In the realm of vector calculus and geometry, the integral of the cross product of the outward unit normal vector n and the position vector r over a closed surface S plays a significant role. This integral, often denoted as $\int_S n\times r dS$, arises in various physical applications, particularly in electromagnetic theory and fluid dynamics. Notably, it has a remarkable property: it always evaluates to zero. This article delves into the proof of this fundamental theorem, exploring the underlying concepts and its implications.
The Significance of $\int_S n\times r dS = 0$
The integral $\int_S n\times r dS$ represents the moment of the surface S with respect to the origin. It measures the tendency of the surface to rotate about the origin, taking into account both the surface area and the distribution of its points. Intuitively, for a closed surface, we can imagine this integral as a measure of the "net torque" experienced by the surface due to its position and shape. The fact that this integral always evaluates to zero implies that the surface, in its entirety, experiences no net tendency to rotate. This finding has profound implications in various fields, such as:
- Electromagnetism: This theorem is crucial in understanding the behavior of electric and magnetic fields. For instance, it underlies the principle of Gauss's Law for magnetic fields, which states that the total magnetic flux through any closed surface is always zero.
- Fluid Dynamics: In fluid dynamics, the integral is related to the angular momentum of a fluid element. The theorem ensures that the total angular momentum of a closed fluid volume remains constant over time.
- Solid Mechanics: In solid mechanics, this integral is related to the moment of inertia of a solid object. The theorem implies that a closed surface experiences no net torque due to its shape and mass distribution.
Proof of $\int_S n\times r dS = 0$
The proof of this theorem relies on Stokes' theorem, a fundamental result in vector calculus. Stokes' theorem relates the integral of a vector field over a surface to the integral of its curl along the boundary of the surface.
Stokes' Theorem: For a smooth surface S with boundary curve C and a continuously differentiable vector field F, we have:
$\int_S (\nabla \times F) \cdot dS = \int_C F \cdot dr$
Proof:
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Choosing a suitable vector field: Let's choose the vector field F to be r × a, where a is an arbitrary constant vector.
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Calculating the curl of F: Using the vector triple product identity, we get:
$\nabla \times (r \times a) = a (\nabla \cdot r) - r (\nabla \cdot a) = 3a$
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Applying Stokes' Theorem: Substituting this result into Stokes' theorem, we obtain:
$\int_S (3a) \cdot dS = \int_C (r \times a) \cdot dr$
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Simplifying the line integral: The line integral on the right-hand side can be simplified using the fact that a is constant:
$\int_C (r \times a) \cdot dr = a \cdot \int_C (r \times dr)$
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Evaluating the line integral: The line integral $\int_C (r \times dr)$ represents the area enclosed by the closed curve C. This is because the cross product r × dr represents a vector perpendicular to the plane containing r and dr, and its magnitude is equal to the area of the parallelogram formed by these vectors. Since C is a closed curve, its initial and final points coincide, resulting in a zero-valued integral.
$\int_C (r \times dr) = 0$
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Final result: Combining all the steps, we get:
$\int_S (3a) \cdot dS = a \cdot 0 = 0$
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Arbitrary constant vector: Since a is an arbitrary constant vector, the above equation holds for any direction. This implies:
$\int_S n \times r dS = 0$
Conclusion: The integral $\int_S n \times r dS$ always evaluates to zero for a closed surface S. This theorem arises from the fundamental connection between surface integrals and line integrals established by Stokes' theorem. Its applications extend across diverse fields, highlighting its importance in understanding various physical phenomena involving surface properties and moments.
