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Evaluate using Stokes' theorem F=(2x−y)i−yz²j−y²zk

Short answer

By Stokes’ theorem, the surface integral equals the line integral around the boundary circle x²+y²=a², z=0. Parametrizing the circle gives πa².

State Stokes' theorem. Apply it to evaluate ∬_S (∇×F)·n ds where F = (2x − y) i − yz² j − y²z k, S is the upper half surface of the sphere x² + y² + z² = a² and C is its boundary.

Statement

If S is an open surface bounded by a simple closed curve C, and F has continuous first partial derivatives, then

∮_C F·dr = ∬_S (∇ × F)·n dS,

where C is traversed anticlockwise as seen from the tip of n.

Application

By Stokes' theorem, ∬_S (∇×F)·n dS equals the line integral round the boundary C: x² + y² = a², z = 0.

F = (2x − y) i − yz² j − y²z k. On C, z = 0, so F·dr = (2x − y) dx. With x = a cos t, y = a sin t:

∮_C (2x − y) dx = ∫₀^2π (2a cos t − a sin t)(−a sin t) dt
                = ∫₀^2π (−2a² sin t cos t + a² sin²t) dt
                = 0 + a² π = π a²

Check using the curl: ∇×F = (−2yz + 2yz) i + 0 j + (0 + 1) k = k. On the hemisphere, (∇×F)·n dS = dx dy, so the integral equals the area of the disc x² + y² ≤ a², that is πa².

Answer: ∬_S (∇×F)·n dS = πa².

✓ Answer checked by Claude Opus 5.5 (high effort)

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