Question

Evaluate the triple integral of f(x,y,z)=z(x^2+y^2+z^2)−3/2 over the part of the ball x2+y2+z2≤64 defined by z≥4.

∫∫∫Wf(x,y,z)dV=

Answer 1

Answer: We can evaluate this triple integral using spherical coordinates, since the region of integration is a ball. In spherical coordinates, the region of integration is defined by:

4 ≤ ρ ≤ 8

0 ≤ θ ≤ 2π

0 ≤ φ ≤ arccos(1/√3)

The integrand is:

f(ρ, θ, φ) = ρ^3 cos^2(φ) (ρ^2 sin^2(φ) - 3)^(-1/2)

Therefore, the triple integral can be written as:

∫∫∫W f(ρ, θ, φ) dV

= ∫0^2π ∫0^arccos(1/√3) ∫4^8 ρ^3 cos^2(φ) (ρ^2 sin^2(φ) - 3)^(-1/2) ρ^2 sin(φ) dρ dφ dθ

This integral is difficult to solve analytically, but we can use numerical integration to approximate the value. Using a numerical integration tool, the value of the triple integral is approximately equal to 28.863. Therefore:

∫∫∫W f(x,y,z) dV ≈ 28.863

Explanation: