Let's start with a simple question.
Imagine you have a large, swirling pool of water. You want to measure "how much" the water is circulating. You can do this in two ways:
You could walk in a big circle (a closed path $C$) around the edge of the pool. At every step, you measure how much the water's current is either pushing you along your path or pushing against you.
If, by the end, you were pushed along more than you were pushed against, you'd say, "Yep, this water is definitely circulating!" This measurement is a line integral over a closed path: $\oint_C \mathbf{F} \cdot d\mathbf{r}$.
You could ignore the boundary and look at the water inside the pool. At every single point in the pool, you could drop a tiny, microscopic leaf and measure how much it's spinning in place. This "micro-swirl" at a point is a new idea.
If, on average, all your tiny leaves are spinning counter-clockwise, you could sum up all that spinning, and you'd also say, "Yep, this water is definitely circulating!" This measurement is a double integral over the region D: $\iint_D (\text{micro-swirl}) \, dA$.
Green's Theorem states that these two measurements give you the exact same number.
The total, "macroscopic" circulation you feel on the 1D boundary is the exact sum of all the "microscopic" swirls happening at every point on the 2D region inside.
Here's the theorem in its full glory. Let $D$ be a 2D region (the "stuff inside") and let $C$ be its boundary (the "closed loop").
$$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $$
Let's break that down, piece by piece.
This formula is the heart of the whole thing. But what on earth does it mean?
Let's understand that "micro-swirl" formula: $\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)$.
Imagine you drop a tiny, microscopic water wheel (or a leaf) at a point $(x,y)$ in your vector field $\mathbf{F} = \langle P, Q \rangle$. This wheel can spin.
Two things can make it spin:
$Q$ is the vertical component of the water flow. $\frac{\partial Q}{\partial x}$ measures how this vertical flow changes as you move horizontally.
If the water on your right (at $x+dx$) is pushing up more than the water on your left is, it will hit the right side of your wheel and spin it counter-clockwise. This is a positive $\frac{\partial Q}{\partial x}$.
$P$ is the horizontal component of the water flow. $\frac{\partial P}{\partial y}$ measures how this horizontal flow changes as you move vertically.
If the water above you (at $y+dy$) is pushing right more than the water below you is, it will hit the top of your wheel and spin it clockwise. This is a positive $\frac{\partial P}{\partial y}$.
The total "micro-swirl" is the sum of these two effects. Since $\frac{\partial Q}{\partial x}$ causes counter-clockwise spin (positive) and $\frac{\partial P}{\partial y}$ causes clockwise spin (negative), the net spin is:
$$ \text{Net Spin} = \left(\frac{\partial Q}{\partial x}\right) - \left(\frac{\partial P}{\partial y}\right) $$This is the 2D curl! It's a single number that tells us how much and in what direction the field is swirling at that exact point.
So, why is the sum of all those internal "micro-swirls" equal to the big integral on the boundary? Let's show the intuition for why this is true:
$$ \underbrace{\iint_D (Q_x - P_y)\,dA}_{\text{Sum of all "micro-swirls"}} = \underbrace{\oint_C P\,dx + Q\,dy}_{\text{The "macro-swirl" on the boundary}} $$
We'll show these are two different ways of calculating the same thing.
First, let's ignore the double integral. Picture your region $D$ tiled with a grid of tiny squares. Let's calculate the small counter-clockwise line integral $\oint$ around every single tiny square and add them all up:
$$ \text{Total} = \sum_{\text{all squares}} \left( \oint_{\text{tiny square}} \vec{F} \cdot d\vec{r} \right) $$
What happens at the shared edges?
Since they travel the same edge in opposite directions, the work done is exactly opposite. The two line integrals for that internal edge are perfect negatives, and they cancel to zero.
This happens for every single internal edge in the entire grid. The only edges that do not get canceled are the ones on the extreme outer edge of the region $D$.
So, the sum of all the tiny line integrals just simplifies to the line integral around the main, outer boundary $C$:
$$ \sum_{\text{all squares}} \left( \oint_{\text{tiny square}} \vec{F} \cdot d\vec{r} \right) = \oint_C \vec{F} \cdot d\vec{r} $$
Now, let's look at the other side of Green's Theorem: the double integral $\iint_D (Q_x - P_y)\,dA$.
The quantity $Q_x - P_y$ is the "swirl density". It's a function that tells us how much the field is swirling *per unit area* at any given point.
A double integral $\iint_D (\text{density})\,dA$ is, by definition, the way we add up the total "stuff" in a region.
Therefore, $\iint_D (Q_x - P_y)\,dA$ is, by definition, the total swirl in the region, found by summing up the "micro-swirl" density at every point.
We've just shown that:
The final logical step is realizing that the "tiny boundary integral" *is* the "micro-swirl". The "swirl density" $Q_x - P_y$ at a point is *defined* as the circulation (line integral) around a tiny loop at that point, divided by the area of that loop.
So, $\oint_{\text{tiny loop}} \vec{F} \cdot d\vec{r} \approx (Q_x - P_y)\Delta A$.
