How to use Green’s theorem to evaluate a line integral

When a line integral follows a closed path, we can sometimes do the work over the region inside it instead. We’ll use Green’s theorem to make that switch and see what the result tells us about the path.

Check the curve and its orientation

Let $C$ be a simple, closed, piecewise smooth curve that bounds a region $R$ . Take $C$ counterclockwise, so the interior stays on your left as you travel. If $P$ and $Q$ have continuous first partial derivatives on an open set containing $R$ and its boundary, Green’s theorem gives

\oint_{C} P d x + Q d y = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A

We differentiate $Q$ with respect to $x$ , subtract the partial derivative of $P$ with respect to $y$ , and integrate over $R$ . For a clockwise curve, the line integral is the negative of the counterclockwise result.

Example: Find the circulation around a square

Take $P (x, y) = 2 x^{2} + 4 y$ and $Q (x, y) = x^{2} - 5 y^{3}$ . The curve goes counterclockwise around the square with both coordinates between $- 1$ and $1$ . The two functions are polynomials, so their first partial derivatives are continuous on and around the square.

On the counterclockwise path, the shaded square stays on your left.

\frac{\partial Q}{\partial x} = 2 x, \frac{\partial P}{\partial y} = 4

The new integrand is $2 x - 4$ . Since it’s constant with respect to $y$ , integrating in $y$ across an interval of length 2 gives $4 x - 8$ .

\begin{array}{l} \iint_{R} (2 x - 4) d A \\ = \int_{- 1}^{1} \int_{- 1}^{1} (2 x - 4) d y d x \\ = \int_{- 1}^{1} (4 x - 8) d x \\ = - 16 \end{array}

The integral of $4 x$ is zero on the symmetric interval, and the constant $- 8$ contributes $- 16$ . That’s the circulation for this vector field. For the opposite direction, the circulation is $16$ .

When does Green’s theorem give area?

To find area, the double integral needs integrand $1$ . We can choose $P = 0$ and $Q = x$ , so $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ . Then the counterclockwise integral $\oint_{C} x d y$ gives the area of $R$ . For this square, the area is $2 \cdot 2 = 4$ square units.

Learn mathKrista KingFebruary 16, 2021math, learn online, online course, online math, calculus 3, calculus iii, calc 3, calc iii, green's theorem, line integrals, multivariable calculus, multivariate calc, multivariate calculus, multivariable calc

How to use Green’s theorem to evaluate a line integral

Check the curve and its orientation

Example: Find the circulation around a square

When does Green’s theorem give area?

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