Question

Set up a triple integral for the volume of the solid. The solid that is the common interior below the sphere \(x^{2}+y^{2}+z^{2}=80\) and above the paraboloid \(z=\frac{1}{2}\left(x^{2}+y^{2}\right)\)

Short answer

The triple integral that represents the volume of the solid within the sphere and above the paraboloid is \(\int_0^{2π}\int_0^{arccos\sqrt{1 - ρ^{2}}}\int_0^{\sqrt{80}} ρ^{2}sin(φ)dρdφdθ\).

Step-by-step solution

Convert the given equations into spherical coordinates

In spherical coordinates, the sphere \(x^{2}+y^{2}+z^{2}=80\) becomes \(ρ^{2}=80\) or \(ρ= \sqrt{80}\). The paraboloid \(z=\frac{1}{2}\left(x^{2}+y^{2}\right)\) becomes \(\cos(φ)=ρsin(φ)\) or \(φ=\arccos\sqrt{1 - ρ^{2}}\). The angle \(θ\) will range over all values, which is [0, 2π]. The volume element in spherical coordinates is \(dV = ρ^{2}sin(φ)dρdφdθ\).

Set up the integral

Now we set up the triple integral. The limits of the integral for \(φ\) are determined by the sphere and the paraboloid, which are: \([0, \arccos\sqrt{1 - ρ^{2}}]\). The limits for \(θ\) are [0, 2π] as \(θ\) covers all the angles in a circle around the positive z-axis. The limits for \(ρ\) are [0, \sqrt{80}] as \(ρ\) ranges from the origin to surface of the sphere. We integrate the volume element \(\int^{\sqrt{80}}_\sigma \int^{2π}_0 \int^{arccos\sqrt{1 - ρ^{2}}} _0 ρ^{2}sin(φ)dφdθdρ\).

Simplify the integral

Simplify the triple integral: in this problem we don't actually carry out the integration because the task only asks for setting up the integral for the volume of the solid.