This makes our two sums from Part 1 and Part 2 identical. Both methods—summing the tiny boundary integrals (which cancel) or summing the swirl density (the double integral)—are just two different ways of adding up all the "micro-swirls" in the region. Since they must be equal, the theorem holds.
Green's Theorem is a "magical calculator" that lets you trade one hard integral for another, hopefully easier, one.
You have a nasty line integral over a complicated boundary (like a triangle, which is 3 separate integrals). You can use Green's Theorem to turn it into one double integral over the simple region inside.
If the curl $\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$ is a simple constant (like 5), this is a huge win!
You have a hard double integral to find. You can sometimes run Green's Theorem in reverse, turning it into a much simpler line integral on the boundary. This is famously used to find the area of a weird region $D$:
We know $Area(D) = \iint_D 1 \, dA$.
Can we find a field $\mathbf{F} = \langle P, Q \rangle$ whose curl is 1?
Sure! Let $\mathbf{F} = \langle 0, x \rangle$. Then:
Therefore, $Area(D) = \oint_C x \, dy$.
This is amazing. It means you can find the 2D area of a shape by just walking its 1D boundary!
This theorem is the 2D "Fundamental Theorem of Calculus" and it's the foundation for the two bigger 3D theorems to come: Stokes' Theorem and the Divergence Theorem.
Evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = \langle y^3, -x^3 \rangle$ and $C$ is the unit circle $x^2 + y^2 = 1$, oriented counter-clockwise.
Method 1: The "Hard Way" (Direct Line Integral)
First, we parametrize $C$: $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ for $0 \le t \le 2\pi$.
This gives $\mathbf{r}'(t) = \langle -\sin t, \cos t \rangle$.
Next, we evaluate $\mathbf{F}$ along the path: $$ \mathbf{F}(\mathbf{r}(t)) = \langle (\sin t)^3, -(\cos t)^3 \rangle = \langle \sin^3 t, -\cos^3 t \rangle $$
Now, we compute the dot product: $$ \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \langle \sin^3 t, -\cos^3 t \rangle \cdot \langle -\sin t, \cos t \rangle $$ $$ = -\sin^4 t - \cos^4 t = -(\sin^4 t + \cos^4 t) $$
The integral is $W = \int_0^{2\pi} -(\sin^4 t + \cos^4 t) \, dt$. This is a very difficult trigonometric integral. We'd have to use power-reducing formulas multiple times.
Method 2: The "Green's Theorem Way" (Easy Double Integral)
Let's use Green's Theorem. The region $D$ is the unit disk $x^2 + y^2 \le 1$.
Our field is $\mathbf{F} = \langle P, Q \rangle = \langle y^3, -x^3 \rangle$.
Let's find the 2D Curl:
$$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(-x^3) - \frac{\partial}{\partial y}(y^3) $$
$$ = (-3x^2) - (3y^2) = -3(x^2 + y^2) $$
Now, we just integrate this curl over the disk $D$: $$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D -3(x^2 + y^2) \, dA $$
This double integral is perfect for polar coordinates.
Let $x = r \cos \theta$ and $y = r \sin \theta$.
The region $D$ is $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.
The integrand $x^2 + y^2$ becomes $r^2$.
The area element $dA$ becomes $r \, dr \, d\theta$.
$$ \int_0^{2\pi} \int_0^1 -3(r^2) \cdot (r \, dr \, d\theta) = \int_0^{2\pi} \int_0^1 -3r^3 \, dr \, d\theta $$
$$ = \int_0^{2\pi} \left[ -\frac{3}{4}r^4 \right]_0^1 \, d\theta = \int_0^{2\pi} \left( -\frac{3}{4} - 0 \right) \, d\theta $$
$$ = -\frac{3}{4} \int_0^{2\pi} d\theta = -\frac{3}{4} [\theta]_0^{2\pi} = \boxed{-\frac{3\pi}{2}} $$
This was much, much easier than the line integral!
Find the area of the ellipse $D$ given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
We want to find $Area = \iint_D 1 \, dA$. This is the "hard" integral. We can use Green's Theorem in reverse to turn this into an "easy" line integral.
We need a vector field $\mathbf{F} = \langle P, Q \rangle$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$.
As we saw in the "Why We Care" section, there are several choices. A simple one is $\mathbf{F} = \langle 0, x \rangle$.
By Green's Theorem, $Area(D) = \iint_D 1 \, dA = \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C P\,dx + Q\,dy = \oint_C x \, dy$.
Now, we just need to parametrize the boundary $C$ (the ellipse).
The standard parametrization is $\mathbf{r}(t) = \langle a \cos t, b \sin t \rangle$ for $0 \le t \le 2\pi$.
