Question

Use Exercise 25 to show that the line integral \(\int_C y dx + xdy + xyzdz\) is not independent of path.

Short answer

The line integral \(\int_C y dx + xdy + xyzdz\) is not independent of path

Step-by-step solution

Given information

The given line integral is \(\int_C y dx + xdy + xyzdz\).

Potential function and condition for vector field to be conservative

When a vector field\({\bf{F}} = P{\bf{i}} + Q{\bf{j}} + R{\bf{k}}\)is conservative, then there exist a potential function such that\({\bf{F}} = \nabla f\)and the integral of vector field is independent of path.

The essential conditions for vector field to be conservative is\(\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}},\frac{{\partial P}}{{\partial z}} = \frac{{\partial R}}{{\partial x}}\), and\(\frac{{\partial Q}}{{\partial z}} = \frac{{\partial R}}{{\partial y}}\).

Show that the line integral \(\int_C y dx + xdy + xyzdz\) is not independent of path

Compare the integrals\(\int_C \nabla f \cdot d{\bf{r}}\)and\(\int_C y dx + xdy + xyzdz\)and obtain the value of\(\nabla f\)as\(\nabla f = y{\bf{i}} + x{\bf{j}} + xyz{\bf{k}}\).

Compare the equation\(\nabla f = y{\bf{i}} + x{\bf{j}} + xyz{\bf{k}}\)with\({\bf{F}} = P{\bf{i}} + Q{\bf{j}} + R{\bf{k}}\). and obtain the value of P, Qand\(R\)as\(P = y,Q = x\)and\(R = xyz\).

Obtain the value of\(\frac{{\partial P}}{{\partial z}}\)as follows.

\(\begin{align} \frac{\partial P}{\partial z} & =\frac{\partial }{\partial z}(y) \\ & =0\,\,\,\,\,\,\left\{ \because \frac{\partial }{\partial t}(k)=0 \right\} \end{align}\)

Obtain the value of\(\frac{{\partial R}}{{\partial x}}\)follows.

\(\begin{aligned}\frac{{\partial R}}{{\partial x}} & = \frac{\partial }{{\partial x}}(xyz)\\ & = yz\frac{\partial }{{\partial x}}(x)\\ & = yz(1)\\ & = yz\end{aligned}\)

Hence\(\frac{{\partial P}}{{\partial z}} \ne \frac{{\partial R}}{{\partial x}}\).

Therefore, the vector field is not a conservative and integral of vector field is not independent of path.

Thus, the line integral \(\int_C y dx + xdy + xyzdz\) is not independent of path.