91Ó°ÊÓ

A scalar potential function is a scalar field from which we can derive a vector field. When dealing with differential forms, finding a scalar potential function can greatly simplify evaluations.

Consider a vector field constructed from a differential form as in the step-by-step solution provided. In the given problem, the vector field is expressed as \[\mathbf{F} = (yz, xz, xy)\,\] where each term corresponds to a part of the differential form. For this vector field to be exact, there must exist a scalar function \( f(x, y, z) \) such that, \[ \frac{\partial f}{\partial x} = yz, \quad \frac{\partial f}{\partial y} = xz, \quad \text{and} \quad \frac{\partial f}{\partial z} = xy. \]

The potential function can often be found by integrating each of these expressions, as seen in the solution where \( f(x, y, z) = xyz \) is identified as consistent across all parts.

This means that to evaluate a line integral of an exact form, we can simply compute the potential function at the endpoints of the path and subtract the values.

Mixed derivatives play a crucial role in verifying whether a differential form is exact. Simply put, mixed derivatives are the partial derivatives of a function taken with respect to different variables, such as first with respect to one variable, then another.
To determine the exactness of a differential form, we must check for certain conditions involving mixed derivatives. In the context of the given exercise, for the differential form to be exact, the following mixed derivative conditions must hold:

• \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)
• \( \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} \)
• \( \frac{\partial R}{\partial x} = \frac{\partial P}{\partial z} \)

In this exercise, each of these partial derivatives was computed and found to be equal:

• \( \frac{\partial P}{\partial y} = z = \frac{\partial Q}{\partial x} \)
• \( \frac{\partial Q}{\partial z} = x = \frac{\partial R}{\partial y} \)
• \( \frac{\partial R}{\partial x} = y = \frac{\partial P}{\partial z} \)

These conditions confirm that the differential form is indeed exact. This check is an essential step before attempting to find a scalar potential function.

Line integrals allow us to integrate over a curve or a path in space, offering powerful applications in physics and engineering by measuring fields' effects along paths. A line integral can be evaluated as an ordinary integral when the function is parametric with respect to the path's curve.
The uniqueness in exact differential forms lies in simplifying line integrals. If we identify a scalar potential function as we did in the problem, the path becomes almost irrelevant, as long as it starts and ends at the specified points.
For the evaluated exercise, the integral \[ \int_{(1,1,2)}^{(3,5,0)} y z \, dx + x z \, dy + x y \, dz \] was calculated using the derived potential function \( f(x, y, z) = xyz \).
By evaluating the potential function at the ending point \((3, 5, 0)\) and starting point \((1, 1, 2)\), we found that:

• \( f(3, 5, 0) = 0 \)
• \( f(1, 1, 2) = 2 \)

Therefore, the line integral evaluated to \[0 - 2 = -2\]. Using potential functions streamlines solving these problems, emphasizing how identifying exact differential forms turns complex line integrals into simpler calculations based directly on endpoints.