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Homework Help: Line Integral Problem

  1. Jul 6, 2013 #1


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    1. The problem statement, all variables and given/known data
    Evaluate ##\int_{(1,1)}^{(4,2)} (x + y)dx + (y - x)dy## along

    (a) the parabola y2 = x
    (b) a straight line
    (c) straight lines from (1,1) to (1,2) and then to (4,2)
    (d) the curve x = 2t2 + t + 1, y = t2 + 1

    3. The attempt at a solution

    (a) is fine.

    For (b), I get two different answers, depending on what I do. Here's what I did.

    The straight line passing through (1,1) and (4,2) is ##y = \frac{1}{3}x + \frac{2}{3}## so ##dy = \frac{1}{3}dx##.

    So we have (replacing y and dy with expressions in terms of x)

    ##\int_{1}^{4} (\frac{4}{3}x + \frac{2}{3})dx + (\frac{2}{3} - \frac{2}{3}x) \cdot \frac{1}{3}dx = \int_{1}^{4} (\frac{10}{9}x + \frac{7}{9})dx = \frac{32}{3}##

    But if, instead, we replace x and dx with expressions in terms of y, we have:

    ##\int_{1}^{2} (12y - 6)dy + (2 - 2y)dy = \int_{1}^{2} (10y - 4)dy = 11##

    and my textbook says that 11 is the correct answer for (b). But why do I get two different answers? Shouldn't they be the same? Does it have something to do with path independence (although aren't I using the same path)?
  2. jcsd
  3. Jul 6, 2013 #2
    It should be [itex]\int_1^4\frac{10}{9}x+\frac{8}{9}\,dx[/itex]
  4. Jul 6, 2013 #3


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    Ah, I see... thanks!
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