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Line Integral + Green Theorem problems

  1. Feb 26, 2007 #1


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    Question 1)
    Integrate y^2 dx + zx dy + dz, along the circle x^2 + y^2 – 2y = 0 to (1, 1, 0)

    I am not sure how to begin on this problem. Would it be beneficial to convert to polar coordinates? The thing that is throwing me off is the fact that the circle’s equation is not the normal circle (x^2 + y^2 = r^2), it has that -2y term and it looks like its radius is growing from 0 to sqrt (2)?

    Question 2)
    Integrate e^x * cos(y) dy – e^x * sin (y) dx,
    Over c, which is the broken line from (ln (2), 0), to (0, 1), to (-ln (2), 0) <-- forms a triangle.
    Where “e” is the natural exponent.

    This is a Green’s Theorem problem I know.
    I can identify that,
    p(x, y) = -e^x * sin (y)
    q(x, y) = e^x * cos (y)

    So I should be able to transform the integral into a double integral over some area with the integrant being,
    The partial derivative of q with respect to x, minus the partial derivative of p with respect to y, all of that integrated by dx dy.

    The partial of q with respect to x = -e^x * sin (y)
    The partial of p with respect to y = -e^x * sin (y)
    So subtracting them appropriately, it ends up being the integral of zero dx dy, which is zero itself.
    But this cannot be correct (I know this to be wrong).
  2. jcsd
  3. Feb 27, 2007 #2


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    Not "polar coordinates" so much as parametric equations for the circle.
    And that certainly is a "normal" circle- it just does not have center at the origin. (And the radius is NOT "growing"!) Complete the square to write the equation as x2+ (y- a)2= a2, a circle of radius a with center at (0,a), for the correct a. Parametric equations for what you call the "normal" circle, with center at (0,0) and radius r, are [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex]. The "translate" the circle to (0, a), just add a to y. Of course, z= 0 at every point of this circle.

    Well, e is the base of the natural exponential function- it clearly is not an exponential here! :rolleyes: (picky, picky)
    However, it is important in mathematics to be very precise as in my next question:

    Is that exactly what the problem says? If so, it is NOT a closed path. You are missing the final leg from (-ln 2, 0) to (ln 2, 0). (and is it really dy first and then dx and not the other way around?) Although Green's theorem is an excellent choice!

    That integral definitely is 0. If, however, the problem is exactly as you stated, it is not a closed path. To use Green's theorem you would have to add the third leg from (-ln(2), 0) to (ln(2), 0). Of course, that also means that the value given by Green's theorem (here 0) is the integral you want plus the integral from (-ln(2), 0) to (ln(2), 0). So that the integral you want is 0 minus that integral: It should be easy to integrate what you have along the x-axis where y= 0. Be careful about the sign- you will want to subtract that integral from 0.

    I asked about whether it is really
    [tex]\int e^x * cos(y) dy – e^x * sin (y) dx[/tex]
    rather than
    [tex]\int e^x * cos(y) dx – e^x * sin (y) dy[/tex]
    because, if it is the former, the answer is still 0!

    (I have no idea where that "8211" is coming from!)
  4. Feb 27, 2007 #3


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    Q 1)
    I completed the square to get a much more familiar looking equation,
    x^2 + (y – 1)^2 + 1 = 0 --> x^2 + (y – 1)^2 = 1,
    r = 1 and it is a circle centered at (0, 1).

    Now the parametric form of the circle would be,
    x = 1 * cos (θ)
    dx = -sin (θ) dθ

    y = 1 * sin (θ) + 1
    y^2 = sin^2(θ) + 2*sin (θ) + 1
    dy = cos (θ) dθ

    z = 0

    Plugging into the equation I am suppose to integrate, …
    (sin^2(θ) + 2*sin(θ) + 1) * (-sin (θ) dθ) + 0 + 0,
    So now I need to take the integral of,
    (-sin^3 (θ) - 2sin^2(θ) – sin(θ)) dθ

    I just need to find how θ changes for the bounds of the integral.
    Would it be from -90° to 0°? Since it is going from “0” (which I assume means (0, 0, 0) to (1, 1, 0). and considering where the circle is centered, it is going from the ‘bottom’ of the circle over to its right side.

    Q 2)
    You are right. In the question’s text it only gives three points to the path, but then in the “hint” below, it says to apply Green’s theorem over the closed curve, meaning that part of the path is “(-ln (2), 0) to (ln (2), 0) as you said.
    Also, you are right about the dx coming before the dy in the integral. Upon double checking the book, it seems that not only did I copy it wrong in my previous post; I have been writing it wrong all through my attempted solutions to this question on paper.
    It is actually,
    Integrate e^2 *cos (y) dx - e^x * sin (y) dy

    So now,
    p(x, y) = e^x * cos (y)
    q(x, y) = -e^x * sin (y),

    The partial derivative of q with respect to x is = -e^x * sin (y)
    The partial derivative of p with respect to y is = -e^x * sin (y)
    Partial q wrt x – partial p wrt y = 0

    I don’t see how integrating zero is going to give me a non-zero answer.

    To apply Green’s theorem, I need to integrate over this triangle boundary which the points above define, going from (ln (2), 0), to (0, 1), to (-ln (2), 0) and back to (ln (2), 0).
    So I can break it up into 4 separate integrals and then add their results (?).
    But I don’t see how that will fix my problem of having to integrate zero.
  5. Feb 27, 2007 #4


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    I went back and read your original post and it just said
    and did not say anything about a starting value. Yes, this is the circle with center at (0,1,0) and radius 1. It includes the point (1, 1, 0) as well as the point (0,0,0). To go from (0,0,0) to (1,1,0) requires that [itex]\theta[/itex] vary from [itex]-\pi /2[/itex] to 0. (It is not a good idea to use "degrees" with trig functions in calculus.)

    That's my point. Your path, lets call it P1, goes from (ln(2),0) to (-ln(2),0). To have a closed path, and so use Green's theorem, you need to add P2, the straight line from (-ln(2), 0) to (ln(2),0). Now you have "integral on P1+ integral on P2= 0" by Greens theorem. Then integral on P1= -integral on P2. Since P2 is on the x-axis where y= 0, that integral, from (-ln(2),0) to (ln(2),0)is easy!
  6. Feb 27, 2007 #5


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    Yes, of course.
    OK, I see what you’re saying.
    In order to use Green’s theorem, we need a closed path. The points given don’t make up a closed path without adding the bottom side of the triangle (so to speak). Once we do this, Green’s theorem tells us the answer over the entire path should be zero. But we don’t care about the entire path; we just care about the top two sides of the triangle. Since the integral over the entire path equals the sum of the integrals over the top two sides + the bottom side, we know the integral over the top two sides (the one we care about) must equal the opposite of the integral over the bottom side. And the integral over the bottom side is easy to do since y is zero over the entire path.

    Letting y = 0, dy= 0,
    Integrate e^x *cos (y) dx - e^x * sin (y) dy
    Integrate e^x dx, with bounds at –ln (2) and +ln (2).
    Integral of e^x = e^x, evaluated at –ln (2) and +ln (2),
    We get
    2 – .5 = 1.5 = 3/2,
    So this means that the integral over the top would be -3/2.

    Thanks for your help.
    I have the second question finished now and I should be able to complete the fist one just as soon as I integrate all those powers of sin() and evaluate.
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