# How do I parametrize this vector so that I can evaluate this line integral?

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.

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

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}