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So I'm asked to evaluate the line integral $$ \int_C F dr $$ where $$ F = yz\cdot\vec{i} + \sin(y)\cdot\vec{j} + \cos(z)\cdot \vec{k} $$ along the curve $$r(t)=t^2\cdot\vec{i}+t\cdot\vec{j}+t^3\cdot\vec{k} \quad \text{ for } 1 \leq t \leq 3 $$

My guess would be to find some kind of parametization for $F$ in terms of t, and then apply the forumla $ \int_C F dr = \int_C F(r(t))\cdot r'(t)dt$, but I can't seem to find such a parametization. Once I do find one, I don't think it'd be too hard to just plug $F(r(t))$ and $r'(t)$ into the forumla.

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  • $\begingroup$ The formula for $r$ means that on $C, x=t^2, y = t, z=t^3$ $\endgroup$ – saulspatz Mar 24 '18 at 21:10
  • $\begingroup$ $r$ is already parametrised, so you can directly apply the formula you have there. $\endgroup$ – Joppy Mar 24 '18 at 21:33
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You don't have to do anything else.

In this case

$$F(x,y,z) = \left(y\cdot z,sin(y),cos(z)\right)$$

and

$$r(t)=(t^2,t,t^3)$$

So $F\left(r(t)\right) = \left(t\cdot t^3,sin(t),cos(t^3)\right)$

And then you can apply the formula that you mentioned.

$\underline{Note}$:

$ \begin{aligned} F(x,y,z) &= yz\cdot \vec{i} + \sin(y)\cdot\vec{j} + \cos(z)\cdot \vec{k} \\ &= yz\cdot (1,0,0) + sin(y)\cdot (0,1,0) + cos(z)\cdot(0,0,1) \\ &= (yz,0,0) + (0,sin(y),0) + (0,0,cos(z))\\ &= \left(y\cdot z,sin(y),cos(z)\right) \end{aligned} $

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