Question

A spherical ballon with a radius r inches has volume V(r)=4/3 pir^3. Find a function that represents the amount of air required to inflate the balloon from a radius inches to a radius of r+1 inches.

Answer 1

The function that represents the amount of air is
`\Delta V =\cfrac 43 \pi (3r^3+3r+1)`

Explanation:

The amount of air here represents the difference between V(r) and V(r+1), so we can start working by finding an expression for the volume at r+1, and then subtract the original volume V(r).

Volume of balloon of radius r+1 inches.

We can replace r with r+1 on the formula and we get:

`V(r+1)=\cfrac43 \pi (r+1)^3`

We can expand
`(r+1)^3` since we will use it to simplify it later on.

So we will have first

`(r+1)^2 = (r+1)(r+1)\\(r+1)^2 =r^2+r+r+1\\(r+1)^2 = r^2+2r+1`

We can multiply that result by (r+1) to get
`(r+1)^3`

`(r+1)^3= (r+1)^2 (r+1)\\(r+1)^3=(r^2+2r+1)(r+1)\\(r+1)^3= r^3+2r^2+r+r^2+2r+1\\(r+1)^3 =r^3+3r^2+3r+1`

Thus the volume equation at r+1 will be

`V(r+1)=\cfrac 43 \pi (r^3+3r^2+3r+1)`

Finding the amount of air required to inflate from r to r+1

The amount required to inflate is the difference of volumes, so we have

`V(r+1)-V(r)=\cfrac 43 \pi (r^3+3r^2+3r+1)  \cfrac 43 \pi r^3`

Combinging both into one term by factor
`\cfrac 43 \pi` give us

`V(r+1)-V(r)=\cfrac 43 \pi (r^3+3r^2+3r+1-r^3)`

Simplifying

`V(r+1)-V(r)=\cfrac 43 \pi (3r^2+3r+1)`

And that function represents the amount required to inflate the balloon from r to r+1 inches.