05 - Area between different coordinate systems

Area between different coordinate systems

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

Then
$A (S) = \iint_{(u, v) \in S} d u d v = \iint_{(x, y) \in R} \frac{\partial ( u , v )}{\partial ( x , y )} d x d y$
Similarly for the other way round
$A (R) = \iint_{(x, y) \in R} d x d y = \iint_{(u, v) \in S} \frac{\partial ( x , y )}{\partial ( u , v )} d u d v$

Proof - Sketch Proof

Consider the small element of area bounded by the coordinate lines
$u = u_{0} u = u_{0} + δ u v = v_{0} v = v_{0} + δ v$
Let $f (x_{0}, y_{0}) = (u_{0}, v_{0})$ and consider small changes $δ x$ and $δy$

Then we have a slightly distorted parallelogram with sides
$a b = f (x_{0} + δ x, y_{0}) - f (x_{0}, y_{0}) \approx \frac{\partial f}{\partial x} (x_{0}, y_{0}) δ x = f (x_{0}, y_{0} + δy) - f (x_{0}, y_{0}) \approx \frac{\partial f}{\partial y} (x_{0}, y_{0}) δy$
The area of this parallelogram in $R^{2}$ with sides $a, b$ is $∣ a \land b ∣$

Hence we have (ignoring higher terms)
$\frac{\partial f}{\partial x} δ x \land \frac{\partial f}{\partial y} δy = \frac{\partial f}{\partial x} \land \frac{\partial f}{\partial y} d x d y$
For $f = (u, v)$ such that $f_{x} = (u_{x}, v_{x})$ and $f_{y} = (u_{y}, v_{y})$ so
$\frac{\partial f}{\partial x} \land \frac{\partial f}{\partial y} = (\frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}) \land (\frac{\partial u}{\partial y}, \frac{\partial v}{\partial y}) = (\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial u}{\partial x}) k$
Hence
$δ A \approx \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x} d x d y = \frac{\partial ( u , v )}{\partial ( x , y )} δ x δy$

Oxford Maths Notes

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05 - Area between different coordinate systems

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