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Spatial symmetry

Spherical coordinates

Material by the IntegralPaso project. Conditions and limits are stated in each guide; external teaching review is pending.

dV=r2sin⁡ϕ dr dϕ dθdV=r^2\sin\phi\,dr\,d\phi\,d\theta

Here φ is the angle from the z axis, between 0 and π; θ is the azimuthal angle in the xy plane. We use x = r sinφ cosθ, y = r sinφ sinθ and z = r cosφ. The Jacobian is r² sinφ. For a full sphere, r runs from 0 to the radius, φ from 0 to π and θ from 0 to 2π.

Common mistake

Confusing the φ and θ convention or using r instead of r² sinφ for the Jacobian.

DEVELOPED GUIDE · 9 min

Choose spherical coordinates for balls and sectors and distinguish polar angle from azimuth.

When to choose this method

  • The region depends on x²+y²+z² or distance from the origin.
  • Fix the convention: here φ is measured from the positive z axis and θ around that axis.

Before calculating

  1. Describe radius, polar angle and azimuth before integrating.
  2. Transform the function and multiply by ρ² sinφ. In the editor the radius is named r.

A complete example, step by step

∭x2+y2+z2≤4(x2+y2+z2) dV\iiint_{x^2+y^2+z^2\le4}(x^2+y^2+z^2)\,dV
  1. The region is a full ball of radius 2. The polar angle spans π, not 2π.

    0≤ρ≤2,0≤ϕ≤π,0≤θ≤2π0\le\rho\le2,\quad0\le\phi\le\pi,\quad0\le\theta\le2\pi
  2. The sum of squares becomes ρ².

    x=ρsin⁡ϕcos⁡θ,y=ρsin⁡ϕsin⁡θ,z=ρcos⁡ϕx=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi
  3. The function ρ² and Jacobian ρ² sinφ produce ρ⁴ sinφ.

    I=∫02π∫0π∫02ρ4sin⁡ϕ dρ dϕ dθI=\int_0^{2\pi}\int_0^\pi\int_0^2\rho^4\sin\phi\,d\rho\,d\phi\,d\theta
  4. Integrate the radial power before touching the angles.

    ∫02ρ4 dρ=325\int_0^2\rho^4\,d\rho=\frac{32}{5}
  5. The polar integral equals 2; the primitive minus sign matters.

    ∫0πsin⁡ϕ dϕ=[−cos⁡ϕ]0π=2\int_0^\pi\sin\phi\,d\phi=[-\cos\phi]_0^\pi=2
  6. Multiply by the azimuthal integral 2π.

    I=325⋅2⋅2π=128π5I=\frac{32}{5}\cdot2\cdot2\pi=\frac{128\pi}{5}

Check the result and domain

For z≥0 replace φ≤π by φ≤π/2. For a full ball the polar angle runs from 0 to π under this convention.

dV=ρ2sin⁡ϕ dρ dϕ dθdV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta
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Now try it yourself

Solve on paper before opening the hint or answer. These are self-assessment activities; to check a typed answer, open Learn.

EXERCISE 1V(x2+y2+z2≤1)V(x^2+y^2+z^2\le1)
Show a hint

Use integrand 1 and Jacobian ρ² sinφ.

Check my result4π3\frac{4\pi}{3}
EXERCISE 2V(x2+y2+z2≤4, z≥0)V(x^2+y^2+z^2\le4,\ z\ge0)
Show a hint

This is a half-ball; φ runs to π/2.

Check my result16π3\frac{16\pi}{3}
EXERCISE 3∭x2+y2+z2≤1(x2+y2+z2) dV\iiint_{x^2+y^2+z^2\le1}(x^2+y^2+z^2)\,dV
Show a hint

The radial power is ρ⁴.

Check my result4π5\frac{4\pi}{5}
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