Question

Approximate the change in the volume of a sphere when its radius changes from r= 5ft to r= 5.1ft (v(r)= 3/4 pi r^3)?

Answer 1

Answer:

18.03 cubic feet

Step-by-step explanation:

Hello,

Step 1

find the volume of the sphere when radius= 5 ft

$v(r)= \frac{3}{4} \pi r^3\\v(5)= \frac{3}{4}\pi (5\ ft)^3\\v(5)= \frac{3}{4}\pi *125\ ft^{3} \\v(5)=294.52\ cubic\ feet$

Step 2

find the volume of the sphere when radius= 5.1 ft

$v(r)= \frac{3}{4}\pi r^3\\v(5.1)= \frac{3}{4} \pi (5.1\ ft)^3\\v(5.1)= \frac{3}{4}\pi *132.651\ ft^{3} \\v(5.1)=312.551\ cubic\ feet$

Step 3

Compare the Volumes to find the change

$\frac{v(r_{2})}{v(r_{1})} =\frac{312.551}{294.52} =1.06$

the volumen of the sphere with radius = 5.1 is 1.06 times bigger than the first one(r=5)

Now, find the change

$change= {v(r_{2})-{v(r_{1})}} \\change=312.551\ cubic\ feet\ -294.52 cubic feet \\ change=18.031\ cubic\ feet$

change=18.03 cubic feet

Have a great day.