Line Integral Problem: Calculating ∫(x+y)dx+(y-x)dy Along Various Curves

[FeDeX_LaTeX]

Homework Statement

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

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)?

 

[szynkasz]

It should be \int_1^4\frac{10}{9}x+\frac{8}{9}\,dx

 

[FeDeX_LaTeX]

Ah, I see... thanks!