The solid below (x^2)+(y^2)+(z^2)=4, above z=sqrt((x^2)+(y^2)), between y=x, and x=0. with y greater than or equal to 0.
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Use spherical coordinates.
x^2 + y^2 + z^2 = 4 ==> ρ = 2
z = √(x^2 + y^2) ==> ρ cos φ = ρ sin φ ==> φ = π/4.
Projecting the region onto the xy-plane yields a portion of the circle x^2 + y^2 = 3 (from intersecting the surfaces given) between x = 0 and y = x ==> θ is in [π/4, π/2].
So, the volume ∫∫∫ 1 dV equals
∫(θ = π/4 to π/2) ∫(φ = 0 to π/4) ∫(ρ = 0 to 2) 1 * (ρ^2 sin φ dρ dφ dθ)
= ∫(θ = π/4 to π/2) dθ * ∫(φ = 0 to π/4) sin φ dφ * ∫(ρ = 0 to 2) ρ^2 dρ
= (π/4) * (1 - √2/2) * (8/3)
= (π/3)(2 - √2).
I hope this helps!
x^2 + y^2 + z^2 = 4 ==> ρ = 2
z = √(x^2 + y^2) ==> ρ cos φ = ρ sin φ ==> φ = π/4.
Projecting the region onto the xy-plane yields a portion of the circle x^2 + y^2 = 3 (from intersecting the surfaces given) between x = 0 and y = x ==> θ is in [π/4, π/2].
So, the volume ∫∫∫ 1 dV equals
∫(θ = π/4 to π/2) ∫(φ = 0 to π/4) ∫(ρ = 0 to 2) 1 * (ρ^2 sin φ dρ dφ dθ)
= ∫(θ = π/4 to π/2) dθ * ∫(φ = 0 to π/4) sin φ dφ * ∫(ρ = 0 to 2) ρ^2 dρ
= (π/4) * (1 - √2/2) * (8/3)
= (π/3)(2 - √2).
I hope this helps!