06. Double Integrals

Lots of examples so you can refer to page $41$+ at the lecture notes

Area

Let $R\subset {\mathbb{R}}^2$ then

Area of $R$ is

$$A(R)={\iint }_{(x,y)\subset R}dxdy$$

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05 - Area between different coordinate systems

Area between different coordinate systems

Let $f:R\to S$ be a bijection between two regions of ${\mathbb{R}}^2$
Suppose $(u,v)=f(x,y)$ and $\frac{\partial (u,v)}{\partial (x,y)}$ is defined and non-zero everywhere

Then

$$A(S)={\iint }_{(u,v)\in S}dudv={\iint }_{(x,y)\in R}\left|\frac{\partial (u,v)}{\partial (x,y)}\right|dxdy$$

Similarly for the other way round

$$A(R)={\iint }_{(x,y)\in R}dxdy={\iint }_{(u,v)\in S}\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv$$

Proof - Sketch Proof

Consider the small element of area bounded by the coordinate lines

$$u=u_{0}u=u_0+\delta uv=v_{0}v=v_0+\delta v$$

Let $f(x_0,y_0)=(u_0,v_0)$ and consider small changes $\delta x$ and $\delta y$

Then we have a slightly distorted parallelogram with sides

$$\begin{array}{rl}\mathbf{a}& =f(x_0+\delta x,y_0)-f(x_0,y_0)\approx \frac{\partial f}{\partial x}(x_0,y_0)\delta x\\ \mathbf{b}& =f(x_0,y_0+\delta y)-f(x_0,y_0)\approx \frac{\partial f}{\partial y}(x_0,y_0)\delta y\end{array}$$

The area of this parallelogram in ${\mathbb{R}}^2$ with sides $\mathbf{a},\mathbf{b}$ is $|\mathbf{a}\wedge \mathbf{b}|$

Hence we have (ignoring higher terms)

$$\left|\frac{\partial f}{\partial x}\delta x\wedge \frac{\partial f}{\partial y}\delta y\right|=\left|\frac{\partial f}{\partial x}\wedge \frac{\partial f}{\partial y}\right|dxdy$$

For $f=(u,v)$ such that $f_x=(u_x,v_x)$ and $f_y=(u_y,v_y)$ so

$$\frac{\partial f}{\partial x}\wedge \frac{\partial f}{\partial y}=\left(\frac{\partial u}{\partial x},\frac{\partial v}{\partial x}\right)\wedge \left(\frac{\partial u}{\partial y},\frac{\partial v}{\partial y}\right)=\left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial u}{\partial x}\right)\mathbf{k}$$

Hence

$$\begin{array}{rl}\delta A& \approx \left|\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right|dxdy\\ & =\left|\frac{\partial (u,v)}{\partial (x,y)}\right|\delta x\delta y\end{array}$$

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