Green’s Theorem

Section 4.6 Green’s Theorem

Motivating Questions

How can we calculate the circulation of a two-dimensional vector field $F$ around a closed curve when $F$ is not path-independent?
What is the meaning of the double integral of the circulation density of a smooth two-dimensional vector field on a region $R$ bounded by a closed curve that does not intersect itself?

We know from Section 4.4 that a vector field is path-independent if and only if the circulation around every closed curve in its domain is

0 .

It is probably not surprising that, given the multitude of names we have for path-independent vector fields, they are important vector fields that arise frequently. However, not every vector field is path-independent, and many times we will want to calculate the circulation around a closed curve in a vector field that is not path-independent. This section explores a connection between line integrals and double integrals that you may find surprising.

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Preview Activity 4.6.1.

🔗

We will consider the vector field

F = ⟨ 2 y, 3 x^{2} y ⟩,

which is defined on the entire plane. Suppose that we want to calculate the circulation of

F

around the circle

C

of radius

2,

centered at

(0, 0),

and oriented counterclockwise.

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(a)

🔗

Verify that

F

is not path-independent by calculating the circulation of

F

around the circle

C .

The SageMath cell below is set up to assist you with this, but you will need to supply a parametrization of

C

on line 3.

xxxxxxxxxx
 
var('r,t,x,y')
F = vector([2*y,3*x^2*y])
C = vector([ , ])# Supply a parametrization of C here inside the [ ]
integral(F(x=C[0],y=C[1]).dot_product(derivative(C,t)),t,0,2*pi)

Answer.

- 8 π

Solution.

If we parametrize

C

r (t) = ⟨ 2 \cos (t), 2 \sin (t) ⟩,

then

\oint_{C} F \cdot d r = - 8 π .

To use the SageMath cell to do this, make line 3 read C = vector([2*cos(t),2*sin(t)]) and execute. Because the circulation around

C

is not

0,

we know that

F

is not path-independent.

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(b)

🔗

Consider

F = ⟨ F_{1} (x, y), F_{2} (x, y) ⟩ .

In Section 4.5, we discussed the circulation density of a two-dimensional vector field at a given point and determined that it can be algebraically determined by

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} .

Calculate the circulation density for this vector field.

Answer.

No;

6 x y - 2 .

Solution.

Here we have that

\partial F_{2} / \partial x = 6 x y

and

\partial F_{1} / \partial y = 2,

so we have that

\partial F_{2} / \partial x - \partial F_{1} / \partial y = 6 x y - 2 .

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(c)

🔗

Sketch the curve

C

and shade the region inside it. Describe the region bounded by

C

in both rectangular and polar coordinates.

Answer.

Polar:

0 \leq r \leq 2,

0 \leq θ \leq 2 π .

Rectangular:

- 2 \leq x \leq 2,

- \sqrt{4 - x^{2}} \leq y \leq \sqrt{4 - x^{2}} .

Solution.

Since we know the radius of

C

2,

the region the circle bounds is covered by all values of

r

with

0 \leq r \leq 2 .

For

θ,

since we have the full disk, the bounds are

0 \leq θ \leq 2 π .

In rectangular coordinates, the circle is

x^{2} + y^{2} = 4 .

The range for

x

- 2 \leq x \leq 2,

and solving the equation of the circle for

y

gives us

- \sqrt{4 - x^{2}} \leq y \leq \sqrt{4 - x^{2}}

as the bounds for

y .

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(d)

🔗

Calculate the double integral of

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}

over the region inside the circle

C .

This integral is not the most fun to do by hand, so a SageMath cell has been provided to assist you.

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integral(integral( FUNCTION HERE , var1, start, stop), var2, start, stop)

Hint.

If you choose polar coordinates, use

θ

for the first variable and

r

for the second variable.

Answer.

- 8 π

Solution.

Denote by

D

the region inside the circle

C .

We can describe

D

readily in polar coordinates with the inequalities

0 \leq r \leq 2

and

0 \leq θ \leq 2 π .

