Question

Use a spherical coordinate integral to find the volume of the the solid between the sphere rho= cosϕ and the hemisphere given by rho=7,z≥0 Show all your work here Please select file(s)

Answer 1

Sure! Let’s find the volume of the solid between the sphere ρ = cos(ϕ) and the hemisphere given by ρ = 7, z ≥ 0 using a spherical coordinate integral.

First, let’s convert the equation of the hemisphere z ≥ 0 into spherical coordinates. Since z = ρcos(ϕ), we have ρcos(ϕ) ≥ 0. Since ρ is always non-negative, this inequality is equivalent to cos(ϕ) ≥ 0, or 0 ≤ ϕ ≤ π/2.

Now, let’s set up the triple integral to find the volume of the solid. The limits of integration for ρ are from cos(ϕ) to 7, the limits of integration for ϕ are from 0 to π/2, and the limits of integration for θ are from 0 to 2π. The volume element in spherical coordinates is given by dV = ρ^2sin(ϕ)dρdϕdθ. Therefore, the volume of the solid is given by:

V=∫02π​∫0π/2​∫cos(ϕ)7​ρ2sin(ϕ)dρdϕdθ

Evaluating this triple integral, we get:

V=∫02π​∫0π/2​[3ρ3​]cos(ϕ)7​sin(ϕ)dϕdθ

V=∫02π​∫0π/2​(3343​−3cos3(ϕ)​)sin(ϕ)dϕdθ

V=∫02π​(3343​∫0π/2​sin(ϕ)dϕ−31​∫0π/2​cos3(ϕ)sin(ϕ)dϕ)dθ

V=∫02π​(3343​−121​)dθ

V=(121029​−121​)[θ]02π​

V=121028​(2π)

V=31714​π≈1798.194

Therefore, the volume of the solid between the sphere ρ = cos(ϕ) and the hemisphere given by ρ = 7, z ≥ 0 is approximately 1798.194 cubic units.