Question

What is the volume of the solid produced by revolving `f(x)=cosx, x in [0,pi]`around the x-axis?

Answer 1

`1/2pi^2 = 4.93` (3sf.)

Explanation:

General formula for volumes of revolution:

`V_x=int_b^apiy^2dx`
`V_y=int_b^apix^2dy`

(the pi can be taken out of the integral in either case to simplify the calculation)

`y=cosx`
`y^2=cos^2x`

The question requires solving:

`piint_0^picos^2xdx`

This is made easier using double angle formulae:

`cos2x=cos^2x-sin^2x = 2cos^2x -1`
`cos^2x=1/2(cos2x+1)`

By taking the `1/2` out of the integral it becomes:

`1/2piint_0^picos2x+1dx`

=`1/2pi[1/2sin2x+x]_0^pi`

=`1/2pi{(1/2sin2pi+pi) -(1/2sin0 + 0)}`

=`1/2pi(pi)`

=`1/2pi^2`