Question

(a) What does it mean to say that \(\int_{\rm{C}} {{\rm{F \times dr}}} \) is independent of path? (b) If you know that\(\int_{\rm{C}} {{\rm{F \times dr}}} \) is independent of path, what can you say about \({\rm{F}}\)?

Short answer

1. Integral value will not change.
2. \({\rm{F}}\) is a conservative vector field if its line integral is independent of path, i.e., there is a scalar function \({\rm{f}}\) that makes \({\rm{F = }}\nabla {\rm{f}}\).

Step-by-step solution

Define line integral

The function to be integrated is determined along a curve in the coordinate system for a line integral. It doesn't matter if the function to be integrated is a scalar or a vector field.

Explanation

1. The integral\(\int_{\rm{C}} {{\rm{F \times dr}}} \)is path independent, which means that the value of the line integral does not change depending on which path between two locations is taken.

Therefore, no change in integral value.

Explanation

1. Let\({\rm{F}}\)be a continuous vector field on an open connected area\({\rm{D}}\), with the line integral,

\(\int_{\rm{C}} {{\rm{F \times dr}}} \)

Which is not affected by the route in\({\rm{D}}\).Then\({\rm{F}}\)is a conservative vector field on\({\rm{D}}\), i.e., a scalar function\({\rm{f}}\)exists such that,

\({\rm{F = }}\nabla {\rm{f}}\)

Therefore, \({\rm{F}}\) is a conservative vector field if the line integral of \({\rm{F}}\) is independent of path, i.e., there exists a scalar function \({\rm{f}}\) such that \({\rm{F = }}\nabla {\rm{f}}\).