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Homework Statement
Let C be the arc [itex]x=t^2, \space y=2t, \space z= \sqrt{4+3t}[/itex] for [itex]t \in [-1,0][/itex]
Evaluate the line integral :
[itex]\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz[/itex]
Homework Equations
[itex]\int_{C} f(P) dx = \int_{a}^{b} f(P(t)) x'(t) dt[/itex] for [itex]t \in [a,b][/itex]
The Attempt at a Solution
So I have not computed one of these before. I know I'm supposed to parametrize ( choose a parameter ) and then compute my according integral. I really only need help setting it up, so here goes nothing :
This question already has a nice parameter t picked out for me on an appropriate interval. Since the curve is an arc, we know it is one to one, i.e there are no self intersections.
So we compute :
[itex]dx = 2tdt, \space dy = 2dt, \space dz = \frac{1}{2} (4+3t)^{-1/2} dt[/itex]
*I'm not sure I should be concerned with smoothness here*.
Thus our integral becomes :
[itex]\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz = \int_{-1}^{0} (6t^2+8t)dt + (2t)dt - (4t^3)dt[/itex]
So after computing everything and setting it all up that's what I think I should get. Then computing the integral is a simple matter. I just want to make sure I'm not jumping too quickly missing anything and if anyone could verify it would be much appreciated.