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

\frac{\partial f_{2}}{\partial x_{1}} - \frac{\partial f_{1}}{\partial x_{2}} = 1 .

One possibility is to choose

(f_{1}, f_{2}) := (0, x_{1}) .

Hence by Green’s theorem

\begin{aligned} area (D) & = \iint_{D} 1 d x_{1} d x_{2} \\ = \iint_{D} \frac{\partial}{\partial x_{1}} x_{1} - \frac{\partial}{\partial x_{2}} 0 d x_{1} d x_{2} \\ = \int_{\partial D} x_{1} d x_{2} . \end{aligned}

Another possibility is to choose

(f_{1}, f_{2}) := (- x_{2}, 0) .

Again Green’s theorem yields

\begin{aligned} area (D) & = \iint_{D} 1 d x_{2} d x_{1} \\ = \iint_{D} \frac{\partial}{\partial x_{1}} 0 + \frac{\partial}{\partial x_{2}} x_{2} d x_{1} d x_{2} \\ = - \int_{\partial D} x_{2} d x_{1} \end{aligned}

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

Suppose that

D

is a domain with a piecewise smooth boundary. Then its surface area is given by

area (D) = \frac{1}{2} \int_{\partial D} x_{1} d x_{2} - x_{2} d x_{1} .

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

Determine the area of the ellipse given by

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1 .

Answer.

The area is

a b π .

Solution.

A possible parametrisation of the boundary of the given ellipse is

(a \cos t, b \sin t), t \in [0, 2 π] .

According to the above formula the area is therefore given by

\begin{aligned} \frac{1}{2} \int_{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1} & x d y - y d x \\ = \frac{1}{2} \int_{0}^{2 π} a \cos t (b \cos t) - b \sin t (- a \sin t) d t \\ = \frac{a b}{2} \int_{0}^{2 π} \cos^{2} t + \sin^{2} t d t \\ = \frac{a b}{2} \int_{0}^{2 π} 1 d t = \frac{a b}{2} 2 π = a b π \end{aligned}

Hence the area of the ellipse is

π a b,

the same we got in Example 5.24 using polar coordinates and the transformation formula.