MATH1011 - Lecture 22

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MATH1011 - Lecture 21
Skeleton Slides

Green’s Theorem

• In ${\mathbb{R}}^2$, Green’s Theorem links circulation of a vector field around a closed curve with double integral over the region enclosed by the curve
  ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D\nabla \times \mathbf{F}dA={\iint }_D\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}dA$
• i.e. it relates circulation to the curl

Using Green’s Theorem can turn a difficult line integral (circulation) into an easier double integral over a region

Finding Area with Green’s Theorem

• Green’s Theorem can be used to find the area of a region $D$ of ${\mathbb{R}}^2$ by using only values on its boundary $C$: $\text{Area}(D)={\iint }_{D}dA={\oint }_C\mathbf{F}\cdot d\mathbf{r}$
  where $\mathbf{F}$ is a field such that $\nabla \times \mathbf{F}=1$
• This is because we know ${\iint }_{D}1dA$ represents area!
  • ${\iint }_{D}1dA={\oint }_C\mathbf{F}\cdot d\mathbf{r}$
• For these questions, you would usually calculate circulation
  • We can solve area of regions with smooth curves as the boundary
  • Reminder: $dr=\dot{\mathbf{r}}(t)$ which is the derivative of $r(t)$
• If you get a negative number, it means $r(t)$ was parametrised clockwise (we want anti-clockwise), take the absolute value of your answer

Green's Theorem with Holes

If the area you want to calculate has a hole/s, add the circulation parametrised in the clockwise direction of the areas you want to remove

This is also the same as just subtracting the anti-clockwise circulation of the holes (i.e. subtracting area of the holes)

If curl is zero, it makes sense that circulation would be zero too, but this is not the case if the region has holes! Even if $\nabla \times F=0$, the line integral can be non-zero!

Green’s Theorem for Flux

• Let $C$ be a piecewise curve in ${\mathbb{R}}^2$, and $D$ the region enclosed by $C$
  ${\oint }_C\mathbf{F}\cdot \mathbf{n}ds={\iint }_D\nabla \cdot \mathbf{F}dA={\iint }_D\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}dA$
• $\nabla \cdot \mathbf{F}$ is not curl but rather divergence

Stokes’ Theorem

• In ${\mathbb{R}}^3$, Stokes’ Theorem links the circulation of a vector field around a closed curve with a flux through the surface enclosed by the curve

A 2D figure is chiral if it cannot be superposed onto its mirror image
i.e. cannot obtain its mirror image through rotation and translation

A surface is orientable if a chiral 2D figure cannot be moved around the surface and back to where it started so that it looks like its mirror image

Orientable Surfaces

An easy way to tell if a surface is orientable is if you can paint one side red and the other side blue, and it’s always clear which is which — no flipping or ambiguity

If $S$ is bounded by $C$, then the orientation of $S$ induces a positive orientation of $C$ as follows:

• If one walks around $C$ in the positive direction with one’s head pointing in the normal direction $\mathbf{n}$ of $S$, then $S$ will always be on the left hand side

Stokes’ Theorem:

• In ${\mathbb{R}}^3$, let $S$ be an oriented piecewise smooth surface, bounded by a simple piecewise smooth closed curve $C$ with positive orientation
  ${\oint }_C\mathbf{F}\cdot dr={\iint }_S(\nabla \times \mathbf{F})\cdot dS$
• i.e. the total circulation of a vector field around the boundary of a surface equals the sum of the curl of the field over the entire surface
  • Stokes’ Theorem implies Green’s Theorem (see slides)

Stokes’ Theorem implies that if surfaces $S_1$ and $S_2$ have the same oriented boundary, then: ${\iint }_{S_1}(\nabla \times \mathbf{F})\cdot dS={\iint }_{S_2}(\nabla \times \mathbf{F})\cdot dS$
We can substitute a simpler surface to make questions easier. e.g. instead of solving the double integral for a dome, just solve for the disk at its base as they have the same boundary. Example in lecture recording.

Gauss’s Theorem

• In ${\mathbb{R}}^3$, let $E$ be a solid region and its boundary has outward orientation
  ${\iint }_S\mathbf{F}\cdot dS={\iiint }_E\nabla \cdot \mathbf{F}dV={\iiint }_E\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}dA$
• Gauss’s Theorem relates the outward flux through a bounding surface with a triple integral
  • i.e. relates outward flux to divergence

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Annotated Slides
End of Unit!