Thus, we need to calculate

\begin{aligned} \iint_{D} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A & = \iint_{D} (6 x y - 2) d A \\ = \int_{0}^{2} \int_{0}^{2 π} (6 r^{2} \cos (θ) \sin (θ) - 2) r d θ d r . \end{aligned}

Using SageMath to calculate this, we obtain

- 8 π .

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A natural question after completing this Preview Activity is if there is something special about the vector field

F

or the curve

C

that led to the results you obtained, and investigating this will be our principal task in this section

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Subsection 4.6.1 Circulation

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In Preview Activity 4.6.1, you integrated the circulation density of a smooth vector field over a disk and found the result was equal to the circulation of the vector field along the region’s circular boundary. This relationship is the theme of this section. To see why this would make sense, consider the region

R

bounded by the curve

C

shown in Figure 4.6.1. We have placed a square grid inside

R

to suggest the idea of breaking

R

up into many smaller regions, most of which are square. This idea should make you think of the methods we have already seen of breaking up a region into smaller and smaller regions for Riemann sums.

described in detail following the image — 🔗
Figure 4.6.1. A region bounded by a curve

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The critical idea here is that if we integrate the circulation density over each of the small regions and add those up, this is the same as integrating the circulation density over the entirety of the region

R

because of the fundamental properties of integrals. Also, integrating circulation per unit area over a two-dimensional region should be related to the total circulation on that region. Now look at Figure 4.6.2 and think of these four square regions as being four of the square regions inside

R

in Figure 4.6.1.

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We orient the boundary

C_{i}

of square region

R_{i}

in the manner suggested by the circular arrows. This means that the vertical boundary in common between

R_{1}

and

R_{2}

is oriented up when we calculate the line integral

\oint_{C_{1}} F \cdot d r

and oriented down when we evaluate

\oint_{C_{2}} F \cdot d r .

Thus, this line segment does not contribute to the sum

\oint_{C_{1}} F \cdot d r + \oint_{C_{2}} F \cdot d r .

The same idea applies to the other three interior line segments so that

\oint_{C_{1}} F \cdot d r + \oint_{C_{2}} F \cdot d r + \oint_{C_{3}} F \cdot d r + \oint_{C_{4}} F \cdot d r

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is equal to the line integral of

F

along the boundary of the large square.

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Returning to Figure 4.6.1, if we find the circulation along the boundary of each of the smaller regions, the line integrals along the boundaries that lie inside the region

R

will all offset. Thus the value should equal

\oint_{C} F \cdot d r .

If we make our grid fine enough, all of the smaller regions into which

R

is divided will be very close to rectangular. Our next activity looks at the relationship between the circulation along a rectangular curve with sides parallel to the coordinate axes and the double integral of the circulation density over the rectangular region inside the curve.

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Activity 4.6.2.

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For this activity, we will consider the region

R

satisfying

a \leq x \leq c

and

b \leq y \leq d .

The rectangular curve

C

will be the boundary of

R,

oriented counterclockwise. Let

F = ⟨ F_{1} (x, y), F_{2} (x, y) ⟩

be a vector field that is smooth on

C

and

R .

We can write

C = C_{1} + C_{2} + C_{3} + C_{4},

where

C_{1}

is the top edge of the rectangle and the edges are numbered counterclockwise around the rectangle.

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(a)

🔗

Sketch the region

R

and curves

C_{1}, C_{2}, C_{3}, C_{4} .

Label the four corners of

R

with the coordinates of the vertices, and be sure to indicate the proper orientation on

C .

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(b)

🔗

Since we know that

\oint_{C} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r + \int_{C_{3}} F \cdot d r + \int_{C_{4}} F \cdot d r,

🔗

it makes sense to parametrize each curve

C_{i}

and write

\int_{C_{i}} F \cdot d r

as a one-variable integral. For example,

C_{1}

can be parametrized as

r (x) = ⟨ c + a - x, d ⟩

with

a \leq x \leq c .

