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Surface integral over a surface F=xi+yj+zk

Short answer

F·n = x²+y²+z² = 1 on the unit sphere, so the surface integral reduces to the area of the upper hemisphere, which is 2π.

Find the surface integral ∬_S F·n ds where F = x i + y j + z k and S is the upper half of the sphere x² + y² + z² = 1.

S is the upper half of the unit sphere x² + y² + z² = 1, z ≥ 0. The outward unit normal at a point of the unit sphere is the position vector itself: n = x i + y j + z k.

F = x i + y j + z k
F·n = x² + y² + z² = 1
∬_S F·n dS = ∬_S dS = area of the hemisphere = ½ (4π(1)²) = 2π

Check with the projection method: dS = dx dy/|n·k| = dx dy/z, F·n = 1, so the integral is ∬ dx dy/√(1 − x² − y²) over the unit disc = ∫₀^2π ∫₀¹ r dr/√(1 − r²) dθ = 2π [−√(1 − r²)]₀¹ = 2π.

Answer: ∬_S F·n dS = 2π (without the base disc).

✓ Answer checked by Claude Opus 5.5 (high effort)

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