Using Stokes' Theorem with a Constant Scalar Field

The problem illustrates the significance of Generalized Stokes' Theorem in relating surface integrals to line integrals, particularly emphasizing that the scalar field $\phi(x)$ must be constant on the boundary curve $C$. This constancy allows for simplification of the line integral, transforming it into a fundamental case where the Fundamental Theorem of Line Integrals applies, ensuring that the integral evaluates to zero. The visual demonstration reinforces that without the boundary condition of a constant scalar field, the integral can yield non-zero results, underscoring the critical role of this condition in achieving a predictable outcome.

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✍️Mathematical Proof

$\gg$Mathematical Structures Underlying Physical Laws

$\complement\cdots$Counselor

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Generalized Stokes' Theorem

This problem highlights a powerful extension of Stokes' theorem. Instead of dealing with the curl of a single vector field, this form relates an integral involving the cross product of two gradients to a line integral of a gradient field.

Constant on a Boundary

The core condition is that the scalar field $\phi(x)$ is constant on the boundary curve $C$. This allows you to pull the value of $\phi$ out of the line integral, turning the problem into a fundamental line integral of a gradient field.

Fundamental Theorem of Line Integrals

This theorem states that the line integral of a gradient field along a closed path is always zero. The reason for this is that a gradient field is a conservative field, and the work done by a conservative force around a closed loop is zero. This principle is key to proving the final result.

🎬Demonstration

The demo visually confirms that the condition of a constant scalar field on the boundary is essential for the surface integral to be zero. When the scalar field $\phi$ is constant on the boundary, the line integral evaluates to zero. When $\phi$ is not constant, the line integral has a non-zero value, and the proof fails. This highlights the importance of the initial condition in the problem statement, which turns a potentially complex integral into a straightforward case with a predictable result.

A constant scalar field leads to a zero integral result