Question

Let \(F_1, F_2,\) and \(F_3\) be functions of \((x, y, z)\). Suppose that for every given pair of points \(A\) and \(B\) in space, the line integral: \[ \int_C \left(F_1 dx + F_2 dy + F_3 dz\right) \] evaluates to the same value along any path \(C\) that starts at \(A\) and ends at \(B\). Then which of the following is/are true?

• A. For every closed path \(\Gamma\), we have: \[ \oint_\Gamma \left(F_1 dx + F_2 dy + F_3 dz\right) = 0. \]
• B. There exists a differentiable scalar function \(f(x, y, z)\) such that: \[ F_1 = \frac{\partial f}{\partial x}, \quad F_2 = \frac{\partial f}{\partial y}, \quad F_3 = \frac{\partial f}{\partial z}. \]
• C. \[ \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} = 0. \]
• D. \[ \frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}, \quad \frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z}, \quad \frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}. \]

Hint:

To determine if a vector field is conservative, check for path independence, verify that the curl is zero, and confirm the existence of a scalar potential function.

Answer 1

The Correct Option is A

Step 1: Path independence and conservative field.
If the line integral of a vector field \(\vec{F} = (F_1, F_2, F_3)\) is path-independent, the field is conservative. This implies there exists a scalar potential function \(f(x, y, z)\) such that: \[ F_1 = \frac{\partial f}{\partial x}, \quad F_2 = \frac{\partial f}{\partial y}, \quad F_3 = \frac{\partial f}{\partial z}. \] Step 2: Line integral over a closed path.
For conservative fields, the line integral over any closed path is zero. Mathematically, this is expressed as: \[ \oint_\Gamma \vec{F} \cdot d\vec{r} = 0. \] Step 3: Curl condition for a conservative field.
A necessary condition for a field to be conservative is that its curl must vanish: \[ {curl } \vec{F} = \nabla \times \vec{F} = 0. \] This leads to the component-wise conditions: \[ \frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}, \quad \frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z}, \quad \frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}. \] Final Answer: \[ \boxed{{(1, 2, 4)}} \]