Green's Theorem (Fully Explained w/ Step-by-Step Examples!)

What if there is a way to link line integrals around a simple closed curve and a double integral over the plane region it encloses.

Jenn (B.S., M.Ed.) of Calcworkshop® teaching green's theorem

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

With Green’s theorem, we can!

Terminology

But before we can understand Green’s theorem and its extraordinary implications, we need to understand some important terminology fully.

A simple curve is a curve that does not intersect itself anywhere, whereas a simple closed curve is a simple curve whose initial and terminal points are the same.

Simple Curve Examples

Therefore, a region $D$ is called simply connected if every closed curve in $D$ can be shrunk to a point by curves staying in $D$ .

Simply Connected Regions

So, simply connected means the region is closed, simple, and without holes.

But now we have a bit of a dilemma. How do we traverse and orient ourselves when discussing a simply connected region?

In Green’s theorem, we use the convention that the positive orientation of a simple closed curve $C$ is the single counterclockwise (CCW) traversal of $C$ .

Positive Negative Orientation Curve

But sometimes, this isn’t always easy to determine, so here’s a little hint!

Imagine walking along the simple closed curve $C$ . As long as your left hand is touching the region $D$ , you are traversing in a positive direction.

Positive Orientation Left Hand Touching Curve

Green’s Theorem

Alright, so now we’re ready for Green’s theorem.

Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane and let $D$ be the region bounded by $C$ . If $P$ and $Q$ have continuous first-order partial derivatives on an open region that contains $D$ , then:

$\int_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A = \iint_{D} (Q_{x} - P_{y}) d A$

And sometimes, we use the following notation to indicate that the line integral is calculated using the positive orientation of the closed curve $C$ .

$\oint_{C} P d x + Q d y or \oint_{C} P d x + Q d y$

So, as long as the curve is closed (this is imperative in order to use Green’s theorem), we can use a double integral to evaluate a line integral!

Example

Let’s see how with an example.

Evaluate the line integral $\oint_{C} y^{2} d x + 3 x y d y$ if the boundary is the semiannular region $D$ in the upper half-plane between the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} = 4$ shown below.

Positive Semi Annular Region

Before we go any further, we must decide how to evaluate, as there are two choices.

If we choose to evaluate using a line integral, we will need to represent $C$ as a piecewise-smooth curve, such that:

Green’s Theorem Semi Annular Region

$\int_{C} P d x + Q d y = \int_{C_{1}} P d x + Q d y + \int_{C_{2}} P d x + Q d y + \int_{C_{3}} P d x + Q d y + \int_{C_{4}} P d x + Q d y$

Ugh! That looks messy and quite tedious.

Thankfully, there’s an easier way.

Because our integration notation $\oint$ tells us we are dealing with a positively oriented, closed curve, we can use Green’s theorem!

$\int_{C} P d x + Q d y = \iint_{D} (Q_{x} - P_{y}) d A$

First, we will find our first partial derivatives.

$\oint \underset{P}{\underset{⏟}{y^{2}}} d x + \underset{Q}{\underset{⏟}{3 x y}} d y$

Thus,

$\oint_{C} y^{2} d x + 3 x y d y = \iint_{D} ((3 y) - (2 y)) d A = \iint_{D} (y) d A$

Next, we will notice that it would be far easier to evaluate our double integral in polar coordinates since our region is circular.

$D = {(r, θ) : 1 \leq r \leq 2, 0 \leq θ \leq π}$

Now, all we have to do is integrate our iterated integral. Don’t forget your Jacobian transformation $r d r d θ$

$\begin{aligned} \oint_{C} y^{2} d x + 3 x y d y & = \iint_{D} (y) d A \\ = \int_{0}^{π} \int_{1}^{2} r \sin θ \cdot r d r d θ \\ = \frac{14}{3} \end{aligned}$

Isn’t Green’s theorem remarkable?

Throughout the lesson, we will show how to apply Green’s theorem to non-simply connected regions. I like to think of them as those “tricky” regions with holes!

And we will quickly discover that the double integral can be way easier to evaluate than a line integral for positively oriented, piecewise smooth, simple, closed curves.

But Green’s theorem does more for us than simply making integration of line integrals easier, as it is one of the most pivotal theorems in vector calculus.

This theorem is useful in finding the amount of work that is done in moving a particle around a curve, which is similar to the way fluid flows around a curve, known as circulation or flux, and will be very important in our future classes.

So, let’s jump right into our lesson and discover how to evaluate line integrals along a positively oriented curve, find the work done by a force, the area of a given region, and so much more.

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

Greens-Theorem-Example

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