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Homework Help: Is this line integral computation correct? (Green's Thm application)

  1. Aug 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the value of the line integral

    [itex]\int[/itex] C(e-x^3 - 3y)dx + (tan y + y4 + x) dy

    where C is the counterclockwise-oriented circle of radius 4 centered at (0,2).

    3. The attempt at a solution
    First off, I didn't think this was path independent since the derivative of the dx term in respect to y equals -3 which doesn't equal the derivative of the dy term with respect to x, 1.

    After getting stuck on some direct approaches, I realized it is probably a Green's Thm application.

    This then translates to [itex]\int[/itex][itex]\int[/itex] 4 dA. So in polar coordinates the path of the circle is

    x = 4 cos [itex]\vartheta[/itex]
    y = 4 sin [itex]\vartheta[/itex] + 2

    This is where I get unsure. To try and get the limits on r I tried to say r = sqrt (x + y) and then plugged in the paramatized equations for x and y. This simplified to r = sqrt (20 + 16 sin [itex]\vartheta[/itex]). So then I thought that r varies from 2 to this general thing and hence

    [itex]\int[/itex] [itex]^{2pi}_{0}[/itex] [itex]\int[/itex][itex]^{\sqrt{20 + 16 sin\vartheta}_{2}}[/itex] 4 r dr d[itex]\vartheta[/itex].

    Is this right? Note that the limits with respect to r is suppose to be 2 to that sqrt expression, the two just kept popping up a little below and to the right...

    In particular is this a correct way to find the respective limits of integration? Thanks!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Aug 20, 2012
  2. jcsd
  3. Aug 20, 2012 #2


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    Gold Member

    Check that you've typed the integral correctly, as it doesn't make much sense as it is.

    Did you mean:
    $$\int_C (e^{-x^3}-3y)dx+(\tan y + y^4+x)dy$$
  4. Aug 20, 2012 #3
    1. The radius of the circle is given directly in the problem statement.

    2. The double integral of a constant is equal to the constant multiplied by the area of the domain of integration, which is a circle of a known radius here.
  5. Aug 20, 2012 #4
    Oops, sorry that was suppose to be a + rather than =. Thanks Voko, I should have seen that. That clears it all up for me.
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