# A company has the task of producing spheres of volume `288pi pm 5` cm^3. Use integration to find the radius of the spheres, writing this in the form `a pm b` cm with b rounded to three decimal...

A company has the task of producing spheres of volume `288pi pm 5` cm^3.

Use integration to find the radius of the spheres, writing this in the form `a pm b` cm

with b rounded to three decimal places.

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The formula for the volume of a sphere is

`V = 4/3pir^3`

Think of this as adding two hemispheres together, where each of those hemispheres is the sum of many many circle slices/discs from `x=0` ro `r` where the radius of the disc at `x ` is (using Pythagoras) `sqrt(r^2-x^2)` . Then, since the area of each disc is `pi(r^2-x^2)` and integrating over `x` we get

`V = 2 times int_0^r pi (r^2-x^2) dx`

`= 2 pi times (xr^2 - 1/3 x^3|_0^r = 2pi times (r^3 - 1/3r^3) = 4/3pir^3`

Now, we have that the volume ` `of the spheres in question is

`V = 288pi pm 5` cm^3

`= 4/3pi(288(3/4) pm 5/pi(3/4)) = 4/3pi(216 pm 15/(4pi))`

Therefore

`r^3 = 216 pm 15/4pi`

The minimum `r` can be is ` ``root(3)(216-15/(4pi)) =5.989`

The maximum `r` can be is `root(3)(216+15/(4pi)) = 6.011`

Therefore

`r = 6 pm 0.011` cm

a = 6 and b = 0.011 to 3dp