Using Green's theorem to find area

Using Green's theorem to find area

Typically we use Green's theorem to calculate as an alternative to calculate some line integral $\int_{{C}}^{}$ F^.ds. If, for example, we are in two dimension, C is a simple closed curve, and F(x, y) is defined everywhere inside C, we can use Green's theorem to convert the line integral into to double integral. Instead of calculating line integral $\int_{{C}}^{}$ F^.ds directly, we calculate the double integral

$\displaystyle \iint_{D}^{}$ $\displaystyle \left(\vphantom{\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}}\right.$ $\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$ $\displaystyle \left.\vphantom{\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}}\right)$ dA

Can we use Green's theorem to go the other direction? If we are given a double integral, can we use Green's theorem to convert the double integral into a line integral and calculate the line integral? If we are given the double integral

$\displaystyle \iint_{D}^{}$ f (x, y)dA,

we can use Green's theorem only if there happens to be a vector field F(x, y) so that

f (x, y) = $\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$ .

However, we haven't learned any method to find such a vector field F. So, we aren't likely to use Green's theorem in this direction very often.

There is one important exception to this rule, however, and that is when we are using a double integral to calculate the area of a region D. Recall that the area of a region D is equal to the double integral of f (x, y) = 1 over D:

Area of D = $\displaystyle \iint_{D}^{}$ dA = $\displaystyle \iint_{D}^{}$ 1 dA.

If f (x, y) = 1, it is easy to find a vector field F so that

$\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$ = f (x, y) = 1.

There are many such vector field F, but we'll pick the vector field F(x, y) = (- y/2, x/2). You can confirm that indeed $\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$ = 1.

In summary, if C is a counterclockwise oriented simple closed curve that bounds a region where you can apply Green's theorem, the area of the region D bounded by C = $\partial$ D is

Area of D = $\displaystyle \int_{{C}}^{}$ F^.ds = $\displaystyle {\frac{{1}}{{2}}}$ $\displaystyle \int_{C}^{}$ x dy - y dx,

where F(x, y) = (- y/2, x/2).

Example

Use Green's Theorem to calculate the area of the disk D defined by x² + y² $\le$ 4.

Solution: Since we know the area of the disk of radius 2 is 4 $\pi$ , we better get 4 $\pi$ for our answer.

The boundary of D is the circle of radius 4. We can parametrized it in a counterclockwise orientation using

c(t) = (2 cos t, 2 sin t), 0 $\displaystyle \le$ t $\displaystyle \le$ 2 $\pi$ .

Then

c'(t) = (- 2 sin t, 2 cos t),

and, by Green's theorem,

area of D	= $\displaystyle \iint$ dA
	= $\displaystyle {\frac{{1}}{{2}}}$ $\displaystyle \int_{C}^{}$ x dy - y dx
	= $\displaystyle {\frac{{1}}{{2}}}$ $\displaystyle \int_{0}^{{2\pi}}$ [(2 cos t)(2 cos t) - (2 sin t)(- 2 sin t)]dt
	= $\displaystyle {\frac{{1}}{{2}}}$ $\displaystyle \int_{0}^{{2\pi}}$ 4(sin²t + cos²t)dt = 2 $\displaystyle \int_{0}^{{2\pi}}$ dt = 4 $\pi$ .

Thankfully, our answer agrees with what we know it should be.

Duane Nykamp
nykamp@math.umn.edu
2005-10-27