Line integral setup

1. The problem statement, all variables and given/known data
A squirrel weighing 1.2 pounds climbed a cylindrical tree by following the helical path

$x = \cos{t}, y = \sin{t}, z = 4t, 0 \leq t \leq 8 \pi$
(distance measured in feet)

How much work did it do?

2. Relevant equations

$\int_{C} \vec{F} \cdot d\vec{r}$

3. The attempt at a solution
I've defined a curve $C$ by the vector

$\vec{r}(t) = \cos{t} \vec{i} + \sin{t} \vec{j} + 4t \vec{k}$,
$0 \leq t \leq 8 \pi$

I'm not sure where to go from here. Specifically, I don't know how to use the weight of the squirrel. Every other problem I've worked on explicitly gave me a vector field to work with.

I know the bounds of the integral will be from 0 to 8π, and that r'(t) will be used.

 Recognitions: Gold Member Homework Help The squirrel's weight points straight down. Try ##\vec F = \langle 0,0,-1.2\rangle##. And remember the line integral gives the work done by the force. You should be able to check your answer by comparing the change in potential energy.
 Got it, thanks. So I should get my answer with the following integral: $W = \int^{8\pi}_{0} (0\vec{i} + 0\vec{j} - 1.2\vec{k}) \cdot (-\sin{t}\vec{i} + \cos{t}\vec{j} + 4\vec{k}) dt ?$ This isn't for a physics course, and we haven't learned anything about potential energy. If the integral's setup is right, though, I can't take it from there.

Line integral setup

I'm not still new at line integrals so take this with a grain of salt.
 Quote by SithsNGiggles So I should get my answer with the following integral: $W = \int^{8\pi}_{0} (0\vec{i} + 0\vec{j} - 1.2\vec{k}) \cdot (-\sin{t}\vec{i} + \cos{t}\vec{j} + 4\vec{k}) dt ?$
Take the dot product inside the integral, and integrate the answer.

 This isn't for a physics course, and we haven't learned anything about potential energy. If the integral's setup is right, though, I can't take it from there.
Potential energy close to the earth is $E_p=mgh$ so $W=mgΔh$ with m being mass, g gravitational acceleration and h is the heigth.
 Sorry, I meant I can take it from there. Thanks though!