Thus, we have

\int_{C_{1}} F \cdot d r = \int_{a}^{c} ⟨ F_{1} (x, d), F_{2} (x, d) ⟩ \cdot ⟨ - 1, 0 ⟩ d x .

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Complete this process to write the integrals for the remaining three sides as integrals with respect to

x

y,

as appropriate.

Answer.

\begin{aligned} \int_{C_{2}} F \cdot d r & = \int_{b}^{d} ⟨ F_{1} (a, y), F_{2} (a, y) ⟩ \cdot ⟨ 0, - 1 ⟩ d y \\ \int_{C_{3}} F \cdot d r & = \int_{a}^{c} ⟨ F_{1} (x, b), F_{2} (x, b) ⟩ \cdot ⟨ 1, 0 ⟩ d y \\ \int_{C_{4}} F \cdot d r & = \int_{b}^{d} ⟨ F_{1} (c, y), F_{2} (c, y) ⟩ \cdot ⟨ 0, 1 ⟩ d y \end{aligned}

Solution.

The parametrizations rely on fixing

x = a

for

C_{2},

y = b

for

C_{3},

and

x = d

for

C_{4} .

\begin{aligned} \int_{C_{2}} F \cdot d r & = \int_{b}^{d} ⟨ F_{1} (a, y), F_{2} (a, y) ⟩ \cdot ⟨ 0, - 11 ⟩ d y \\ \int_{C_{3}} F \cdot d r & = \int_{a}^{c} ⟨ F_{1} (x, b), F_{2} (x, b) ⟩ \cdot ⟨ 1, 0 ⟩ d y \\ \int_{C_{4}} F \cdot d r & = \int_{b}^{d} ⟨ F_{1} (c, y), F_{2} (c, y) ⟩ \cdot ⟨ 0, 1 ⟩ d y \end{aligned}

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(c)

🔗

Write

\oint_{C} F \cdot d r

as the sum of two one-variable integrals (one with respect to

x

and one with respect to

y

Answer.

\oint_{C} F \cdot d r = \int_{b}^{d} (F_{2} (c, y) - F_{2} (a, y)) d y + \int_{a}^{c} (F_{1} (x, b) - F_{1} (x, d)) d x

Solution.

Because the limits of integration for the

d x

integrals match and the limits of integration for the

d y

integrals match, we can add and obtain

\oint_{C} F \cdot d r = \int_{b}^{d} (F_{2} (c, y) - F_{2} (a, y)) d y + \int_{a}^{c} (F_{1} (x, b) - F_{1} (x, d)) d x .

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(d)

🔗

We would like to see if the equality of the line integral and double integral in Preview Activity 4.6.1 was a coincidence or holds for more general vector fields. (Soon we will expand our curves beyond rectangles, but for the moment, we focus on the curve

C

and region

R .

) To do this, we have to find some way to take our answer to the previous part of this activity, which includes two single integrals and turn them into a double integral over the rectangle

R .

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We have assumed that

F

is smooth on

C

and

R,

so the partial derivatives

\partial F_{2} / \partial x

and

\partial F_{1} / \partial y

are continuous functions on

C

and

R .

Rewrite the expression you have for

\oint_{C} F \cdot d r

in the previous step as the sum of two double integrals.

Hint.

With

x

fixed, the integrand in your

d x

integral matches what you see on the left-hand side of the Total Change Theorem

activecalculus.org/single/sec-4-4-FTC.html#wtR

, but now instead of prime notation for the derivative, you will have a partial derivative with respect to

y .

Answer.

\oint_{C} F \cdot d r = \int_{b}^{d} \int_{a}^{c} \frac{\partial F_{2}}{\partial x} (x, y) d x d y + \int_{a}^{c} \int_{d}^{b} \frac{\partial F_{1}}{\partial y} (x, y) d y d x

Solution.

