- #1
Hannisch
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Homework Statement
Calculate the curve integral
[tex]\int_{\gamma}(x^2+xy)dx+(y^2-xy)dy[/tex]
where [tex]\gamma[/tex] is the line segments from (0,0) to (2,0) and from (2,0) to (2,2).
Homework Equations
[tex]\int_{\gamma} P(x,y)dx+Q(x,y)dy= \int^{\beta}_{\alpha}(P(g(t),h(t))g'(t)+Q(g(t),h(t))h'(t))dt[/tex]
where
x=g(t)
y=h(t)
The Attempt at a Solution
I started by splitting the line up in two parts,
[tex]\gamma _{1}[/tex] = the segment from (0,0) to (2,0)
[tex]\gamma _{2}[/tex] = the segment from (2,0) to (2,2)
I then stated that for the first segment, the parametric equation would be
(x,y)=(t,0) and [tex]0 \leq t \leq 2[/tex]
and for the other segment:
(x,y)=(0,t) and [tex]0 \leq t \leq 2[/tex]
Both of these segments are in the positive direction, so
[tex]\int _{\gamma}(x^2+xy)dx+(y^2-xy)dy = \int _{\gamma_{1}}(x^2+xy)dx+(y^2-xy)dy + \int _{\gamma{2}}(x^2+xy)dx+(y^2-xy)dy = \int^{2}_{0} t^2 dt + \int^{2}_{0} t^2 dt = 16/3[/tex]
But this is incorrect and I haven't the faintest idea where and why I went wrong. It's driving me crazy (it should be such a simple problem) but I can't figure it out. I thought this was what the did in the book, but alas.. According to the book the answer is 4/3.