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What is the mathematical statement of Gauss's Divergence Theorem relating surface and volume integrals?

Gauss's Divergence Theorem states that the flux of a vector field F⃗\vec{F} through a closed surface SS equals the triple integral of the divergence of F⃗\vec{F} over the volume VV enclosed by SS. Mathematically, it is expressed as ∬SF⃗⋅n⃗dS=∭V(∂Fx∂x+∂Fy∂y+∂Fz∂z)dV\iint_S \vec{F} \cdot \vec{n} dS = \iiint_V (\frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}) dV.

Conditions

  • SS is a piecewise smooth closed surface bounding volume VV
  • n⃗\vec{n} is the outward unit normal vector on SS
  • F⃗\vec{F} is continuously differentiable in VV

Reasoning, step by step

  1. Identify the closed surface SS and the enclosed volume VV.
  2. Compute the dot product of the vector field F⃗\vec{F} and the outward normal n⃗\vec{n} on the surface.
  3. Integrate this product over the surface area dSdS to find total outward flux.
  4. Alternatively, compute the divergence ∇⋅F⃗\nabla \cdot \vec{F} at each point inside the volume.
  5. Integrate the divergence over the volume element dVdV.
  6. Equate the two results as per the theorem.

Example

The video displays the explicit expression: ∬SF⃗⋅n⃗dS=∭V(∂Fx∂x+∂Fy∂y+∂Fz∂z)dV\iint_S \vec{F} \cdot \vec{n} dS = \iiint_V (\frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}) dV.

Common misconceptions

  • Believing the theorem applies to open surfaces without closing boundaries.
  • Confusing the direction of the normal vector; it must be outward for the standard formulation.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.