$
Which becomes:
$\underset{D}{\iint }|\mathbf{N}(u,v)|dudv$

• Integral does not depend on the choice of parametrisation for $S$

Lecturer uses a speedy way of doing the cross product where he finds the determinant of a $3\times 3$ matrix with the top row being $i,j,k$ (the unit vector) and the other rows being made up of ${\mathbf{S}}_{\mathbf{u}}(u,v)$ and ${\mathbf{S}}_{\mathbf{v}}(u,v)$ arranged horizontally. $i,j,k$ are treated as being $1$ in our calculations.

e.g. $\left[\begin{array}{c}1\\ 0\\ 3\end{array}\right]\times \left[\begin{array}{c}2\\ 1\\ 4\end{array}\right]=\det \left[\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& 3\\ 2& 1& 4\end{array}\right]=\left[\begin{array}{c}0\cdot 4-3\cdot 1\\ 3\cdot 2-1\cdot 4\\ 1\cdot 1-0\cdot 2\end{array}\right]=\left[\begin{array}{c}-3\\ 2\\ 1\end{array}\right]$

This determinant can still be solved by excluding the top row from calculation (see the determinant). i.e. cross out top row and first column, multiply diagonal pairs and subtract 2nd pair from 1st pair, repeat with each column crossed out to get a vector with three components. May need to watch lecture again.

Surface Integrals

• A surface integral will calculate the area beneath a surface:
  $\underset{D}{\iint }f(\mathbf{S}(u,v))|\mathbf{N}(u,v)|dudv$
• $f(\mathbf{S}(u,v))$ just means $f(x,y,z)$ in terms of $u,v$

Vector Fields

• Vector fields assign to each point in ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$ a vector
• The various entries of the vector might describe different properties that the location possesses
  • The properties could be different at different locations
  • Thus, properties are dependent on the location

Circulation Along a Curve

• Lets say we want to calculate the work done by the wind in pushing a boat along the Swan river

  • We only want to consider the force that is tangent to the direction of $C$ at each point on the river
  • $C$ is the path of the boat

• Lets say that for a given point $\mathbf{r}(t)$ ($C$ in parametric form), the tangent component of $\mathbf{F}$ at $\mathbf{r}(t)$ is given by:
  $\mathbf{F}(\mathbf{r}(t))\cdot \mathbf{T}=\frac{\mathbf{F}(\mathbf{r}(t))\cdot \dot{\mathbf{r}}(t)}{|\dot{\mathbf{r}}(t)|}$

• $\dot{\mathbf{r}}(t)$ is the derivative of $\mathbf{r}(t)$

• To find out how much the wind force acts to push the boat along the curve, integrate the tangent component of the force along the curve $\underset{C}{\int }\mathbf{F}\cdot \mathbf{T}ds={\int }_a^b\mathbf{F}(\mathbf{r}(t))\cdot \dot{\mathbf{r}}(t)dt$

  • We describe this as the circulation of $\mathbf{F}$ along $C$
  • See slides for full simplification of expression

• In the context of our earlier scenario, this integral gives us work done

Closed Curve Convention

If $C$ is a closed curve in ${\mathbb{R}}^2$, we parametrise it anti-clockwise and write:
$\underset{C}{\oint }\mathbf{F}\cdot d\mathbf{r}$

If $C$ is given by a piecewise function, i.e. is made up of separate edges, the circulation of $\mathbf{F}$ along $C$ can still be found by integrating each function or ‘edge’ separately

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MATH1011 - Lecture 20
Annotated Slides