## 17Calculus - Green's Theorem

Relationships Between Green's Theorem, Stokes' Theorem and the Divergence Theorem

Green's Theorem has two forms, the circulation form and the divergence form. Green's Theorem is in two dimensions, While Stokes' Theorem is the three-dimensional form of the circulation form of Green's Theorem. Similarly, the Divergence Theorem is the three-dimensional form of the divergence form of Green's Theorem. The following table summarizes these relationships.

2-dim

3-dim

Green's Theorem (circulation form)

Stokes' Theorem

Greens' Theorem (divergence form)

Divergence Theorem

If you want a complete lecture on this topic, here is a good video.

### Prof Leonard - Green's Theorem [1hr-45min-53secs]

video by Prof Leonard

Green's Theorem

IF WE HAVE

- a simply-connected region $$\mathcal{R}$$ which is bounded by a smooth curve $$\mathcal{C}$$, oriented anticlockwise (counter-clockwise) and traversed once, and
- a vector field $$\vec{F} = M\hat{i}+N\hat{j}$$ which is continuously differentiable on the region $$\mathcal{R}$$.

THEN THIS EQUATION HOLDS

$$\displaystyle{ \oint_{\mathcal{C}}{M~dx+N~dy} = \iint\limits_{\mathcal{R}}{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dA } }$$

Notation - - The circle on the integral sign clarifies that the curve $$\mathcal{C}$$ is a closed curve. It is not necessary and not all books and instructors use it.

Green's Theorem relates the calculation of line integrals to double integrals over area in the special case when the line integral completely encloses a single region in the plane. This enclosed region must be simply-connected, i.e. the enclosing line must not cross itself and it must consist of one simple closed curve. This theorem is very powerful and allows us to evaluate line integrals using double integrals.

Let's get started with a video. This video is excellent in explaining Green's Theorem, including several examples. It covers the idea of simply-connected regions as well.

### Dr Chris Tisdell - What is Green's Theorem? [47mins-49secs]

The last example in this video has an incorrect answer. The final answer should be $$14/3$$.

video by Dr Chris Tisdell

Alternative Forms of Green's Theorem

Okay, so the theorem at the top of the page is the basic form of Green's Theorem. There are two other alternative forms that are equivalent to this basic form. These forms were touched on in the previous video.

1. circulation or curl ( in 3-dim called Stokes' Theorem )

$$\displaystyle{ \oint_{\mathcal{C}}{ \vec{F} \cdot \vec{T} ~ds} = }$$ $$\displaystyle{ \iint_{\mathcal{R}}{(curl ~\vec{F}) \cdot \hat{k} ~ dA} }$$

$$\displaystyle{ \oint_{\mathcal{C}}{\vec{F} \cdot d\vec{r} } = }$$ $$\displaystyle{ \iint_{\mathcal{R}}{ \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA} }$$

2. flux or divergence ( in 3-dim called the Divergence Theorem )

$$\displaystyle{ \oint_{\mathcal{C}}{ \vec{F} \cdot \vec{n} ~ds} = }$$ $$\displaystyle{ \iint_{\mathcal{R}}{div~\vec{F}~dA} }$$

$$\displaystyle{ \iint_{\mathcal{R}}{ \left( \frac{\partial N}{\partial y} + \frac{\partial M}{\partial x}\right) dA} }$$

In the table above, $$\vec{T}$$ is the unit tangent vector and $$\vec{n}$$ is the outward unit normal vector.

This next video explains these alternative forms in more detail and has some examples.

### Dr Chris Tisdell - Green's Theorem [39mins-13secs]

video by Dr Chris Tisdell

Another by-product of Green's Theorem is that we can now use a line integral to calculate area. Here is the equation. $A = \frac{1}{2} \oint_{\mathcal{C}}{x~dy - y~dx}$ The previous video shows the derivation of this equation. If you skipped it, now is the time to go back and watch it.

Meaning of Green's Theorem

So, now that you know how to use Green's Theorem, what is really going on here? This next video gives a quick review of the Theorem and then explains the interpretation and meaning of it.

### Michael Hutchings - Multivariable calculus 4.4.1: Review and interpretation of Green's theorem [7mins-25secs]

Proofs of Green's Theorem

Here are several video proofs of Green's Theorem. Each instructor proves Green's Theorem differently. So it will help you to understand the theorem if you watch all of these videos.

