Answer: To evaluate the given triple integral, we'll use spherical coordinates, since the region is bounded by a sphere. In spherical coordinates, the volume element is given by:
dV = ρ^2 sin(φ) dρ dφ dθ
where ρ is the radial distance, φ is the polar angle (measured from the positive z-axis), and θ is the azimuthal angle (measured from the positive x-axis in the xy-plane).
The limits of integration for ρ, φ, and θ are as follows:
0 ≤ ρ ≤ √3 (since the sphere has a radius of √3)
0 ≤ φ ≤ π/2 (since the region is in the positive octant)
0 ≤ θ ≤ π/2 (since the region is in the positive octant)
Now, let's evaluate the integral:
∭
√(x^2 + y^2 + z^2) dV
= ∫₀^(π/2) ∫₀^(π/2) ∫₀^√3 (ρ^3 sin(φ) dρ dφ dθ)
= ∫₀^(π/2) ∫₀^(π/2) [ρ^4/4]₀^√3 sin(φ) dφ dθ
= ∫₀^(π/2) ∫₀^(π/2) (3/4)√3 sin(φ) dφ dθ
= ∫₀^(π/2) [-3/4 √3 cos(φ)]₀^(π/2) dθ
= ∫₀^(π/2) (3/4) √3 dθ
= (3/4) √3 [(π/2) - 0]
= (3/4) (π/2) √3
= (3/8) π √3
Therefore, the value of the given triple integral is (3/8) π √3.
Explanation: