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Cylindrical vector to spherical

Short answer

The vector A = ρ sinφ a_z at (1, 45°, 2) has only a z-component of 0.7071. Convert to spherical: A_r = A_z cosθ = 0.6325, A_θ = -A_z sinθ = -0.3162, A_φ = 0. So A = 0.632 a_r - 0.316 a_θ.

Transform vector A = ρ sinφ a_z at point (1, 45°, 2) in cylindrical co-ordinate system to a vector in spherical co-ordinate system.

Given: A = ρ sinφ a_z at the point (ρ = 1, φ = 45°, z = 2). Find the spherical components.

Step 1: Location in spherical coordinates

r = √(ρ² + z²) = √(1 + 4) = √5 = 2.236
θ = tan⁻¹(ρ/z) = tan⁻¹(1/2) = 26.57°
φ = 45° (same in both systems)
sinθ = ρ/r = 0.4472,  cosθ = z/r = 0.8944

Step 2: Value of A at the point

Only A_z exists: A_z = ρ sinφ = 1 × sin45° = 0.7071 (A_ρ = A_φ = 0).

Step 3: Cylindrical to spherical

A_r = A_ρ sinθ + A_z cosθ = 0 + 0.7071(0.8944) = 0.6325
A_θ = A_ρ cosθ − A_z sinθ = 0 − 0.7071(0.4472) = −0.3162
A_φ = A_φ = 0

Check: √(0.6325² + 0.3162²) = √(0.4 + 0.1) = 0.7071 ✓

Answer: A = 0.632 a_r − 0.316 a_θ at (1, 45°, 2) (A_φ = 0).

(General form: A = ρ sinφ cosθ a_r − ρ sinφ sinθ a_θ, with ρ = r sinθ.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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