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If A = (3x$^2$-6yz)$\mathbf{i}$ + 2(y+3xz)$\mathbf{j}$ + (1-4xyz$^2$)$\mathbf{k}$ how do I evaluate

$\int$ A $\bullet$ d$\mathbf{r}$ from (0,0,0) to (1,1,1)

along the path x=t, y=t$^2$, z=t$^3$?

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First find out the interval of the parameter that sweeps out this path. Technically you do this by solve the vector equations $(t, t^{2},t^{3}) = (0,0,0)$ and $(t, t^{2},t^{3}) = (1,1,1)$ but in this case it's obvious that, in the first case, $t=0$ and in the second $t=1$.

Next, find the gradient of $A$, i.e. vector of partial derivatives,

$(6x, 2y, -8xyz)$

and substitute the parameterization into this,

$(6t, 2t^{2}, -8tt^{2}t^{3})$ (note that I haven't combined terms so you can see where things came from)

Next, plug in the parameterization into $A$, which gives you $A(r(t))$, namely,

$A(r(t)) = (3t^{2}-6t^{2}t^{3}, 2(t^{2}+3tt^{3}), 1-4tt^{2}[t^{3}]^{2})$

Next form the dot-product of this with the gradient found earlier:

$(3t^{2}-6t^{2}t^{3}, 2(t^{2}+3tt^{3}), 1-4tt^{2}[t^{3}]^{2}) \cdot (6t, 2t^{2}, -8tt^{2}t^{3})$

When you compute this, find the integral of it from 0 to 1 with respect to $t$.

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