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Evaluate Stokes' theorem for G field

Short answer

∮G·dl = ∫(∇×G)·dS. For G=10 sinθ a_φ on r=3, 0≤θ≤90°, 0≤φ≤90°: left side integrates over equator arc only, giving 30×(π/2)=15π. Right side: (∇×G)_r = 20 cosθ/r, flux = 15π. Both sides equal 15π ≈ 47.1.

State Stokes' theorem. Evaluate both sides of Stokes' theorem for the field G = 10 sinθ a_φ and the surface r = 3, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 90°. Let the surface have the a_r direction.

Stokes' theorem: The circulation of a vector field around a closed path equals the flux of its curl through any surface bounded by that path: ∮ G · dl = ∫ (∇ × G) · dS.

Assumption: A polar angle θ cannot exceed π, so the range is taken as 0 ≤ θ ≤ 90° (the usual form of this problem). The surface is the spherical patch r = 3, 0 ≤ θ ≤ 90°, 0 ≤ φ ≤ 90° (one octant of a sphere), with dS along a_r.

          z
          │   patch: r = 3
          │ ╭──╮
          │╱    ╲  boundary = equator arc (θ = 90°, φ: 0→90°)
          ●──────── y         + two meridian arcs (φ = 0 and φ = 90°)
         ╱
        x

Left side: closed line integral

G = 10 sinθ a_φ has only an a_φ component.

  • On the two meridian arcs (φ = 0, φ = 90°), dl is along a_θ, so G · dl = 0.
  • On the equator arc (r = 3, θ = 90°), G_φ = 10 sin90° = 10 and dl = r sinθ dφ a_φ = 3 dφ a_φ.
∮ G · dl = ∫₀^(π/2) 10 (3) dφ = 30 (π/2) = 15π = 47.12

Right side: surface integral of curl

(∇ × G)_r = (1/(r sinθ)) ∂(sinθ G_φ)/∂θ = (1/(r sinθ)) (10 · 2 sinθ cosθ) = 20 cosθ / r
(∇ × G)_θ = −(1/r) ∂(r G_φ)/∂r = −10 sinθ / r

With dS = r² sinθ dθ dφ a_r only the a_r component contributes:

∫ (∇ × G) · dS = ∫₀^(π/2) ∫₀^(π/2) (20 cosθ / 3) (9 sinθ) dθ dφ
               = 60 ∫₀^(π/2) sinθ cosθ dθ · ∫₀^(π/2) dφ
               = 60 (1/2)(π/2) = 15π = 47.12

Answer: Both sides equal 15π ≈ 47.1 (in the units of G). Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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