Chapter 10: Problem 4
Let \(U\) be an open set in \(\mathbb{R}^{3}\), and let \(S\) be a smooth oriented surface in \(U\) with orienting normal vector field \(\mathbf{n}\). Let \(\mathbf{F}\) be a continuous vector field on \(U\). If \(\mathbf{F}(\mathbf{x})=\mathbf{n}(\mathbf{x})\) for all \(\mathbf{x}\) in \(S,\) prove that \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}=\operatorname{Area}(S)\)
Short Answer
Step by step solution
Understanding the Problem
Expressing the Surface Integral
Substituting Vector Field Expression
Simplifying the Dot Product
Computing the Integral of 1 Over the Surface
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Fields
- Imagine a wind map where arrows at different points show the wind direction and speed.
- Each arrow represents the vector at that point.
- The length denotes the wind's speed, and the direction of the arrow shows the wind's direction.
Normal Vector
- The normal vector determines how a surface "points" in space.
- For any smooth oriented surface \(S\), the normal vector fields are smoothly varying vectors that define directions perpendicular to \(S\).
Integration
- For a vector field on a surface, surface integrals help compute work done by the field across the surface.
- In this problem, the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) represents the total effect of the field \(\mathbf{F}\) across the surface \(S\).
- Since \(\mathbf{F}(\mathbf{x}) = \mathbf{n}(\mathbf{x})\), the integral simplifies to \(\iint_{S} \mathbf{n} \cdot \mathbf{n} \, dS\).
Smooth Surfaces
- Smooth surfaces ensure that normal vectors can be consistently defined.
- Allow for reliable mathematical operations like integration, because there are no sharp edges or discontinuities to complicate calculations.
- On a smooth surface, the magnitude of the normal vectors is consistent, making calculations more predictable.