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

2023
Views:

119

Pages:

3
Author:

Shahbaz A.

Ex 15.7 Page 884 Q24 THOMAS’ CALCULUS twelfth Edition shahbaz ahmed July 2024 Spherical Coordinates In spherical coordinates position of a point is described by ρ, r, θ as in figure below 1 Relevant Formulas n−1 sinn ax dx = − sin R ax cos ax na + n−1 n R sinn−2 ax dx Putting n = 3, a = 1, x = ϕ R R 2 sin3 ϕ dϕ = − sin ϕ3cos ϕ + 23 sin ϕ dϕ R 2 sin3 ϕ dϕ = − sin ϕ3cos ϕ − 32 cos ϕ + C Applying the limit iπ h Rπ 3 sin2 ϕ cos ϕ 2 − 3 cos ϕ + C 0 sin ϕ dϕ = − 3 0 h i Rπ 3 sin2 π cos π 2 sin2 0 cos 0 2 − 3 cos π + C − [− 3 − 3 cos 0 + C] 0 sin ϕ dϕ = − 3 2 Rπ 3 2 sin ϕ dϕ = − (−1) + C − [− (1) + C] 0 3 3 4 Rπ 3 sin ϕ dϕ = 0 3 Q24 Evaluate the spherical coordinate integral Z 3π 2 Z π Z 0 3π 2 5ρ3 sin3 ϕ dρ dϕ dθ = 5 0 0 0 1 Z Z 0 π Z 3π 2 Z 0 1 5 5ρ sin ϕ dρ dϕ dθ = 4 3 3 2 Z 0 3π 2 1 Z 0 0 Z π ρ3 dρ sin3 ϕ dϕ dθ 0 Z 0 π 4 1 3 ρ 0 sin ϕ dϕ dθ Z 0 3π 2 π Z 0 1 Z 5 5ρ3 sin3 ϕ dρ dϕ dθ = 4 0 Z 3π 2 π Z 0 sin3 ϕ dϕ 0 Putting 4 Rπ 3 sin ϕ dϕ = 0 3 Z 3π 2 π Z 0 0 π Z 3π 2 Z π Z Z 0 3π 2 π 0 3π 2 dθ 0 3π 5 [θ]02 3 5ρ3 sin3 ϕ dρ dϕ dθ = 5 3π × 3 2 1 0 Z Z 5ρ3 sin3 ϕ dρ dϕ dθ = 0 0 0 3 1 Z 0 0 Z 5 4 5ρ sin ϕ dρ dϕ dθ = × 4 3 3 0 3π 2 Z 1 Z Z 1 5ρ3 sin3 ϕ dρ dϕ dθ = 0 3 5π 2 dθ