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

18.03 cubic feet

Explanation:

Hello,

Step 1

find the volume of the sphere when radius= 5 ft

`v(r)= (3)/(4) \pi r^3\\v(5)= (3)/(4)\pi (5\ ft)^3\\v(5)= (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)= (3)/(4)\pi r^3\\v(5.1)= (3)/(4) \pi (5.1\ ft)^3\\v(5.1)= (3)/(4)\pi *132.651\ ft^(3) \\v(5.1)=312.551\ cubic\ feet\\\\`

Step 3

Compare the Volumes to find the change

`(v(r_(2)))/(v(r_(1))) =(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.