Using Green's Theorem for line integral

[Calpalned]

Homework Statement

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 $x^2 + y^2 =9$

Homework Equations

n/a

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!

 

[HallsofIvy]

Homework Statement

Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

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

1) Is the statement above the same as finding the area enclosed?

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.

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 $x^2 + xy^2 = 2$

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.

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 $x^2 + y^2 =9$

Homework Equations

n/a

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

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) $\int_0^2 \int_0^5 (\sin y(2x+1)) dxdy$

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)

3) $\int \int -2\sin ^3 \theta d \theta$

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

4) $\int_0^{2\pi} \int_2^3 3r^3 drd \theta$

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

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!

 

[STEMucator]

#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.

 

[Calpalned]

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?

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

 

[Calpalned]

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?

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