Question 9.9.7: (a) Show that ∫C F · dr, where F(x, y) = (y² - 6xy + 6)i + (......
(a) Show that \int_{C} \mathrm{F} \cdot d \mathrm{r}, where \mathrm{F}(x, y)=\left(y^{2}-6 x y+6\right) \mathrm{i}+\left(2 x y-3 x^{2}-2 y\right) \mathrm{j}, is independent of the path C between (-1,0) and (3,4).
(b) Find a potential function \phi for \mathrm{F}.
(c) Evaluate \int_{(-1,0)}^{(3,4)} \mathrm{F} \cdot d \mathrm{r}.
(a) Identifying P=y^{2}-6 x y+6 and Q=2 x y-3 x^{2}-2 y yields
\frac{\partial P}{\partial y}=2 y-6 x=\frac{\partial Q}{\partial x}.
The vector field \mathrm{F} is conservative because (6)
\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} (6)
holds throughout the x y-plane and as a consequence the integral \int_{C} \mathrm{F} \cdot d \mathrm{r} is independent of the path between any two points A and B in the plane.
(b) Because \mathrm{F} is conservative there is a potential function \phi such that
\frac{\partial \phi}{\partial x}=y^{2}-6 x y+6 \text { and } \frac{\partial \phi}{\partial y}=2 x y-3 x^{2}-2 y . (8)
Employing partial integration on the first expression in (8) gives
\phi=\int\left(y^{2}-6 x y+6\right) d x=x y^{2}-3 x^{2} y+6 x+g(y), (9)
where g(y) is the “constant” of integration. Now we take the partial derivative of (9) with respect to y and equate it to the second expression in (8):
\frac{\partial \phi}{\partial y}=2 x y-3 x^{2}+g^{\prime}(y)=2 x y-3 x^{2}-2 y.
From the last equality we find g^{\prime}(y)=-2 y. Integrating again gives g(y)=-y^{2}+C, where C is a constant. Thus
\phi=x y^{2}-3 x^{2} y+6 x-y^{2}+C. (10)
(c) We can now use Theorem 9.9.2 and the potential function (10) (without the constant) to evaluate the line integral:
\begin{aligned} \int_{C} \mathrm{F} \cdot d \mathrm{r} & \left.=\int_{(-1,0)}^{(3,4)} \mathrm{F} \cdot d \mathrm{r}=\left(x y^{2}-3 x^{2} y+6 x-y^{2}\right)\right]_{(-1,0)}^{(34)} \\ & =(48-108+18-16)-(-6)=-52 . \end{aligned}