Question

Use spherical coordinates to find the volume of the solid. The solid between the spheres \(x^{2}+y^{2}+z^{2}=a^{2}\) and \(x^{2}+y^{2}+z^{2}=b^{2}, b>a,\) and inside the cone \(z^{2}=x^{2}+y^{2}\)

Short answer

The volume of the solid that is bounded by the two spheres and inside the cone is given by \[ V = (2π/√2) [[b^{3}-a^{3}]/3 - (b^{3}-a^{3})/3 √2]] \]

Step-by-step solution

Conversion to spherical coordinates

The equation of the sphere can be convert to spherical coordinates as \(r^{2}=a^{2} => r=a\) and \(r^{2}=b^{2} => r=b\). And the equation of the cone in spherical coordinates is \(ρ=cosφ\). In spherical coordinates the three variables we will integrate over are ρ (the radial distance from the origin), φ (the angle with the positive z-axis), and θ (the angle counterclockwise from the positive x-axis).

Set up the triple integral

The volume element in spherical coordinates is \(ρ^{2} sinφ dρdφdθ\). The limits for ρ are from a to b (\(a<= ρ <=b\)), the limits for φ are from 0 to \(π/4\) (\(0<= φ <=π/4\)), and the limits for θ are from 0 to \(2π\) (\(0<= θ <=2π\)). So the volume can be expressed as: \[ ∫_{0}^{2π} ∫_{0}^{π/4} ∫_{a}^{b} ρ^{2} sinφ dρdφdθ\]

Evaluate the integral

This is a straightforward triple integral. After integration, plugging in the limits and simplifying gives us the volume. \[ V = 2π ∫_{0}^{π/4} sinφ dφ ∫_{a}^{b} ρ^{2} dρ \] \[ = 2π [ -cosφ ]_{0}^{π/4} [1/3 ρ^{3}]_{a}^{b} \] \[ = 2π [(1/√2)-(1)] [(1/3) b^{3} - (1/3) a^{3}] \] \[ = 2π/√2 [[b^{3}-a^{3}]/3 - (b^{3}-a^{3})/3 √2]]