Question

Let \( \vec{F} \) be an irrotational vector function. If \( C \) is the closed curve which is the boundary of an open surface \( S \), then: \[ \oint_C \vec{F} \cdot d\vec{R} = ? \]

• A. \( \iint_E \left( \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} \right) dx\,dy \), \( \phi \) is a scalar function and \( E \) is the region of \( C \)
• B. \( \iiint_S \text{div } \vec{F} \, dv \)
• C. \( \int_C (\text{grad } \phi) \cdot d\vec{R} = 0 \), for some scalar function \( \phi \)
• D. divergence of \( \vec{F} \)

Hint:

In irrotational fields, \( \vec{F} \) is conservative and can be written as \( \nabla \phi \). Line integrals over closed paths in conservative fields are always zero.

Answer 1

The Correct Option is C

To solve this problem, we need to utilize the properties of an irrotational vector field. An irrotational vector field \( \vec{F} \) means that its curl is zero, i.e., \( \nabla \times \vec{F} = 0 \). According to Green's theorem, if \(\vec{F}\) is irrotational over the open surface \(S\) with boundary \(C\), then the line integral around the closed curve \(C\) is zero. Mathematically, this is expressed using the following: \[ \oint_C \vec{F} \cdot d\vec{R} = 0 \] Since \(\vec{F}\) is irrotational, there exists a scalar potential function \(\phi\) such that \(\vec{F} = \nabla \phi\). Therefore, substituting back, we have the line integral of the gradient: \[ \oint_C \vec{F} \cdot d\vec{R} = \oint_C (\nabla \phi) \cdot d\vec{R} = 0 \] This is because the line integral of a gradient over a closed path is zero, as it corresponds to the net change in the scalar field \(\phi\), and over a closed loop, that change is zero.
Hence, the correct answer is: \(\int_C (\text{grad } \phi) \cdot d\vec{R} = 0\), for some scalar function \(\phi\).