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

An irrotational vector field means: \[ \nabla \times \vec{F} = 0 \Rightarrow \vec{F} = \nabla \phi \] for some scalar function \( \phi \). Therefore, \[ \oint_C \vec{F} \cdot d\vec{R} = \oint_C \nabla \phi \cdot d\vec{R} = 0 \] by the fundamental theorem for line integrals, since \( C \) is closed.