Triple integrals in Cartesian, cylindrical and spherical coordinates

Try

A solid box has 0≤x≤1, 0≤y≤2 and 0≤z≤3. Its volume is 6. For density ρ=z+1, integrate first in z: ∫₀³(z+1)dz=15/2. Multiplying by the xy-area 2 gives mass 15. A triple integral accumulates density over small volume pieces; the extra variable does not change the logic of accumulation, but it makes an accurate description of the solid more important.

Predict

Predict suitable coordinates for three shapes: a rectangular box, a circular cylinder and a ball. Cartesian coordinates match flat coordinate-aligned faces. Cylindrical coordinates keep z and replace x,y by r,θ. Spherical coordinates use distance ρ from the origin, azimuth θ around the z-axis and polar angle φ measured down from the positive z-axis. State this convention explicitly: some books interchange the angular symbols, and memorizing a formula without its convention can reverse a region.

Experiment

For the cylinder x²+y²≤4, 1≤z≤3, cylindrical bounds are 0≤r≤2, 0≤θ≤2π and 1≤z≤3. The volume element is r dr dθ dz, giving volume 8π. For a ball of radius 2, use 0≤ρ≤2, 0≤φ≤π and 0≤θ≤2π. Its volume is ∫ρ²dρ ∫sin φdφ ∫dθ=(8/3)(2)(2π)=32π/3.

Compare these results with elementary geometry. Then change to the upper half-ball by restricting φ to [0,π/2]; the volume halves because the polar-angle integral becomes 1 rather than 2.

Understand

Cylindrical coordinates are x=r cos θ, y=r sin θ, z=z, with volume factor r. Spherical coordinates are x=ρsin φcos θ, y=ρsin φsin θ and z=ρcos φ, with volume factor ρ²sin φ. The three local lengths scale as dρ, ρdφ and ρsin φdθ. Their product explains the factor geometrically.

For a cone z=√(x²+y²), the upper sheet corresponds to φ=π/4. Points above it satisfy 0≤φ≤π/4, not an inequality in θ. For an upper hemisphere the angular condition is φ≤π/2. These comparisons let the bounds encode shape rather than rely on a remembered template.

Apply

Set up the volume below z=4−x²−y² and above z=0. In cylindrical coordinates the roof is 4−r², which is nonnegative for 0≤r≤2. Thus volume is ∫₀²π∫₀²∫₀^(4−r²)r dz dr dθ=2π∫₀²(4r−r³)dr=8π. The radial endpoint comes from intersecting the roof and floor, not from guessing the plot’s width.

Repair “replace x²+y²+z² by r² in cylindrical coordinates.” Cylindrical r is horizontal radius, so that expression becomes r²+z². Only spherical ρ represents full distance from the origin. Keep the two radius symbols distinct in calculations.

For a nonuniform solid, the center of mass uses three first moments divided by total mass. For example, symmetry about both coordinate planes x=0 and y=0 forces x̄=ȳ=0 when density has the same symmetries; it does not determine z̄ without a further calculation.

Continue

Optional depth examines quadrics through coordinate slices. The surface x²+y²−z²=1 has circular horizontal slices of radius √(1+z²), while x²+y²+4z²=1 bounds an ellipsoid. A sketch of these slices helps determine whether a proposed solid is bounded and which coordinates are useful.

Save one mass setup where density varies with height and one volume setup with a curved boundary. Include units and explicitly distinguish the transformed density from the volume factor. The next lesson generalizes these familiar coordinate systems to arbitrary regular changes of variables and makes their determinant-based scaling precise.