### David Metzler - Greens Theorem Proof [16mins-22secs]

video by David Metzler

### Khan Academy - Green's Theorem Proof (2) [19mins-26secs]

Okay, you are now ready for some practice problems. Next, you need to understand how to parameterize surfaces before you go on to surface integrals and the three dimensional versions of Green's Theorem, Stokes' Theorem and the Divergence Theorem.

Practice

Unless otherwise instructed, use Green's Theorem to evaluate these integrals.

Basic

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ (y^2 ~dx + 3xy ~dy) } }$$ where $$\mathcal{C}$$ is the boundary of the semiannular region D in the upper half plane between the circles $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$.

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ (y^2 ~dx + 3xy ~dy) } }$$ where $$\mathcal{C}$$ is the boundary of the semiannular region D in the upper half plane between the circles $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$.

Solution

### 1569 video

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

Calculate the outward flux of $$\vec{F} = -x\hat{i} + 2y\hat{j}$$ over the square with corners $$(\pm 1, \pm 1)$$ where the unit normal is outward-pointing.

Problem Statement

Calculate the outward flux of $$\vec{F} = -x\hat{i} + 2y\hat{j}$$ over the square with corners $$(\pm 1, \pm 1)$$ where the unit normal is outward-pointing.

Solution

### 1570 video

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ y^2~dx+x^2~dy } }$$ where $$\mathcal{C}$$ is the triangle bounded by the lines $$x=0, x+y=1, y=0$$

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ y^2~dx+x^2~dy } }$$ where $$\mathcal{C}$$ is the triangle bounded by the lines $$x=0, x+y=1, y=0$$

Solution

### 1571 video

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

Let $$\mathcal{C}$$ be the circle $$x^2 + y^2 = 4.$$ Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ \left[ (3y-e^{\tan^{-1}x})~dx + (7x + \sqrt{y^4+1}) ~dy \right] } }$$

Problem Statement

Let $$\mathcal{C}$$ be the circle $$x^2 + y^2 = 4.$$ Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ \left[ (3y-e^{\tan^{-1}x})~dx + (7x + \sqrt{y^4+1}) ~dy \right] } }$$

Solution

### 1572 video

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{ x~dx - x^2y^2~dy } }$$ where $$\mathcal{C}$$ is the triangle with vertices $$(0,0)$$, $$(0,1)$$ and $$(1,1)$$.

Problem Statement

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{ x~dx - x^2y^2~dy } }$$ where $$\mathcal{C}$$ is the triangle with vertices $$(0,0)$$, $$(0,1)$$ and $$(1,1)$$.

Solution

### 1573 video

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ P~dx + Q~dy } }$$ where $$\mathcal{C}$$ is the square with corners $$(\pm 1, \pm 1)$$; $$P(x,y) = x + y^2$$ and $$Q(x,y) = y + x^2$$.

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ P~dx + Q~dy } }$$ where $$\mathcal{C}$$ is the square with corners $$(\pm 1, \pm 1)$$; $$P(x,y) = x + y^2$$ and $$Q(x,y) = y + x^2$$.

Solution

### 1574 video

video by Krista King Math

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ P~dx+Q~dy } }$$ where $$\mathcal{C}$$ is the triangle with corners $$(0,0)$$, $$(1,1)$$, $$(2,0)$$, $$P(x,y)=y+e^x$$ and $$Q(x,y)=2x^2+\cos y$$.

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ P~dx+Q~dy } }$$ where $$\mathcal{C}$$ is the triangle with corners $$(0,0)$$, $$(1,1)$$, $$(2,0)$$, $$P(x,y)=y+e^x$$ and $$Q(x,y)=2x^2+\cos y$$.

Solution

### 1575 video

video by Krista King Math

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}} }$$ where $$\mathcal{C}$$ is the circle $$x^2+y^2=4$$ and $$\vec{F}=\langle -y, x \rangle$$.

Problem Statement

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}} }$$ where $$\mathcal{C}$$ is the circle $$x^2+y^2=4$$ and $$\vec{F}=\langle -y, x \rangle$$.

Solution

### 1576 video

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}$$ where $$\mathcal{C}$$ is along $$y=x^2/2$$ and $$y=x$$ and $$\vec{F}=\langle y^2, x^2 \rangle$$.

Problem Statement

Evaluate $$\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}$$ where $$\mathcal{C}$$ is along $$y=x^2/2$$ and $$y=x$$ and $$\vec{F}=\langle y^2, x^2 \rangle$$.

