Consider the vector field
F=\begin{pmatrix} -3z^2sinx \\[0.3em] 8y^3z \\[0.3em] 2y^4+6zcosx \end{pmatrix}
By evaluating the path integral, compute $\int\limits_C F.t $ ds where the path C is a straight line that begins at $(3\pi/2,2,-1)$ and ends at the point $(X,Y,Z)$ where $X,Y,Z$ are fixed values.
Now Ive had a go, parametrised the curve to get
$x=3\pi/2(1-t)+Xt$,
$y=2(1-t)+Yt$
$z=-1(1-t)+Zt$
I was wondering where I go from here, simply evaluating $F(x(t),y(t),z(t))$, at least the way I have approached it seems like a lot of work with what I have (plugging in x,y and z) .
So I guess I am asking: Am I correct thus far and what is the best way to continue?
What you have done is correct, but a simple way of doing this
For a scalar field $$S=2y^4z+3z^2cosx$$
F is gradient of S i.e., $F=\bigtriangledown S$
Hence $\int Fdt=\int\bigtriangledown S=S(X,Y,Z)-S(\frac{3\pi}{2}, -2,1)=S(X,Y,Z)-32$