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Chapter 1: Problem 22

Evaluate (if possible) the function at each specified value of the independent variable and simplify. \(V(r)=\frac{4}{3} \pi r^{3}\) (a) \(V(3)\) (b) \(V\left(\frac{3}{2}\right)\) (c) \(V(2 r)\)

Short Answer

Expert verified
For the given values of \(r\), the volumes are \(V(3)=36\pi\), \(V\left(\frac{3}{2}\right)=\frac{9\pi}{2}\), and \(V(2r)\) is eight times the volume of a sphere of radius \(r\).

Step by step solution

01

Substitute \(r = 3\) into Volume Function

Place \(r = 3\) into the equation: \(V(3)=\frac{4}{3}\pi(3)^3\)
02

Simplify \(V(3)\)

Evaluate the expression to obtain \(V(3)=36\pi\)
03

Substitute \(r = \frac{3}{2}\) into Volume Function

Place \(r = \frac{3}{2}\) into the equation: \(V\left(\frac{3}{2}\right)=\frac{4}{3}\pi\left(\frac{3}{2}\right)^3\)
04

Simplify \(V\left(\frac{3}{2}\right)\)

Evaluate the expression to get \(V\left(\frac{3}{2}\right)=\frac{9\pi}{2}\)
05

Substitute \(r = 2r\) into Volume Function

Place \(r = 2r\) into the equation: \(V(2r)=\frac{4}{3}\pi(2r)^3\)
06

Simplify \(V(2r)\) and Recognize Relationship

Evaluate the expression to get \(V(2r)=32r^3\pi\). This is eight times the volume of a sphere with radius \(r\) thus, \(V(2r)=8V(r)\)

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