Evaluation of Triple integral by changing to Spherical Polar co-ordinates
1. Why Spherical
When is a ball, hemisphere, spherical shell, cone, or integrand involves , spherical polar simplifies. Definitions (mathematical vs physics swapped — use the convention your notes use; here = azimuth in , = colatitude from ):
with , , and Jacobian (volume element):
If your text swaps , the stays with the angle from ; the factor remains.
2. Limits
- Full ball : .
- Ball octant : .
- Hemisphere : .
- Between spheres : .
- Cone ⇒ (since ).
3. Worked Examples
(a) Volume of sphere radius a: V= ∫_{0}^{2π}∫_{0}^{π}∫_{0}^{a} r² sinφ dr dφ dθ = 4π a³/3 ✓
(b) ∭_{x²+y²+z²≤a²} (x²+y²+z²) dV = ∫_{0}^{2π}∫_{0}^{π}∫_{0}^{a} r²· r² sinφ dr dφ dθ = 4π·a⁵/5? Actually ∫ r⁴ dr = a⁵/5, angular integral 4π ⇒ 4π a⁵/5 ✓
(c) ∭_{hemisphere z≥0} z dV : = ∫_{0}^{2π}∫_{0}^{π/2}∫_{0}^{a} (r cosφ)·r² sinφ dr dφ dθ = 2π·(a⁴/4)·(1/2)·? Compute: ∫ cosφ sinφ dφ 0→π/2 =1/2, ∫ r³ dr =a⁴/4 ⇒ = π a⁴/4 ✓
(d) Mass of ball with density ρ = k r (proportional to distance from origin): Mass= ∭ k r dV = ∫∫∫ k r·r² sinφ =4π k a⁴/4? Actually ∫ k r³ dr =k a⁴/4 ×4π ⇒ π k a⁴ ✓
(e) I = ∫_{0}^{a}∫_{0}^{√(a²-x²)}∫_{0}^{√(a²-x²-y²)} (x²+y²+z²)^{3/2} dz dy dx (octant): convert ⇒ ∫_{0}^{π/2}∫_{0}^{π/2}∫_{0}^{a} r³·r² sinφ dr dφ dθ = (π/2)(1)·a⁶/6? Actually r³·r²= r⁵? Wait (r²)^{3/2}=r³, times Jacobian r² sinφ ⇒ r⁵ sinφ ⇒ ∫ r⁵= a⁶/6 × (π/2) ✓
4. Tips
- Do not forget — most errors drop it.
- Keep order consistent with you write.
- For , each .
5. Exam Points
- Convert over ball/cone to — 7 marks, very frequent.
- Evaluate , over sphere/hemisphere — 7 marks, frequent.
- Volume via spherical — 5 marks, frequent.