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Using Green's Theorem

  1. Apr 28, 2015 #1
    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. jcsd
  3. Apr 28, 2015 #2

    HallsofIvy

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    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 [itex]\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[/itex]
    Since area is given by [itex]\int\int_D dxdy[/itex] that will give the area if and only if [itex]\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}= 1[/itex] 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 [itex]Q(x,y)= x^2 sin(y)[/itex] So [itex]Q_x- P_y= 2x sin(y)- 1[/itex] NOT [itex]sin(y)(2x+ 1)[/itex]

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

    This one is set up correctly. What did you get for the actual integrals?
     
  4. Apr 28, 2015 #3

    Zondrina

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    Homework Helper

    #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
  5. Apr 28, 2015 #4
    Oops, number three should have been ##x^2+ 2y^2= 2##
     
  6. Apr 28, 2015 #5
    I solved number two again and I still got the same integral. Additionally, Zondrina says I'm right
     
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