Integrals in Spherical Coordinates

University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

14

Pages:

3
Author:

Shahbaz A.

Integrals in Spherical Coordinates shahbaz ahmed July 2024 Relevant Formulas R n−1 sinn ax dx = − sin ax cos ax na + n−1 n R sinn−2 ax dx Putting n = 4, a = 1, x = ϕ R R 3 sin4 ϕ dϕ = − sin ϕ4cos ϕ + 34 sin2 ϕ dϕ R R 3 sin4 ϕ dϕ = − sin ϕ4cos ϕ + 34 [− sin ϕ2cos ϕ + 21 dϕ] R 3 sin4 ϕ dϕ = − sin ϕ4cos ϕ − 83 sin ϕ cos ϕ + 38 ϕ + C Applying the limit Rπ 4 sin3 ϕ cos ϕ sin ϕ dϕ = [− − 83 sin ϕ cos ϕ + 38 ϕ + C]π0 0 4 Rπ 4 0 sin ϕ dϕ = h iπ sin3 π cos π 3 3 sin3 0 cos 0 3 3 − − 8 sin π cos π + 8 π + C − [− 4 − 8 sin 0 cos 0 + 8 (0) + C] 4 0 Rπ 4 3 3 3 3 0 sin ϕ dϕ = −0 − 8 (0) + 8 π + C − [−(0) − 8 (0) + 8 (0) + C] 3 Rπ 4 sin ϕ dϕ = π + C − C 0 8 Rπ 4 3 0 sin ϕ dϕ = 8 π 1 Also in spherical coordinates position of a point is described by ρ, r, θ as in figure below 1 Ex 15.7 Page 884 Q21 THOMAS’ CALCULUS twelfth Edition Q21 R π R π R 2 sin ϕ 0 0 0 R π R π R 2 sin ϕ ρ2 sin ϕ dρ dϕ dθ R π R π R 2 sin ϕ 2 ρ dρ] sin ϕ dϕ dθ 0 0 0 0 0 [ 0 R R R π R π R 2 sin ϕ 2 1 π π 3 2 sin ϕ sin ϕ dϕ dθ ρ sin ϕ dρ dϕ dθ = 0 0 0 3 0 0 [ρ ]0 R π R π R 2 sin ϕ 2 R R 3 8 π π ρ sin ϕ dρ dϕ dθ = 0 0 0 3 0 0 [sin ϕ sin ϕ dϕ] dθ R π R π R 2 sin ϕ 2 R R 4 8 π π ρ sin ϕ dρ dϕ dθ = 0 0 0 3 0 0 [sin ϕ dϕ] dθ ρ2 sin ϕ dρ dϕ dθ = 2 Using the formula R R n−1 ax cos ax sinn ax dx = − sin na + n−1 sinn−2 ax dx n Rπ 4 3 0 sin ϕ dϕ = 8 π R π R π R 2 sin ϕ 2 R 8 π 3 ρ sin ϕ dρ dϕ dθ = 0 0 0 3 0 8 π dθ Rπ R π R π R 2 sin ϕ 2 3 8 × π ρ sin ϕ dρ dϕ dθ = 0 dθ 0 0 0 3 8 R π R π R 2 sin ϕ 2 ρ sin ϕ dρ dϕ dθ = π[θ]π0 = π[π − 0] = π 2 0 0 0 3