Green’s Theorem—Applications
Learning Goals: Green’s theorem is used to compute area; the tangent vector to the boundary is
rotated 90° clockwise to become the outward-pointing normal vector, to derive the divergence
form of Green’s theorem.
Theorem: Let D be a region to which Green’s Theorem applies, and let C be its positively
oriented boundary. Then ∫ x dy is the area of the region.
C
∫
Proof: applying Green’s Theorem,
The same also applies to
∫
−y dx ,
C
C
x dy = ∫
∂x
dA = ∫ dA = area.
D ∂x
D
1
x dy − y dx , and indeed any
2 ∫C
∫
C
P dx + Q dy where
∂Q ∂P
−
is equal to one. It is quite interesting that you can measure the area of a region just by
∂x ∂y
walking around it, measuring how far you move in the x-direction per unit moved in the ydirection.
Of course, this shouldn’t come as a big surprise.
What if you have a function f(x) defined on an
interval from a to b? Isn’t
∫
b
a
f (x)dx the area
underneath it? Well, if we go around the
boundary of this region, integrating –y dx we
will get 0 along all the straight sides (dx = 0
along the left and right edges, and y = 0 on the
bottom), while the last piece of the line integral
would parameterize the curve as (x, f(x)) as x
runs from b down to a, so
∫
C
a
−y dx = − ∫ f (x)dx
b
which is our regular one-variable integral.
Example: try computing the area of a circle.
Solution: let’s parameterize the circle as (cos(t), sin(t)), 0 ≤ t ≤ 2π. Then
computed as
∫
2π
0
∫
C
x dy would be
cos(t)cos(t)dt = π .
Example: If you take a circle of radius a and roll a smaller circle of radius a/4 around its inside,
tracing the trajectory of a point on the small circle as it rolls around, you obtain the hypocycloid
of four cusps. It happens to have equation x2/3 + y2/3 = a2/3. In parametric form, it has the
parameterization (a cos3(t), a sin3(t)). Find its area.
Solution: For reasons that will become clear, we will
1
use the form ∫ x dy − y dx . We compute:
2 C
dy = 3a sin2(t) cos(t) dt
dx = –3a cos2(t) sin(t) dt
x dy – y dx = 3a2 sin2(t) cos4(t) dt
+ 3a2 cos2(t) sin4(t) dt
Now we factor out 3a2 cos2(t) sin2(t), leaving
cos2(t) + sin2(t) = 1. Using the double-angle formula
for sine, this becomes 3a2/4 sin2(2t). So our job is to
3a 2 2 π 2
sin (2t)dt . The integral is a simple
integrate
8 ∫0
double-angle trick with cosine, and we get 3a2π/8.
Now let’s change the integral we are trying to compute. So far, we have been computing the
flow of a vector field F along a curve. It might be important in a problem instead to compute the
flow of the vector field across the curve. That is, we might want to find how much a vector field
is flowing into or out of the region bounded by the curve.
It might make more sense to switch back to path integrals here, because the outward flow doesn’t
care which direction we traverse the boundary. So we will want the path integral ∫ OF ds where
C
OF is the outward flow of the vector field per unit length of the curve. How can we find the
outward flow per unit length? Simple! We break F into the component along C and the
component perpendicular to C. The component along C was F⋅σ′(t) / ||σ′(t)||, dotting F with the
unit tangent vector. So it stands to reason that the component perpendicular to C would be
dotting F with the unit normal vector. Since we want outward flow, let’s use the outwardpointing unit normal.
Now if we walk along the boundary of a region in the positive direction, the region is on the left,
so we will need to rotate the unit tangent vector 90° to the right—that is 90° clockwise—to give
us the unit normal we desire. If we have a vector ai +bj, we can rotate it 90° clockwise by using
the vector bi – aj. So instead of dotting with 1//||σ′(t)|| (x′i + y′j) we will need the dot product
with 1//||σ′(t)|| (y′i – x′j). In other words, the outward flow of a vector field is given by
y′i − x ′j
∫C F ⋅ σ ′(t) ds .
That right there is an important formula—you can use it to calculate the outward flow of a vector
field from a region (or indeed, the flow across any curve, even if it is not a closed loop). But
things get interesting when we convert it back to a line integral.
Let F(x, y) = P(x, y)i + Q(x, y)j as usual. Computing:
y′i − x ′j
y′i − x ′j
∫C F ⋅ σ ′(t) ds = ∫C F ⋅ σ ′(t) σ ′(t) dt = ∫C F ⋅( y′ dti − x ′ dtj) and plugging in F gives
∫
C
Py′ − Qx ′ dt = ∫ P dy − Q dx .
C
This line integral is the more convenient form in which to compute the flow across a curve—and
especially the flow out of a region.
What’s really nice is when we apply Green’s Theorem to this line integral, for we get
∂P ∂Q
∫C P dy − Q dx = ∫C −Q dx + P dy = ∫D ∂x + ∂y dA . The reason this is so nice is that we recognize
the thing inside the area integral as the divergence of F!
Green’s Theorem (Divergence Theorem in the Plane): if D is a region to which Green’s
Theorem applies and C its positively oriented boundary, and F is a differentiable vector field,
then the outward flow of the vector field across the boundary equals the integral of the
divergence across the entire regions: ∫ −Q dx + P dy = ∫ ∇ ⋅ F dA .
C
D
Example: let F be the vector field xi + yj. Verify the divergence theorem for the region that is
the unit square.
Solution: we don’t actually have to do any integrals here. Along the x-axis part of the square,
the vector field points along the x-axis, so does not cross this edge of the square at all. Same
goes for the y-axis portion of the boundary. Along the edge where x = 1, the x-component of F is
one, while the y-component is irrelevant since only x is flowing across this part of the
boundary—y is flowing along it. So we always have 1 unit outflow per unit length, and since
this side of the square is one unit long, this contributes one to the outflow. The exact same
applies for the top edge of the square, only now it’s just the y-component that matters. So the
total outflow is 2.
On the other hand, the divergence ∇⋅(x, y) = 1 + 1 = 2, and integrating 2 over the unit square
gives 2. So they are in fact equal.
Recall that divergence is exactly net outflow per unit volume (area in the plane). It is expected
that if we integrate this over a region, we should get the net outflow of the vector field from the
region. The divergence theorem confirms this.
Finally, the divergence theorem is yet another version of the Fundamental theorem—we are
integrating a function around the boundary, and some kind of derivative around the inside, and
they are equal. It seems almost like we should be able to guess that these kinds of things are true
by now!