Curve integral

  • Thread starter 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.
 

Answers and Replies

  • #2
116
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Okay, never mind, I realised my mistake. I'd stared myself blind in my tiredness x)
 

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