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So this may seem a little strange, but I'm not entirely sure how to solve a problem like this.

$P(x,y) = x+2y$, $Q(x,y) = 2x - y$. $C$ consists of line segments from $(3,2)$ to $(3,-1)$ and from $(3,-1)$ to $(-2,-1)$.

I want to do $$\int_C P(x,y) dx + Q(x,y)dy.$$

I understand how to do this on a normal curve - I parametrize the region, by parametrizing $x$,$y$ and $z$, I convert $dx$ and $dy$ to $dt$ and integrate in one parameter. However, what I don't get is how to do in this specific case in which parametrization seems hard.

I originally thought of integrating $P(x,y)$ over -2 to 3 and $Q(x,y)$ between -1 and 2, because that's when $dx$ and $dy$ are not equal to 0 respectively, however, if I do that, I'm left with a $y$ term in the former and a $x$ term in the latter.

So how should I approach this ?

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1 Answer

up vote 0 down vote accepted

Usually if the curve is closed, then Green's Theorem could come into play (well not always) and simplify things.

In your case the curve isn't closed, so split the integral into two paths

So $C = \{(3,2-3t) : 0 \le t \le 1 \} \cup \{(3-5s,-1): 0 \le s \le 1 \}$

Does this help?

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Ah yes it does- I got it- thanks! –  praks5432 Nov 11 '12 at 23:23
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