# Consider F and C below. F(x, y, z) = (2xz + y^2) i + 2xy\space j + (x^2 + 6z^2) k C: x = t^2,...

## Question:

Consider {eq}\vec{F} {/eq} and {eq}C {/eq} below.

{eq}\vec{F}(x, y, z) = (2xz + y^2) \hat{i} + 2xy\space \hat{j} + (x^2 + 6z^2) \hat{k} {/eq}

{eq}C: x = t^2, y = t + 2, z = 2t -1, 0 \leq t \leq 1 {/eq}

(a) Find a function {eq}f {/eq} such that {eq}\vec{F} = \nabla f. f(x, y, z) =\square {/eq}

(b) Use part (a) to evaluate {eq}\displaystyle \int \nabla f \cdot d\vec{r} {/eq} along the given curve {eq}C{/eq}.

The Gradient Theorem is the Fundamental Theorem of Calculus for Line Integrals. The former name is a bit more revealing, as it implies the restriction that it is valid only for gradient vector fields. Like the original theorem, it works only because of path independence. Also, in order to use it we must find a potential function for the field. The Gradient Theorem says

{eq}\begin{align*} \int_C \vec F \cdot d\vec r &= \int_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)} \nabla f \cdot d\vec r = f(x_2,y_2,z_2) - f(x_1,y_1,z_1) \end{align*} {/eq}