Question 9.9.8: (a) Show that the line integral ∫C (y + yz)dx + (x + 3z³ + x......

(a) With the identifications

we see that the equalities

\frac{\partial P}{\partial y}=1+z=\frac{\partial Q}{\partial x}, \quad \frac{\partial P}{\partial z}=y=\frac{\partial R}{\partial x}, \quad \text { and } \quad \frac{\partial Q}{\partial z}=9 z^{2}+x=\frac{\partial R}{\partial y}

hold throughout 3-space. From (12) we conclude that \mathrm{F} is conservative and therefore the integral is independent of the path.

\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \quad \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \quad \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}   (12)

(b) The path C shown in FIGURE 9.9.6 represents any path with initial and terminal points (1,1,1) and (2,1,4). To evaluate the integral we again illustrate how to find a potential function \phi(x, y, z) for \mathrm{F} using partial integration.

First we know that

\frac{\partial \phi}{\partial x}=P, \quad \frac{\partial \phi}{\partial y}=Q, \quad \text { and } \quad \frac{\partial \phi}{\partial z}=R

Integrating the first of these three equations with respect to x gives

\phi=x y+x y z+g(y, z)

The derivative of this last expression with respect to y must then be equal to Q :

\frac{\partial \phi}{\partial y}=x+x z+\frac{\partial g}{\partial y}=x+3 z^{3}+x z.

Hence,

\frac{\partial g}{\partial y}=3 z^{3} \text { implies } g=3 y z^{3}+h(z).

Consequently, \phi=x y+x y z+3 y z^3+h(z).

The partial derivative of this last expression with respect to z must now be equal to the function R :

\frac{\partial \phi}{\partial z}=x y+9 y z^{2}+h^{\prime}(z)=9 y z^{2}+x y-1.

From this we get h^{\prime}(z)=-1 and h(z)=-z+K. Disregarding K, we can write

\phi=x y+x y z+3 y z^{3}-z .   (13)

Finally, from (11) and the potential function (13) we obtain

\int_C F \cdot d r =\int_C \nabla \phi \cdot d r =\phi(x(b), y(b), z(b))-\phi(x(a), y(a), z(a))=\phi(B)-\phi(A) .   (11)