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*}
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.
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*}
We use the above formula because it is invariant under rotations of the coordinate system, whereas the other two are not.
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*}
Answer.
The area is \(ab\pi\text{.}\)
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.
