   Chapter 14.7, Problem 20E

Chapter
Section
Textbook Problem

Volume In Exercises 15-20, use cylindrical coordinates to find the volume of the solid.Solid inside the sphere x 2 + y 2 + z 2 = 4 and above the upper nappe of the cone z 2 = x 2 + y 2

To determine

To calculate: Find out the Volume of Solid confined below the cone bysphere x2+y2+z2=4 and also find the volume above the upper nappe of the cone defined by the equation z2=x2+y2.

Explanation

Given:

Volume of Solid confined below the cone by sphere x2+y2+z2=4 and above its upper nappe

Upper nappe by the equation z2=x2+y2.

Formula used:

By triple integration of cylindrical coordinate’s volume of solid can be defined as

V=Qf(x,y,z)dV=Qf(rcosθ,rsinθ,z)rdzdrdθ

Rectangular conversion equations of cylindrical coordinates are,

x=rcosθy=rsinθz=z

Relation between x, y, and r is given as,

x2+y2=r2

As, sin2θ+cos2θ=1

Trigonometry formula,

sin3θ=3sin1θ4sin3θ

Calculation:

Deliberate the solid confined by the sphere by x2+y2+z2=4 and above the upper nappe of cone by z2=x2+y2.

Consider the available equation,

x2+y2+z2=4 …… (1)

And,

z2=x2+y2 …… (2)

Use equation (2) in equation (1) as,

x2+y2+z2=4z2+z2=42z2=4z=±2

Also,

x2+y2+z2=4r2+z2=4z2=4r2z=4r2

Use polar components in eq. (2)

x2+y2=z2r2=z2

Limits on z is,

rz4r2

Limits of r is,

0r2

Limits on θ is,

0θ2π

Therefore, volume of required solid is given as,

V=Qf(x,y,z)dV=02π02r4r2rdzdrdθ=02π02r[z]r4r2

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