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The question reads: Evaluate $$ \int_\gamma f(z)dz$$ where $$f(z)=x^2: x,y \in \mathbb{R} $$ and $\gamma$ is the parabola $y=2x^2$ from $x=0$ to $x=2$.

This is the first question I've encountered where $f(z)$ is given in the form $u(x,y)$, and I've been lost as to how I should go about solving it. I started by parameterizing the parabola as $\gamma(t)=t+i2t^2$ but I can't see what to do from here. Any tips will be greatly appreciated, thanks!.

Forcefedglas
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  • parametrize the contour $\gamma(t) = t+i 2t^2, t \in [0,2]$, $\gamma'(t) = 1+ 4i$, then change of variable $z = \gamma(t), dz = \gamma'(t) dt = (1+4i)dt$, and $\int_\gamma f(z) dz = \int_0^2 f(\gamma(t)) \gamma'(t) dt = \ldots$ the main thing you'll have to prove on contour integrals is that $\int_\gamma f(z) dz \overset{def}= \int_a^b f(\gamma(t)) \gamma'(t) dt$ doesn't depend on the (piecewise $C^1$) parametrization you chose . – reuns May 01 '16 at 01:15
  • I integrated $\int (t+2it^2)(1+4it)$ from 0 to 2 and got -30+16i? The imaginary part of the answer matches up but the real part seems to be different. (Real part should be 8/3) – Forcefedglas May 01 '16 at 01:48
  • $f(x+iy) = x^2$, hence $f(\gamma(t)) = f(t+2i t^2) = t^2$ – reuns May 01 '16 at 01:52
  • Ah ok that makes sense. Why does the imaginary part of the integral stay the same though? – Forcefedglas May 01 '16 at 02:08
  • Nevermind I think I got it, thanks! – Forcefedglas May 01 '16 at 02:52

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