Triple integral (x²+y²+z²) over a ball

This problem asks to evaluate a triple integral over a ball using the change to spherical coordinates.

Problem

Evaluate the triple integral:

\underset{V}{∭} (x^{2} + y^{2} + z^{2}) d x d y d z

V : x^{2} + y^{2} + z^{2} \leq 1

Solution

Step 1

The region V is the unit ball bounded by the sphere x² + y² + z² = 1. For such a region it is convenient to change to spherical coordinates:

Why choose spherical coordinates?

The integration region is a ball, and the integrand depends only on r² = x² + y² + z². Spherical coordinates turn the ball into the rectangle [0; 1] × [0; π] × [0; 2π] in (r, θ, φ), and the integrand simplifies to r². This significantly simplifies the computation.

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ

Step 2

Write the volume element in spherical coordinates and the limits of integration:

d x d y d z = r^{2} sin θ d r d θ d φ

0 \leq r \leq 1, 0 \leq θ \leq π, 0 \leq φ \leq 2 π

Step 3

The integrand in spherical coordinates takes a simple form:

x^{2} + y^{2} + z^{2} = r^{2}

Step 4

Pass to an iterated integral:

I = \int_{0}^{2 π} d φ \int_{0}^{π} sin θ d θ \int_{0}^{1} r^{2} \cdot r^{2} d r

Step 5

Evaluate the inner integral over r:

\int_{0}^{1} r^{4} d r = \frac{1}{5}

Step 6

Evaluate the middle integral over θ:

\int_{0}^{π} sin θ d θ = {[- cos θ]}_{0}^{π} = 2

Step 7

Evaluate the outer integral over φ:

\int_{0}^{2 π} d φ = 2 π

Step 8

Multiply the results:

I = 2 π \cdot 2 \cdot \frac{1}{5} = \frac{4 π}{5}

Answer

I = 4π/5.

Verification via the average value

The volume of the unit ball is 4π/3. The average value of the function x² + y² + z² over the ball can be found as 3 · (1/5) · R² = 3/5 for R = 1 (by symmetry, each coordinate contributes R²/5 on average). Then the integral equals V · ⟨f⟩ = (4π/3) · (3/5) = 4π/5. The result matches.

Frequently asked questions

How do I evaluate a triple integral over a ball?

For a ball, spherical coordinates are most convenient: x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. The Jacobian of the change is r² sin θ, so the volume element is dx dy dz = r² sin θ dr dθ dφ. The limits: 0 ≤ r ≤ R, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π.

What are spherical coordinates?

Spherical coordinates are a coordinate system in which the position of a point in space is given by three numbers: r — the distance from the origin, θ — the angle from the Oz axis, φ — the angle from the Ox axis in the Oxy plane. They are especially convenient for regions with spherical symmetry: a ball, a spherical layer, a spherical sector.

What is the difference between spherical and cylindrical coordinates?

In cylindrical coordinates (r, φ, z), the third coordinate is the height z (Cartesian). In spherical coordinates (r, θ, φ), r is the distance from the origin and θ is the angle from the Oz axis. Spherical coordinates are convenient for balls and spheres; cylindrical coordinates are convenient for cylinders and regions with axial symmetry.

Why is the Jacobian equal to r² sin θ?

The Jacobian of the change from Cartesian to spherical coordinates is the determinant of the Jacobian matrix. It equals r² sin θ. This factor reflects the stretching of space under the change. The factor r² comes from radial stretching; sin θ comes from the angular dependence.