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

2023
Views:

272

Pages:

3
Author:

Shahbaz A.

Ex 15.7 Page 884 Q28 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 Q28 Evaluate the spherical coordinate integral π 3 Z π 6 Z 2 cos ϕ csc ϕ Z 2π ρ2 sin ϕ dθ dρ dϕ 0 Solution π 3 Z 2 cos ϕ Z 2π Z ρ π 6 π 3 Z π 6 π 6 csc ϕ π 3 Z π 6 π 3 Z ρ2 sin ϕ dθ dρ dϕ = 2π π 3 Z 2π csc ϕ 2π 3 ρ2 sin ϕ dθ dρ dϕ = 0 2π [θ]0 ρ2 sin ϕ dρ dϕ csc ϕ "Z π 6 Z dθ ρ2 sin ϕ dρ dϕ 2 cos ϕ Z π 6 0 2 cos ϕ Z ρ2 sin ϕ dθ dρ dϕ = 2π Z 0 csc ϕ 0 2 cos ϕ Z 2π Z sin ϕ dθ dρ dϕ = 2π Z 2 cos ϕ Z π 6 csc ϕ π 3 Z 2 cos ϕ π 3 Z 0 csc ϕ Z 2 2 cos ϕ # ρ2 dρ sin ϕ dϕ csc ϕ π 3 Z π 6 3 2 cos ϕ ρ csc ϕ sin ϕ dϕ 3 2 cos ϕ ρ csc ϕ = [8 cos3 ϕ − csc3 ϕ] π 3 Z Z π 6 π 3 Z Z 2 cos ϕ Z csc ϕ 2 cos ϕ 2π Z π 3 Z π 6 csc ϕ Z csc ϕ π 3 Z π 6 2 0 2 cos ϕ Z ρ2 sin ϕ dθ dρ dϕ = 0 ρ π 6 2π 16π sin ϕ dθ dρ dϕ = 3 2π ρ2 sin ϕ dθ dρ dϕ = − 0 Z 2 cos ϕ csc ϕ Z 16π 3 2 π 3 Z ρ2 sin ϕ dθ dρ dϕ = − 0 2π cos ϕ sin ϕ dϕ − 3 3 π 6 2π [8 cos3 ϕ − csc3 ϕ]] sin ϕ dϕ π 6 π 3 Z π 3 Z 2π 3 π 6 cos3 ϕ d(cos ϕ) + π 3 Z 2π 3 csc2 ϕ dϕ π 6 π 3 Z d(cot ϕ) π 6 π π 16π 1 2π × [cos4 ϕ] π3 + [cot ϕ] π3 6 6 3 4 3 π 3 Z Z π 6 2 cos ϕ π 6 π 3 ρ2 sin ϕ dθ dρ dϕ = − 0 csc ϕ [cos4 ϕ] π3 = cos4 2π Z − cos4 π 6 = 3 3 6 π 3 Z Z 2 cos ϕ 2π Z ρ π 6 1 16 √ π [cot ϕ] π3 = cot π3 − cot π6 = 2 0 csc ϕ π 3 Z π 6 Z π π 4π 2π [cos4 ϕ] π3 + [cot ϕ] π3 6 6 3 3 9 8 − 16 = − 16 √ √ − 3 = −233 √ 4π 8 2π 2 3 sin ϕ dθ dρ dϕ = (− ) × (− ) + × (− ) 3 16 3 3 2 cos ϕ csc ϕ Z 2π ρ2 sin ϕ dθ dρ dϕ = 0 3 √ 6−4 3 π 9