See image ### Spherical Coordinates and Volume Calculation**Problem Statement:**Use spherical coordinates.Find the volume of the region bounded by the sphere \( \rho = 2 \cos \phi \) and the hemisphere \(\rho = 1; z \geq 0\).---**Diagram Explanation:**The diagram illustrates a three-dimensional volume in spherical coordinates. It shows a sphere defined by the equation \( \rho = 2 \cos \phi \) intersected with a hemisphere of radius 1. The \( z \)-axis is vertical, while the \( x \)- and \( y \)-axes lie in the horizontal plane, forming part of the boundary conditions for the defined volume.- **Axes:** - \( z \)-axis: Vertical direction. - \( x \)- and \( y \)-axes: Orthogonal horizontal directions. The shaded region represents the volume to be determined using spherical coordinate integration. It visually highlights where the two spherical surfaces intersect above the plane \(z = 0\).

Name: See image ### Spherical Coordinates and Volume Calculation**Problem St
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: See image ### Spherical Coordinates and Volume Calculation**Problem Statement:**Use spherical coordinates.Find the volume of the region bounded by the sphere \( \rho = 2 \cos \phi \) and the hemisphere \(\rho = 1; z \geq 0\).---**Diagram Explanation:

The objective here is to find the volume bounded by the sphere ρ=2cosϕ and the hemisphere ρ=1,z≥0 The required volume can be find using the spherical coordinates. First note that the intersection of given sphere and hemisphere is given by ρ=2cosϕ=1⇒cosϕ=12⇒ϕ=π3 According to spherical coordinates: ρ2=x2+y2+z2x=ρsinϕcosθ,y=ρsinϕsinθ,z=ρcosϕ, anddxdydz=ρ2sinϕdρdϕdθNow, the volume bounded by the given sphere and hemisphere can be calculated by finding the volume of the given sphere that is not inside the given hemisphere, as: V=∫02π∫0π3∫01ρ2sinϕdρdϕdθ+∫02π∫π3π2∫02cosϕρ2sinϕdρdϕdθ=∫02π∫0π3sinϕρ3301dϕdθ+∫02π∫π3π2ρ3302cosϕdϕdθ=∫02π∫0π313sinϕdϕdθ+∫02π∫π3π2sinϕ2cosϕ33dϕdθ=13∫02π−cosϕ0π3dθ+∫02πdθ∫π3π2sinϕ2cosϕ33dϕ=13∫02π−cosπ3+cos0dθ+θ02π∫π3π2sinϕ2cosϕ33dϕ=13θ02π−12+1+2π∫π3π2sinϕ2cosϕ33dϕ=2π3×12+2π∫π3π2sinϕ2cosϕ33dϕ=π3+16π∫π3π2sinϕ2cosϕ33dϕThe second integral in above part can be calculated using substitution, as cosϕ=t−sinϕdϕ=dtThus,∫sinϕcosϕ33dϕ=∫−t33dt=13−t44=−t412Thus,∫π3π2sinϕcosϕ33=−cosϕ412π3π2=−cosπ2412−cosπ3412=−1120−124=1192Thus, the required volume is: V=π3+16π1192=π3+π12=5π12