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In order to solve a line integral, I need to establish a smooth parametrization of the curve over which it is supposed to be integrated.

The curve, $D$, is the intersection of the surfaces $x^2 + y^2 = 1$ and $z=x^2$.

To me, a logical parametrization is: $$r(t)=(t, (1-t^2)^{1/2}, t^2)$$ because if $x = t$, then $$y = (1-t^2)^{1/2},$$ and $$z = t^2$$

Is this right? I appreciate any help.

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2 Answers

No, your parametrization only gives you positive values of $y$. Since $x^2 + y^2 = 1$ is a circle in the $x-y$ plane, you might try $x = \cos(t)$, $y = \sin(t)$.

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awesome, worked like a charm. Thanks! –  William K. Aug 3 '11 at 19:41
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r(t)=

so r'(t) = <-sin(t), cos(t)> and you can easily calculate F(r(t)) from this to find the parametrization of F

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Is part of your post missing? It says $r(t) =$, but ends there. Did it get cut off? Regards –  Amzoti May 1 '13 at 20:30
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