Solution of (i). The work done by the force (7.24a) in moving the particle
from (0, 0) to (1, a) on any path is given by (7.23):

\mathscr{W}=\int_{\left(x_1, y_1, z_1\right)}^{\left(x_2, y_2, z_2\right)}\left(F_x d x+F_y d y  +  F_z d z\right)               (7.23)

\mathscr{W}=\int_{\mathscr{C}} \mathbf{F} \cdot d \mathbf{x}=\int_{(0,0)}^{(1, a)} b x y d x+c y d y.                     (7.24b)

To integrate the first term in this equation, it is evident that we shall need to know how y is related to x. This means that the path must be specified, and hence the force (7.24a) is a nonconservative force.

For the parabolic path y = ax² shown in Fig. 7.5a, the path integral in (7.24b) becomes

\mathscr{W}=\int_{(0,0)}^{(1, a)} a b x^3 d x  +  c y d y=\int_0^1 a b x^3 d x  +  \int_0^a c y d y.                      (7.24c)

Hence, the work done by the nonconservative force (7.24a) in moving the  particle along the parabolic path y = ax² from (0, 0) to (1, a) is

\mathscr{W}=\frac{a b}{4}  +  \frac{c a^2}{2}                   (7.24d)

Solution of (ii). Consider the work done by the same force (7.24a) acting over the path defined by the lines x = 0 and y = a joining the same end points, as shown in Fig. 7.5b. The path  \mathscr{C}  consists of two parts  \mathscr{C}_1  and  \mathscr{C}_2 ;  hence , (7.21) is written as

\mathscr{W}=\int_{\mathscr{C}} \mathbf{F}(\mathbf{x}) \cdot d \mathbf{x}=\int_{\mathbf{x}_1}^{\mathbf{x}_2} \mathbf{F}(\mathbf{x})\cdot d \mathbf{x}                                (7.21)

\mathscr{W}=\int_{\mathscr{C}=\mathscr{C}_1 \cup \mathscr{C}_2} \mathbf{F} \cdot d \mathbf{x}=\int_{\mathscr{C}_1} \mathbf{F} \cdot d \mathbf{x}  +  \int_{\mathscr{C}_2} \mathbf{F} \cdot d \mathbf{x}                       (7.25a)

wherein each integrand has the form of (7.24b). Now, x = 0 and y is variable on   \mathscr{C}_1;  hence, (7.24b) applied to  \mathscr{C}_1  yields

\int_{\mathscr{C}_1} \mathbf{F} \cdot d \mathbf{x}=\int_{(0,0)}^{(0, a)} c y d y=\int_0^a c y d y=\frac{c a^2}{2}               (7.25b)

Similarly, y = a, dy = 0 and x is variable on  \mathscr{C}_2;  hence, application of (7.24b) to  \mathscr{C}_2  gives

\int_{\mathscr{C}_2} \mathbf{F} \cdot d \mathbf{x}=\int_{(0,0)}^{(1, a)} a b x d x=\int_0^1 a b x d x=\frac{a b}{2}.                    (7.25c)

Therefore, the total work done by the force (7.24a) in moving the particle over  the path   \mathscr{C}=\mathscr{C}_1 \cup \mathscr{C}_2  between the same end states from (0, 0) to (1, a), by  (7.25a) , is

\mathscr{W}=\frac{a b}{2}  +  \frac{c a^2}{2}                 (7.25d)

To conclude, let the reader consider the following additional example.

Exercise 7.4. Show that the work done by the force (7.24a) in moving the particle over the straight line path y = ax joining (0, 0) to (1 , a) is given by

\mathscr{W}=\frac{a b}{3}  +  \frac{c a^2}{2}                 (7.25e)

Solution of (iii). Finally, we wish to determine the condition to be satisfied in order that (7.24a) may be conservative. The solutions (7.24d), (7.25d), and (7.25e) show that the work done by the same force will be the same for all paths considered above only if b = 0. This is the condition needed for the force (7.24a) to be conservative . Indeed, conversely, let us consider the force

\mathbf{F}=c y \mathbf{j}.                     (7.26a)

Then the work done by F acting over a path  \mathscr{C}  joining (0, 0) to (1, a) is given by

\mathscr{W}=\int_{\mathscr{C}} \mathbf{F} \cdot d \mathbf{x}=\int_{(0,0)}^{(1, a)} c y d y=\int_0^a c y d y=\frac{c a^2}{2}.               (7.26b)

Note that in this integration there is no need to mention a specific path  \mathscr{C}.  This result is independent of the path; it depends only on values at the end points.  The force (7.26a) is a conservative force. We thus find that the work done by the force (7.24a) is independent of the particle’s path, when and only when b =0.