Calculus. Function Spherical radius?

Question

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

Please show me how to do this problem with detailed steps so I can understand how to come up with the answer. I already have the final answer, but I would like to know how to do the problem. Thank you!

Answer 1

Hi,

You want to find the difference in the volume with a radius of r+1 and the volume with a radius of r.

That will be 4/3π(r + 1)³ - 4/3πr³ =

4/3π(r³ + 3r² + 3r + 1)³ - 4/3πr³ =

4/3πr³ + 4πr² + 4πr + 4/3π - 4/3πr³ =

4πr² + 4πr + 4/3π or 4π(r² + r + 1/3)

I hope that helps!! :-)

Answer 2

All you have to do is write out what the two volumes are:
v(r) = (4/3)pi r^3
v(r+1) = (4/3)pi (r+1)^3

And now subtract them:
v(r+1) - v(r) = (4/3)pi (r+1)^3 - (4/3)pi r^3

This gives you the increase in volume of the balloon when the radius changes from r to r+1.

You can also integrate if that is what you are looking for. Use an infinitesimal shell for the element of volume to integrate.

Surface area = s = 4pi r^2
Element of volume = 4pi r^2 dr
Now integrate the element of volume from r to r+1 and you get the same as above:
(4/3)pi (r+1)^3 - (4/3)pi r^3

Answer 3

V = (4/3) * pi * r^3 V[2] = (4/3) * pi * (r + a million)^3 V[a million] = (4/3) * pi * (r)^3 V[2] - V[a million] = (4/3) * pi * ((r + a million)^3 - r^3) = (4/3) * pi * (r^3 + 3r^2 + 3r + a million - r^3) = (4/3) * pi * (3r^2 + 3r + a million)