Answer to Question #143570 in Calculus for Promise Omiponle

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 x^2+y^2+z^2≤16 defined by z≥2.

Answer 1

Converting the triple integrals to spherical coordinates, we have;

$f(ρ,ϕ,θ)=ρcos(ϕ)ρ^{−3/2}\\ f(ρ,ϕ,θ)=cos(ϕ)ρ^{−1/2}$

Then for the bounds:

$ρ^2≤16\\ ρ≤4$

and;

$ρcos(ϕ)≥2\\ \dfrac{2}{cos(ϕ)}≤ρ$

So:

$\dfrac{2}{cos(ϕ)}≤ρ≤4$

Then since the plane crosses the sphere at z=2 where ρ=4, and z = ρcos(ϕ):

$4cos(ϕ)=2\\ ϕ=π/3$

So:

$0≤ϕ≤π/3$

Since it is a sphere I have: $0≤θ≤2π$

So my bounds are:

$\frac{2}{cos(ϕ)}≤ρ≤4\\ 0≤θ≤2π$

Over:

$f(ρ,ϕ,θ)=cos(ϕ)ρ^{−1/2}$

With $dV=ρ^2sin(ϕ)dρdϕdθ$,

The integral can be written as:

$∫_0^{2π}∫_0^{π/3}∫_{\frac{2}{cos(ϕ)}}^4ρ^{−1/2}cos(ϕ)ρ^2sin(ϕ)dρdϕdθ=\\$

$∫_0^{2π}∫_0^{π/3}∫_{\frac{2}{cos(ϕ)}}^4ρ^{-3/2}cos(ϕ)sin(ϕ)dρdϕdθ=$

$∫_0^{2π}dθ∫_0^{π/3}cos(ϕ)sin(ϕ)dϕ∫_{\frac{2}{cos(ϕ)}}^4ρ^{−3/2}dρ=$

$2π× \frac{√3}{4}×∫_{\frac{2}{cos(ϕ)}}^4ρ^{−3/2}dρ=$

$\frac{π√3}{2} × \frac{2}{√3} = π$