By the Total Change Theorem

activecalculus.org/single/sec-4-4-FTC.html#wtR

, with

y

fixed, we have that

\begin{aligned} F_{2} (c, y) - F_{2} (a, y) & = \int_{a}^{c} \frac{\partial F_{2}}{\partial x} (x, y) d x . \end{aligned}

Similarly,

\begin{aligned} F_{1} (x, b) - F_{1} (x, d) & = \int_{d}^{b} \frac{\partial F_{1}}{\partial y} (x, y) d y . \end{aligned}

Therefore, we have

\oint_{C} F \cdot d r = \int_{b}^{d} \int_{a}^{c} \frac{\partial F_{2}}{\partial x} (x, y) d x d y + \int_{a}^{c} \int_{d}^{b} \frac{\partial F_{1}}{\partial y} (x, y) d y d x .

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(e)

🔗

Rewrite your sum of two double integrals in the previous step above as a double integral over the rectangle

R,

thereby showing that the line integral of

F

along the curve

C

is equal to a double integral of the region inside

C .

Answer.

\oint_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A

Solution.

After interchanging the order of integration so that both integrals are done

d x d y,

the other step we must do to add them is to make the integral of

\partial F_{1} / \partial y

have lower limit

b

and upper limit

d

(with respect to

y

). We do this by multiplying by

- 1

and then can add the integrals to obtain

\oint_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A .

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Now we can see that our result in Preview Activity 4.6.1 was not a coincidence and integrating the circulation density of a smooth vector field on a rectangular region gives the same result as calculating the circulation along the boundary of the region.

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The phenomenon we verified for rectangular regions in Activity 4.6.2 is a more general result that will be the subject of our next section.

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Subsection 4.6.2 Green’s Theorem

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So far in this section, we have restricted ourselves to relatively nice closed curves when thinking about circulation. While the main theorem of this section will not allow us to consider arbitrary closed curves, it does cover more varied curves than we have discussed so far. A simple closed curve is a closed curve that does not cross itself, and these are the curves to which our next theorem applies.

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Theorem 4.6.3. Green’s Theorem.

🔗

Let

C

be a simple closed curve in the plane that bounds a region

R

with

C

oriented in such a way that when walking along

C

in the direction of its orientation, the region

R

is on our left. Suppose that

F = ⟨ F_{1}, F_{2} ⟩

is vector field with continuous partial derivatives on the region

R

and its boundary

C .

The circulation of

F

along

C

may be calculated as

\oint_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A .

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In Activity 4.6.2, we showed that Green’s Theorem holds when the region

R

is a rectangle with sides parallel to the coordinate axes. The discussion that introduced the previous subsection may have you convinced that Green’s Theorem holds in its full form. Although our argument there was not a rigorous proof, we will not say more about why this theorem is true in full generality.

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At first glance, it may seem that Green’s Theorem is of purely intellectual interest. However, we have already encountered situations where parametrization a curve can be complicated. This is particularly true when the curve has “corners” that require us to give separate parametrizations for several pieces of the curve, such as with the rectangular curve pictured in Figure 4.6.4. However, as with this rectangle, is often the case that a curve that is difficult to parametrize bounds a region that is not too complex to describe using rectangular or polar coordinates. For instance, the rectangular region in Figure 4.6.4 can be described as

1 \leq x \leq 4

and

2 \leq y \leq 4,

rather than parametrization each of the four sides of the rectangle separately. Additionally, the integrand of the double integral in Green’s Theorem involves partial derivatives that can sometimes result in an integrand that is very easy to work with. The purpose of Green’s Theorem is, at its core, to allow you to exchange one type of integration problem (a line integral) for another type of integration problem (a double integral). This will be a recurring theme as this chapter continues.

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The next activity asks you to apply Green’s Theorem.

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Activity 4.6.3.

🔗

For each of the curves described below, find the circulation of the given vector field around the curve. Do this both by calculating the line integral directly as well as by calculating the double integral from Green’s Theorem.

🔗

(a)

🔗

The curve

C_{1}

is the circle of radius

3

centered at the point

(2, 1)

(oriented counterclockwise) and the vector field is

F = ⟨ y^{2}, 5 x + 2 x y ⟩ .

Answer.

45 π

Solution.

Here we find that

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = (5 + 2 y) - 2 y = 5 .

Therefore, the integrand in Green’s Theorem is the constant

5 .

Thus, the circulation of

F

around

C

5

times the area of the region inside

C .

That area is

π \cdot 3^{2} = 9 π,

which means that the circulation is

45 π .

🔗

(b)

🔗

The curve

C_{2}

is the triangle with vertices

(0, 0),

(3, 0),

and

(3, 3)

(oriented counterclockwise) and the vector field is

G = ⟨ y^{2}, 3 x y ⟩ .

Answer.

27 / 6

Solution.

Here we find that

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = 3 y - 2 y = y .

The region inside

C_{2}

can be described by

0 \leq x \leq 3

and

0 \leq y \leq x .

Therefore, by Green’s Theorem, we have

\oint_{C_{2}} G \cdot d r = \int_{0}^{3} \int_{0}^{x} y d y d x = \int_{0}^{3} {\frac{y^{2}}{2} |}_{0}^{x} d x = \int_{0}^{3} \frac{x^{2}}{2} d x = {\frac{x^{3}}{6} |}_{0}^{3} = \frac{27}{6} .

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Generally, Green’s Theorem is useful for allowing us to calculate line integrals by instead calculating a double integral. Going the other direction is harder, since finding a vector field

⟨ F_{1}, F_{2} ⟩

so that

\partial F_{2} / \partial x - \partial F_{1} / \partial y

is equal to the integrand in the double integral can be difficult. However, the exercises will explore some situations where calculating a suitable line integral is a viable alternative to the double integral.

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Subsection 4.6.3 What happens when vector fields are not smooth?

🔗

Notice that the assumptions in Green’s Theorem require that the region

R

be bounded by a simple closed curve

C

and that the vector field have continuous partial derivatives on

R

and

C .

In Exercise 2, we will explore what happens when the region

R

cannot be bounded by a simple closed curve, as sometimes multiple applications of Green’s Theorem can be used in those circumstances. Now, however, we will take a look at what happens in some cases where the vector field

F

is not smooth.

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Activity 4.6.4.

🔗

Consider the vector field

F = \frac{- y}{x^{2} + y^{2}} i + \frac{x}{x^{2} + y^{2}} j .

Notice that

F

is smooth everywhere in the plane other than at the point

(0, 0) .

This vector field is plotted in Figure 4.6.5, but we have not plotted the vectors close to the origin as their magnitudes get so large that they make it hard to interpret the figure. Here Green’s Theorem applies to any simple closed curve

C

that neither passes through

(0, 0)

nor bounds a region containing

(0, 0) .

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(a)

🔗

Find the circulation density of

F

(i.e., the integrand of the double integral in Green’s Theorem).

Hint.

Make sure to note any exceptional points at which your formula is not valid.

Answer.

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = 0

for

(x, y) \neq (0, 0)

Solution.

For

(x, y) \neq (0, 0),

we have

\begin{aligned} \frac{\partial F_{2}}{\partial x} & = - \frac{2 x^{2}}{{(x^{2} + y^{2})}^{2}} + \frac{1}{x^{2} + y^{2}} = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} \\ \frac{\partial F_{1}}{\partial y} & = \frac{2 y^{2}}{{(x^{2} + y^{2})}^{2}} - \frac{1}{x^{2} + y^{2}} = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} . \end{aligned}

Thus,

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = 0 .

🔗

(b)

🔗

Suppose that

C

is the unit circle centered at the origin. Without doing any calculations, what can you say about

\oint_{C} F \cdot d r ?

what does this tell you about if

F

is path-independent?

Answer.

Positive;

F

is not path-independent

Solution.

Notice that

F

is tangent to

C .

Thus,

\oint_{C} F \cdot d r

must be positive. (In fact, the circulation of

F

along

C

is the length of

C

2 π

because

F

consists of unit vectors and the vector field is is tangent to

C .

) Since there is a closed curve with nonzero circulation,

F

is not path independent.

🔗

(c)

🔗

What would you get if you integrated the circulation density of

F

over the region bounded by

C ?

Answer.

0

Solution.

The circulation density is

0

everywhere inside

C

except at the origin, where it is undefined. That single point does not impact the value of the double integral of the circulation density, which is thus

0 .

🔗

(d)

🔗

Do the previous two parts contradict Green’s Theorem? Explain your reasoning.

Answer.

No.

Solution.

There is no contradiction because Green’s Theorem does not apply here. The region

R

inside the curve

C

includes the point

(0, 0),

and

F

is not defined at that point. Green’s Theorem requires that the vector field be smooth at every point of

R .

🔗

(e)

🔗

Is the vector field

G = \frac{x}{x^{2} + y^{2}} i + \frac{y}{x^{2} + y^{2}} j,

which is shown in Figure 4.6.6, path-independent? Why or why not?

Answer.

Yes

Solution.

Let

g (x, y) = \frac{1}{2} \ln (x^{2} + y^{2}) .

Note that

\nabla g = G,

and the domain of

g

is the same as the domain of

G

(all of

R^{2}

except

(0, 0)

). Thus,

G

is a gradient vector field and is therefore path-independent.

🔗

(f)

🔗

Suppose that

C

is the unit circle centered at the origin. Find

\oint_{C} G \cdot d r .

Can you do this using Green’s Theorem?

Answer.

0;

Solution.

We have already shown that

G

is path-independent, and therefore

\oint_{C} G \cdot d r = 0

since

C

is a closed curve. Because

G

is not defined at the origin, Green’s Theorem does not apply to this line integral.

🔗

We can now see that Green’s Theorem is a powerful tool. However it cannot be used for all line integrals. In particular, the restriction that the vector field be smooth on the entire region bounded by a simple closed curve

C

creates limitations. This can occur even in cases where holes in the vector field’s domain or points where the vector field is not continuously differentiable lie away from

C .

🔗

Subsection 4.6.4 Computing area as a line integral

🔗

Recall that the area of a region

D

in the plane can be computed by integrating

1

over

D .

So if we can find a vector field

F

such that

\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = 1,

then the double integral in Green’s Theorem becomes the area of

D .

Luckily, there are many such vector fields

F :

e.g. F = ⟨ 0, x ⟩, F = ⟨ - y, 0 ⟩, F = ⟨ - y / 2, x / 2 ⟩

🔗

Using Green’s Theorem to compute area.

🔗

D

is a region in the plane with boundary

\partial D,

then the area of

D

can be computed as any of the following line integrals around

\partial D .

\begin{aligned} area (D) & = \iint_{D} 1 d A \\ = \oint_{\partial D} x d y \\ = \oint_{\partial D} - y d x \\ = \frac{1}{2} \oint_{\partial D} x d y - y d x \end{aligned}

🔗

Example 4.6.7.

🔗

Compute the area of the ellipse

{(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1.

Solution.

The ellipse is parametrized as follows.

\begin{matrix} x (t) = a \cos (t) \\ y (t) = b \sin (t) \end{matrix}, 0 \leq t \leq 2 π

Using Green’s Theorem, we compute the area of the ellipse as a line integral of

F = ⟨ 0, x ⟩

around the boundary of the ellipse. (Any of the other vector fields listed above would work just as well.)

\begin{aligned} \iint_{E} 1 d A & = \oint_{\partial E} x d y \\ = \int_{t = 0}^{2 π} (a \cos (t)) (b \sin (t)) d t \\ = a b \int_{0}^{2 π} \cos^{2} (t) d t \\ = \frac{a b}{2} \int_{0}^{2 t} 1 + \cos (2 t) d t \\ = \frac{a b}{2} {[t + \frac{1}{2} \sin (2 t)]}_{0}^{2 π} \\ = \frac{a b}{2} (2 π) = a b π \end{aligned}

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Subsection 4.6.5 Summary

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Green’s Theorem tells us that we can calculate the circulation of a smooth vector field along a simple closed curve that bounds a region in the plane on which the vector field is also smooth by calculating the double integral of the circulation density instead of the line integral.
Integrating the circulation density of a smooth vector field on a region bounded by a simple closed curve gives the same value as calculating the circulation of the vector field around the boundary of the region with suitable orientation.