Solution

### 1577 video

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}$$ where $$\vec{F}=\langle y^3, 6xy^2 \rangle$$ and $$\mathcal{C}$$ is the curve with the 3 segments:
1. $$y=0, 0 \leq x \leq 4$$,
2. $$x=4, 0 \leq y \leq 2$$,
3. $$y=\sqrt{x}, 0 \leq x \leq 4$$

Problem Statement

Evaluate $$\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}$$ where $$\vec{F}=\langle y^3, 6xy^2 \rangle$$ and $$\mathcal{C}$$ is the curve with the 3 segments:
1. $$y=0, 0 \leq x \leq 4$$,
2. $$x=4, 0 \leq y \leq 2$$,
3. $$y=\sqrt{x}, 0 \leq x \leq 4$$

Solution

### 1578 video

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}} }$$ where $$\mathcal{C}$$ is boundary between the two circles $$x^2+y^2=1$$ and $$x^2+y^2=3$$ that lies in the first quadrant and $$\vec{F} = \langle e^x+2xy, 4x^2+\sin y \rangle$$.

Problem Statement

Evaluate $$\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}} }$$ where $$\mathcal{C}$$ is boundary between the two circles $$x^2+y^2=1$$ and $$x^2+y^2=3$$ that lies in the first quadrant and $$\vec{F} = \langle e^x+2xy, 4x^2+\sin y \rangle$$.

Solution

### 1579 video

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

Determine the flux of the field $$\vec{F} = \langle -x, -y \rangle$$ across $$x^2+y^2=16$$.

Problem Statement

Determine the flux of the field $$\vec{F} = \langle -x, -y \rangle$$ across $$x^2+y^2=16$$.

Solution

### 1580 video

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

Calculate the outward flux of $$\vec{F} = -x\hat{i} + 2y\hat{j}$$ over the square with corners $$(\pm 1, \pm 1)$$ where the unit normal is outward-pointing.

Problem Statement

Calculate the outward flux of $$\vec{F} = -x\hat{i} + 2y\hat{j}$$ over the square with corners $$(\pm 1, \pm 1)$$ where the unit normal is outward-pointing.

4

Problem Statement

Calculate the outward flux of $$\vec{F} = -x\hat{i} + 2y\hat{j}$$ over the square with corners $$(\pm 1, \pm 1)$$ where the unit normal is outward-pointing.

Solution

### 1984 video

video by Dr Chris Tisdell

4

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ 2xy~dx + (x^2+y^2)~dy } }$$ where $$\mathcal{C}$$ is $$4x^2 + 9y^2 = 36$$.

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ 2xy~dx + (x^2+y^2)~dy } }$$ where $$\mathcal{C}$$ is $$4x^2 + 9y^2 = 36$$.

Solution

### 2200 video

video by Michel vanBiezen

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{(x^2-y^2)~dx + 2xy~dy} }$$ on the curve which bounds the region $$\mathcal{R}: \{(x,y): 0 \leq x \leq 1, 2x^2 \leq y \leq 2x \}$$

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{(x^2-y^2)~dx + 2xy~dy} }$$ on the curve which bounds the region $$\mathcal{R}: \{(x,y): 0 \leq x \leq 1, 2x^2 \leq y \leq 2x \}$$

Solution

### 1581 video

Log in to rate this practice problem and to see it's current rating.

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ 2y~dx-3x~dy } }$$ on the unit circle.

Problem Statement

Evaluate $$\displaystyle{ \oint_{\mathcal{C}}{ 2y~dx-3x~dy } }$$ on the unit circle.

Solution

### 1582 video

Log in to rate this practice problem and to see it's current rating.

Intermediate

Verify Green's theorem for the line integral $$\displaystyle{ \oint_{\mathcal{C}}{ xy~dx + x~dy } }$$ where $$\mathcal{C}$$ is the unit circle.

Problem Statement

Verify Green's theorem for the line integral $$\displaystyle{ \oint_{\mathcal{C}}{ xy~dx + x~dy } }$$ where $$\mathcal{C}$$ is the unit circle.

Solution

### 1568 video

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

You CAN Ace Calculus

 vectors double integrals vector fields line integrals

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

 The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.
 Theorem Relationships Green's Theorem Explanation Alternative Forms Meaning Proofs Practice

The Practicing Mind: Developing Focus and Discipline in Your Life - Master Any Skill or Challenge by Learning to Love the Process Save 20% on Under Armour Plus Free Shipping Over \$49! Shop Amazon - Rent Textbooks - Save up to 80% As an Amazon Associate I earn from qualifying purchases.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.