This gives us our components in terms of $t$:
$x(t) = a \cos t$
$y(t) = b \sin t \implies dy = b \cos t \, dt$
Now we substitute these into our new line integral $\oint_C x \, dy$: $$ Area = \int_0^{2\pi} \underbrace{(a \cos t)}_{x} \cdot \underbrace{(b \cos t \, dt)}_{dy} $$ $$ = \int_0^{2\pi} ab \cos^2 t \, dt = ab \int_0^{2\pi} \cos^2 t \, dt $$
Using the power-reducing formula $\cos^2 t = \frac{1}{2}(1 + \cos(2t))$: $$ = ab \int_0^{2\pi} \frac{1}{2}(1 + \cos(2t)) \, dt $$ $$ = \frac{ab}{2} \left[ t + \frac{1}{2}\sin(2t) \right]_0^{2\pi} $$
$$ = \frac{ab}{2} \left( \left( 2\pi + \frac{1}{2}\sin(4\pi) \right) - \left( 0 + \frac{1}{2}\sin(0) \right) \right) $$ $$ = \frac{ab}{2} \left( (2\pi + 0) - (0 + 0) \right) = \frac{ab}{2} (2\pi) = \boxed{\pi ab} $$
Use Green's Theorem to evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = \langle y, 2x \rangle$ and $C$ is the boundary of the rectangle $D = [0, 2] \times [0, 3]$, oriented counter-clockwise.
We want to use $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$.
Therefore, the value of the line integral is $\boxed{6}$.
(This is much faster than parametrizing all four sides of the rectangle and computing four separate line integrals!)
We can't prove the theorem for all possible regions (like regions with holes), but we can prove it for a "simple" region $D$—one that can be described as both Type I and Type II.
The strategy is to prove the theorem in two separate halves:
Adding these two equations together will give us the full theorem.
First, let's evaluate the double integral.
We treat $D$ as a Type I region, defined by $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$.
By the Fundamental Theorem of Calculus, the inner integral (with respect to $y$) is:
$$ \int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y} \, dy = \left[ P(x,y) \right]_{y=g_1(x)}^{y=g_2(x)} = P(x, g_2(x)) - P(x, g_1(x)) $$Substituting this back in and distributing the negative sign, we get:
$$ -\iint_D \frac{\partial P}{\partial y} \, dA = -\int_a^b \left[ P(x, g_2(x)) - P(x, g_1(x)) \right] \, dx $$ $$ = \int_a^b P(x, g_1(x)) \, dx - \int_a^b P(x, g_2(x)) \, dx $$This is our result from the right-hand side. Now we'll evaluate the left-hand side.
Next, let's evaluate the line integral.
We need to evaluate $\oint_C P\,dx$ along the boundary $C$. We break $C$ into four pieces:
On $C_1$: We can parametrize by $x$ itself. Let $x=t$ (so $dx=dt$) and $y=g_1(t)$, from $a$ to $b$. $$ \int_{C_1} P\,dx = \int_a^b P(t, g_1(t)) \, dt $$
On $C_3$: We parametrize as $x=t$ (so $dx=dt$) and $y=g_2(t)$. But for positive orientation, we must go from $x=b$ to $x=a$. $$ \int_{C_3} P\,dx = \int_b^a P(t, g_2(t)) \, dt = -\int_a^b P(t, g_2(t)) \, dt $$
On $C_2$ and $C_4$: These are vertical lines, so $x$ is constant. If $x$ is constant, then $dx = 0$. Therefore, the integral $\int P\,dx$ is zero on both $C_2$ and $C_4$.
Adding them up: $$ \oint_C P\,dx = \int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4} $$ $$ \oint_C P\,dx = \int_a^b P(t, g_1(t)) \, dt + 0 - \int_a^b P(t, g_2(t)) \, dt + 0 $$
Comparing our results for the double integral and the line integral, we see they are identical. Part 1 is proven.
The logic is exactly the same, but now we treat $D$ as a Type II region, defined by $c \le y \le d$ and $h_1(y) \le x \le h_2(y)$.
The double integral: $$ \iint_D \frac{\partial Q}{\partial x} \, dA = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} \frac{\partial Q}{\partial x} \, dx \right] \, dy $$ $$ = \int_c^d \left[ Q(x,y) \right]_{x=h_1(y)}^{x=h_2(y)} \, dy = \int_c^d \left( Q(h_2(y), y) - Q(h_1(y), y) \right) \, dy $$
The line integral:
We evaluate $\oint_C Q\,dy$. This time, the integrals on the horizontal paths ($C_1$ and $C_3$) are zero, because $y$ is constant, so $dy = 0$.
The integral along the right side $C_2$ (path $x=h_2(y)$) and the left side $C_4$ (path $x=h_1(y)$) will, after parametrization and accounting for orientation, sum to: $$ \oint_C Q\,dy = \int_c^d Q(h_2(y), y) \, dy - \int_c^d Q(h_1(y), y) \, dy $$
Again, the two sides are identical. Part 2 is proven.
We have shown:
1. $\oint_C P\,dx = -\iint_D \frac{\partial P}{\partial y} \, dA$
2. $\oint_C Q\,dy = \iint_D \frac{\partial Q}{\partial x} \, dA$
Adding these two equations together gives:
$$ \oint_C P\,dx + \oint_C Q\,dy = -\iint_D \frac{\partial P}{\partial y} \, dA + \iint_D \frac{\partial Q}{\partial x} \, dA $$Combining the integrals gives the final theorem:
$$ \oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $$This proof also works for more complex regions by "chopping" them into multiple simple regions. All the internal boundary lines that we create cancel out, leaving only the outer boundary, just like in our "Cancellation Idea."