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Exercises 4.6.6 Exercises

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1.

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Sometimes, the value in Green’s Theorem is in converting a double integral into a line integral. Recall that the area of a region

R

in the plane can be found by calculating

\iint_{R} 1 d A .

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(a)

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Find a smooth vector field

F

such that the circulation density of

F

1

everywhere.

Hint.

There are several ways to choose

F = ⟨ F_{1}, F_{2} ⟩

so that

\partial F_{2} / \partial x - \partial F_{1} / \partial y = 1 .

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(b)

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Write out both integrals in Green’s Theorem for the vector field you selected in part a.

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(c)

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For positive real numbers

a, b,

the ellipse

x^{2} / a^{2} + y^{2} / b^{2} = 1

can be parametrized as

r (t) = ⟨ a \cos (t), b \sin (t) ⟩

with

0 \leq t \leq 2 π .

Find the area of this ellipse by calculating a line integral.

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(d)

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Find another vector field

G

(different from the one you found in part a) that has circulation density

1

everywhere, and calculate the area of the ellipse in part c by calculating the circulation of

G

along the ellipse.

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(e)

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How many possible vector fields are there with circulation density

1

everywhere? Why?

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2.

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The types of regions in

R^{2}

to which Green’s Theorem applies are formally called simply connected regions. To be precise, a simply connected region

D

R^{2}

is a set so that there is a path between every pair of points in

D

that stays inside

D

and any simple closed curve in

D

can be shrunk to a point while remaining inside

D .

We can think of a simply connected region as being a region that does not have any “holes”. There are many instances where we can find a way to apply Green’s Theorem multiple times to work with regions that are not simply connected, however.

Figure 4.6.8. A region that is not simply connected

Explain why the region $R$ in Figure 4.6.8 is not simply connected.
If we let $C = C_{1} + C_{2},$ then the orientation of $C$ is as required by Green’s Theorem—when walking along the pieces of $C,$ the region $R$ is always on the left-hand side. Assume that $F$ is a vector field that is smooth on $R$ and $C .$ Using Figure 4.6.9 as a guide, write $\oint_{C} F \cdot d r$ as a sum of double integrals, one over $R_{1}$ and the other over $R_{2} .$
In Activity 4.6.4, we considered the vector field $F = \frac{- y}{x^{2} + y^{2}} i + \frac{x}{x^{2} + y^{2}} j$ and found that the circulation of $F$ around the unit circle centered at the origin (oriented counterclockwise) was $2 π .$ Show that for every simple closed curve $C$ in the plane that bounds a region containing the origin, $\oint_{C} F \cdot d r = 2 π .$

Hint.

For part c, create a region

R

as in Figure 4.6.8 using the curve

C

in place of

C_{1} .

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3.

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This exercise presents another occasion where Green’s Theorem can be used to convert a double integral into a line integral.

Recall from Section 3.4 that the centroid of a lamina $D$ of area $A$ is given by

$\overset{―}{x} = \frac{1}{A} \iint_{D} x d A and \overset{―}{y} = \frac{1}{A} \iint_{D} y d A .$

Find vector fields $F$ and $G$ so that

$\overset{―}{x} = \oint_{C} F \cdot d r and \overset{―}{y} = \oint_{C} G \cdot d r,$

where $C$ is the boundary of the lamina $D .$
Find the centroid of the triangle with vertices $(0, 0),$ $(a, 0),$ and $(0, b)$ for real numbers $a, b > 0 .$

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