Application: The Area of a Domain.

Section 10.3 Application: The Area of a Domain.

Choosing an appropriate vector field we can derive a formula to compute the area of domain \(D\) with a piecewise boundary. The idea is to choose a vector field \((f_1,f_2)\) in such a way that

\begin{equation*} \frac{\partial f_2}{\partial x_1}-\frac{\partial f_1}{\partial x_2}=1\text{.} \end{equation*}

One possibility is to choose \((f_1,f_2):=(0,x_1)\text{.}\) Hence by Green’s theorem

\begin{align*} \area(D)\amp=\iint_D1\,dx_1\,dx_2\\ \amp=\iint_D\frac{\partial}{\partial x_1}x_1 -\frac{\partial}{\partial x_2}0\,dx_1\,dx_2\\ \amp=\int_{\partial D}x_1dx_2\text{.} \end{align*}

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Another possibility is to choose \((f_1,f_2):=(-x_2,0)\text{.}\) Again Green’s theorem yields

\begin{align*} \area(D)\amp=\iint_D1\,dx_2\,dx_1\\ \amp=\iint_D\frac{\partial}{\partial x_1}0 +\frac{\partial}{\partial x_2}x_2\,dx_1\,dx_2\\ \amp=-\int_{\partial D}x_2dx_1 \end{align*}

Adding these two formulas and dividing by two we obtain the following formula for the area of a domain.

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Proposition 10.9. Area of region via Green’s theorem.

Suppose that \(D\) is a domain with a piecewise smooth boundary. Then its surface area is given by

\begin{equation*} \area(D)=\frac{1}{2}\int_{\partial D}x_1\,dx_2-x_2\,dx_1\text{.} \end{equation*}

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We use the above formula because it is invariant under rotations of the coordinate system, whereas the other two are not.

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Example 10.10. Area via Green’s theorm.

Determine the area of the ellipse given by

\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\text{.} \end{equation*}

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Answer.

The area is \(ab\pi\text{.}\)

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Solution.

A possible parametrisation of the boundary of the given ellipse is

\begin{equation*} (a\cos t,b\sin t),\qquad t\in[0,2\pi]\text{.} \end{equation*}

According to the above formula the area is therefore given by

\begin{align*} \frac 12\int_{\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1}\amp x\,dy-y\,dx\\ &=\frac 12\int_0^{2\pi}a\cos t(b\cos t)-b\sin t(-a\sin t)\,dt\\ &=\frac {ab}2\int_0^{2\pi}\cos^2 t+\sin^2 t\,dt\\ &=\frac {ab}2\int_0^{2\pi}1\,dt =\frac {ab}2 2\pi =ab\pi \end{align*}

Hence the area of the ellipse is \(\pi ab\text{,}\) the same we got in Example 5.24 using polar coordinates and the transformation formula.

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