$ F (x,y,z) = ( zy + sinx , zx - 2y , yx-z ) $ is the vector field.
Find the line integral of F which has curve C given as $ x = y = z^2 $ between (0,0,0) and (1,1,1)
I first did this:
Take $ z=t $ so $ z^2= t^2 $ and hence, $ x=y=t $ given parametrisation as
$ r(t) = ( t, t, t^2 ) $ and $ r'(t) = ( 1 , 1 , 2t ) $
But I am not so keen on this because after finding F[ r(t) ] I get a long expression which is then calculated with r'(t) via dot product. I do NOT think this is correct simply because the question is not worth many marks so shouldn't require too much effort.
Is the parametrisation $ r (t) = ( 1 , 1 , t) $ by any chance? Could you kindly explain how to parametrise the curve, thanks in advance!
$$z = t \implies x = y = t^2 \implies F(x,y,z) = F(t^2, t^2, t) = (t^3 + sint, t^3 - 2t^2, t^4 - t)$$ Now $$\mathbf r'(t) = (2t, 2t, 1)$$
And the line integral is calculated as usual with the dot product. Unfortunately, I don't know of a more effortless way.
Edit:
Actually, after thinking for a bit, it occurred to me that there is a quicker way in the case of this particular vector field. Notice that $zy + sinx, zx - 2y, yx - z$ are the partial derivatives w.r.t. $x$, $y$ and $z$ respectively of $G(x,y,z)=xyz - cosx - y^2 - \frac 12 z^2$. The gradient theorem therefore applies, and the integral equals $G(1,1,1) - G(0,0,0)$.