# Similar volumes of revolution

1. Aug 5, 2012

### HalcyonStorm

1. The problem statement, all variables and given/known data
For an assignment, I'm required to design three water bottle using 3 different polynomial functions. I've used a linear, quadratic and cubic. The first bottle needs to be 600ml, the second 300ml, and the third 1L.

In order to 'create' the bottles, we are to calculate the volume (really the area) between the function and the x-axis as if the bottle were 3D, by rotating it around the axis.

I had no trouble with the first part; I've created my 600ml bottle using the following functions across the given intervals:

f(x) = (-1/24)*(x-6)^2 + 4, from x = 0 to x = 6

2. Linear function
f(x) = 4, from x = 6 to x = 10

3. Cubic function
f(x) = (-3/128)*(x-10)^3 + 4, from x = 10 to x = 14

When these three functions are rotated, the calculated area is approximately 603ml, which is spot on (we are allowed to have a 10% variation from 600ml).

Now, here's the part where I am having trouble. In order to create the new bottles, I wish to scale the functions that I already have down or up, to create bottles of 300ml and 1L.

2. Relevant equations

Volume of revolution: V = ∏$\int^{b}_{a}y^{2}dx$

3. The attempt at a solution
I've realised that, obviously, to halve the volume to 300ml I will need to take half of V = ∏$\int^{b}_{a}y^{2}dx$. However, I have not been able to work out how to retrograde this change to my original functions. I'm starting to get frustrated!

Any help would be greatly appreciated. Thanks!

2. Aug 5, 2012

### HallsofIvy

Staff Emeritus
Since volume is proportional to a length cubed, to reduce a volume to 1/2 you will have to multiply lengths by $1/\sqrt[3]{2}$.