Calculate Triple Integral of a Half Sphere- Step-by-Step
What You're Actually Calculating
A triple integral of a half sphere gives you the volume of that shape. That's it. Half a sphere, integrated three times, and you get a number back.
Most textbooks throw cylindrical coordinates at you. Some push spherical coordinates. The truth? Spherical coordinates are faster here. Cylindrical work, but they add steps you don't need.
Setting Up the Bounds
A half sphere sits above the xy-plane. The bottom flat face is at z = 0. The curved top follows the equation:
x² + y² + z² = R²
In spherical coordinates (ρ, φ, θ):
- ρ goes from 0 to R
- φ (angle from positive z-axis) goes from 0 to π/2 — this is what makes it a half sphere
- θ goes from 0 to 2π — full rotation around the z-axis
The Integral Setup
Your triple integral looks like this:
∫∫∫ f(x,y,z) dV
With spherical coordinates, dV = ρ² sin(φ) dρ dφ dθ
For volume (where f = 1), the integral becomes:
∫₀²π ∫₀π/2 ∫₀ᴿ ρ² sin(φ) dρ dφ dθ
Working Through It — Step by Step
Integrate with respect to ρ first:
∫₀ᴿ ρ² dρ = [ρ³/3]₀ᴿ = R³/3
Now integrate with respect to φ:
∫₀π/2 sin(φ) dφ = [-cos(φ)]₀π/2 = 1
Finally, integrate with respect to θ:
∫₀²π dθ = 2π
Multiply everything together:
V = (R³/3) × 1 × 2π = 2πR³/3
That matches the known volume of a half sphere. ✓
Spherical vs. Cylindrical — Quick Comparison
| Coordinate System | Integration Steps | Bounds Complexity | Best For |
|---|---|---|---|
| Spherical | 3 straightforward integrals | Simple ρ, φ, θ limits | Volume, density problems |
| Cylindrical | More nested integration | z depends on r | Problems with symmetry about z-axis AND specific z-limits |
Common Mistakes That Waste Time
- Wrong φ range — Using 0 to π instead of 0 to π/2 gives you a full sphere
- Forgetting sin(φ) in the Jacobian — this kills the problem
- Setting ρ incorrectly — ρ must go to R, not to a function of φ
- Using cylindrical when spherical is cleaner
How To Actually Do This on a Test
- Identify the region: half sphere means φ ∈ [0, π/2]
- Write the Jacobian: ρ² sin(φ)
- Set bounds: ρ from 0 to R, φ from 0 to π/2, θ from 0 to 2π
- Integrate innermost to outermost
- Check your answer against V = 2πR³/3
When f(x,y,z) Isn't 1
If you're calculating mass, moment of inertia, or some other property, the integrand changes but the bounds stay the same. For a density function ρ(ρ,φ,θ), you get:
∫₀²π ∫₀π/2 ∫₀ᴿ f(ρ,φ,θ) · ρ² sin(φ) dρ dφ dθ
The geometry doesn't change. Only what you're measuring changes.