Answer to Question #218905 in Calculus for Nkululeko

Question

Let D be the region in R^3 that lies inside the cone z=sqr(x^2+y^2) above the plane z=1 and below the hemisphere z=sqr(4-x^2-y^2). (c) Express the volume of D as a triple integral using spherical coordinates. Do not evaluate the integral.

Answer 1

Solution

In spherical coordinates ρ, ϕ, θ where 0≤ ϕ ≤ π and 0≤ θ ≤ 2π given surfaces may by described by equations:

·        cone z=sqr(x^2+y^2):  ϕ = π/4

·        sphere z=sqr(4-x^2-y^2): ρ = 2

·        plane z=1: from equality z = ρ cos ϕ =>  ρ = z/cos ϕ and for z = 1  ρ = 1/cos ϕ

So

$V\ =\ \iiint_{D}{dV\ =\ }\ \iiint_{D}{dV\ =\ }\ \iiint_{D}{\rho^2sin\varphi d\rho d\varphi d\theta}=\int_{0}^{2\pi}{\int_{0}^{\pi/4}{\int_{1/cos \varphi}^{2}d\rho d\varphi}d\theta}$