How compute with cylindrical shells the volume of a sphere of radius r?

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to evaluate the volume of sphere of radius r, using cylindrical shells method, such that:

`V = int_a^b 2pi*x*f(x)dx`

You need to use the equation of circle of radius `r` to evaluate `f(x)` , such that:

`x^2 + y^2 = r^2 => y^2 = r^2 - x^2 => y = sqrt(r^2 - x^2)`

Replacing `sqrt(r^2 - x^2)` for ` f(x)` yields:

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

`int_0^r x*sqrt(r^2 - x^2)dx = int_0^r x*(r^2 - x^2)^(1/2)dx`

You should come up with the following substitution, such that:

`r^2 - x^2 = t => -2xdx = dt => 2xdx = -dt`

Changing the limits of integration yields:

`x = 0 => t = r^2`

`x = r => t = 0`

`2pi*int_0^r 2x*(r^2 - x^2)^(1/2)dx = 2pi*int_(r^2)^0 t^(1/2)*(-dt)`

`2pi*int_(r^2)^0 t^(1/2)*(-dt) = 2pi*int_0^(r^2) t^(1/2)*dt`

`2pi*int_0^(r^2) t^(1/2)*dt = 2pi*(t^(1/2+1))/(1/2+1)|_0^(r^2)`

Using the fundamental theorem of calculus yields:

`2pi*int_0^(r^2) t^(1/2)*dt = (4pi)/3*(r^2)^(3/2)`

`2pi*int_0^(r^2) t^(1/2)*dt = (4pi)/3*r^3`

Hence, evaluating the volume of the sphere of radius r, using cylindrical shells method, yields `V = (4pi)/3*r^3.`