# Using Green's Theorem

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1. Apr 28, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

1) Is the statement above the same as finding the area enclosed?
2) $\int_C \cos ydx + x^2\sin ydy$, C is the rectangle with vertices (0,0) (5,0) (5,2) and (0,2).
3) $\int_C y^4 dx + 2xy^3dy$, C is the ellipse $x^2 + xy^2 = 2$
4) $\int_C {(1-y^3)dx + (x^3+e^\psi)dy}$, where $\psi = y^2$ and C is the boundary of the region between the circles $x^2 + y^2 =4$ and $x^2 + y^2 =9$

2. Relevant equations
n/a

3. The attempt at a solution
I solved all of these questions, with the exception of the first one. Unfortunately, these do not come with answers, so I would like to check if my answers are correct. For questions 2 to 3, I just wrote out Green's double integral.

1) Yes
2) $\int_0^2 \int_0^5 (\sin y(2x+1)) dxdy$
3) $\int \int -2\sin ^3 \theta d \theta$
4) $\int_0^{2\pi} \int_2^3 3r^3 drd \theta$

I think I got #1 and #3 wrong. If they are wrong, I will put up my original solution attempt. .

Note to self: Stewart, page 1090, 1 refers to question 6, 2 = 8 and 3 = 10.

Thanks everyone!

Last edited: Apr 28, 2015
2. Apr 28, 2015

### HallsofIvy

Staff Emeritus
To evaluate what line integral? Do you mean each of 2, 3, 4, below? (So this does not apply to 1.)

Green's Theorem says that $\oint_C P(x,y)dx+ Q(x,y)dy= \int\int_D \left(\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}\right)dxdy$
Since area is given by $\int\int_D dxdy$ that will give the area if and only if $\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}= 1$ and that will depend upon what P and Q are.

This is not an ellipse. At first I thought you meant just $x^2+ y^2= 2$ but that would be called a circle, not ellipse.

I would have thought Green's Theorem would be a "relevant equation"!

In (2), P(x,y)= y and $Q(x,y)= x^2 sin(y)$ So $Q_x- P_y= 2x sin(y)- 1$ NOT $sin(y)(2x+ 1)$

Assuming the "ellipse" was actually the circle $x^2+ y^2= 2$ then you are missing a factor of $\sqrt{2}$

This one is set up correctly. What did you get for the actual integrals?

3. Apr 28, 2015

### Zondrina

#1 is correct. Green's theorem allows you to find the area of a region $D$ with a positively oriented, piecewise-smooth, simple closed curve $C$ making up the boundary.

#2 is correct.

#3 needs some work. First of all, the boundary curve $C$ is not $x^2 + xy^2 = 2$, but is actually $x^2 + 2y^2 = 2$ according to Stewart question #8. Now that we actually have an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$, it can be parametrized by means of the equations:

$$x = a \text{cos}(\theta) = \text{cos}(\theta)$$
$$y = b \text{sin}(\theta) = \frac{1}{\sqrt{2}} \text{sin}(\theta)$$

Now, what happens to $-2y^3$? What will be the limits of integration?

#4 is correct.

Last edited: Apr 28, 2015
4. Apr 28, 2015

### Calpalned

Oops, number three should have been $x^2+ 2y^2= 2$

5. Apr 28, 2015

### Calpalned

I solved number two again and I still got the same integral. Additionally, Zondrina says I'm right