Work of a Vector Field along a path

[NaN-Gram]

I need to find the work of the vector field

F= 3xi+6y^2j+4zk

along the path

r(t)=ti+cos(t)j+3sin(t)k

from the point (3,0,0) to (pi,0,1).

I know the fundamental theorem of calculus states to take a function differentiable to the vector field, insert the given endpoints, and take the difference.

I'm having trouble finding that function. I know I can take the derivative of r(t) and multiply it with the integral of the field, and though that would be equivalent, I'm having trouble determining how to know that I'm on the right track.

r'(t)=i-sin(t)j+3cos(t)k

I'm stumped as to where to go from here.

 

[HallsofIvy]

First, only the component of force tangent to the path contributes to the work done. That is given by the dot product of the force vector and the tangent vector to the path.

Here the force vector is F= 3xi+6y^2j+4zk and the path is given by r(t)=ti+cos(t)j+3sin(t)k so the tangent vector is dr= (i- sin(t)j+ 3cos(t)kdt.

The dot product of those two vectors is (3x- 6y^2sin(t)+ 12zcos(t))dt.

And, of course, on this path x= t, y= cos(t), z= 3sin(t) so that becomes (3t- 6cos^2(t)sin(t)+ 36sin(t)cos(t))dt.

However, you have a serious problem- since sin(x) and cos(x) are never 0 for the same x, the point (3, 0, 0) is NOT on that path! Also since cos(pi)= -1 and sin(pi)= 0, neither is (pi, 0, 1)!

The work is the integral of that from