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

2023
Views:

247

Pages:

3
Author:

Shahbaz A.

Ex 15.7 Page 884 Q29 THOMAS’ CALCULUS twelfth Edition 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 = 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 h i π4 Rπ sin2 ϕ cos ϕ 3 2 4 − 3 cos ϕ + C 0 sin ϕ dϕ = − 3 0 h i 2 π π Rπ sin cos 3 2 π sin2 0 cos 0 2 4 4 4 − 3 cos 4 + C − [− 3 − 3 cos 0 + C] 0 sin ϕ dϕ = − 3 h i Rπ sin2 π4 cos π4 3 2 π 2 4 − 3 cos 4 + C − [− 3 + C] 0 sin ϕ dϕ = − 3 h i Rπ sin2 π4 cos π4 3 2 π 2 4 − 3 cos 4 + 3 0 sin ϕ dϕ = − 3 1 Rπ 4 0 Rπ 4 0 h √ sin ϕ dϕ = − 122 − 32 × 3 3 sin ϕ dϕ = √1 2 + 2 3 i √ 8−5 2 12 Also in spherical coordinates position of a point is described by ρ, r, θ as in figure below 1 Ex 15.7 Page 884 Q29 THOMAS’ CALCULUS twelfth Edition Q29 Z 1Z 0 0 π π 4 Z 12ρ sin3 ϕ dϕ dθ dρ 0 2 solution Z 1Z π π 4 Z Z 1Z 3 π π 4 Z 12ρ sin ϕ dϕ dθ dρ = 12 0 0 sin ϕ dϕ ρ dθ dρ 0 0 Using the formula R n−1 ax cos ax + sinn ax dx = − sin na √ Rπ 8−5 2 3 4 sin ϕ dϕ = 0 12 π Z 1Z 0 0 Z 1Z π 4 Z 0 π R 0 0 sinn−2 ax dx √ Z 1Z π 8 − 5 2 ρ dθ dρ 12ρ sin3 ϕ dϕ dθ dρ = 12 × 12 0 0 π 4 Z n−1 n 3 √ 3 Z 1 Z π 12ρ sin ϕ dϕ dθ dρ = (8 − 5 2) 0 0 0 Z 1Z π 0 0 π 4 0 Z 1Z 0 Z 1 √ 12ρ sin3 ϕ dϕ dθ dρ = (8 − 5 2)π ρ dρ 0 π Z 0 π 4 √ 1 12ρ sin3 ϕ dϕ dθ dρ = (8 − 5 2) π [ρ2 ]10 2 0 0 Z 1Z 0 π 4 πZ 0 0 ρ dρ √ Z 1 π [θ]0 ρ dρ 12ρ sin ϕ dϕ dθ dρ = (8 − 5 2) 3 0 Z 1Z 0 dθ 0 0 Z π Z 0 π 4 √ 1 12ρ sin3 ϕ dϕ dθ dρ = (8 − 5 2) π 2 3