Question

What is the volume of the solid produced by revolving `f(x)=cotx, x in [pi/4,pi/2]`around the x-axis?

Answer 1

`V=pi-1/4pi^2`

Explanation:

The formula for finding the volume of a solid produced by revolving a function `f` around the `x`-axis is

`V=int_a^bpi[f(x)]^2dx`

So for `f(x)=cotx`, the volume of its solid of revolution between `pi"/"4` and `pi"/"2` is

`V=int_(pi"/"4)^(pi"/"2)pi(cotx)^2dx=piint_(pi"/"4)^(pi"/"2)cot^2xdx=piint_(pi"/"4)^(pi"/"2)csc^2x-1dx=-pi[cotx+x]_(pi"/"4)^(pi"/"2)=-pi((0-1)+(pi/2-pi/4))=pi-1/4pi^2`

Answer 2

`"Area of revolution around"` `x"-axis"=0.674`

Explanation:

`"Area of revolution around"` `x"-axis"=piint_a^b(f(x))^2dx`

`f(x)=cotx`
`f(x)^2=cotx`

`int_(pi/4)^(pi/2)cot^2xdx=int_(pi/4)^(pi/2)csc^2x-1dx`
`color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[-cotx-x]_(pi/4)^(pi/2)`
`color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[(-cot(pi/2)-pi/2)-(-cot(pi/4)-pi/4)]`
`color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[(-0-pi/2)-(-1-pi/4)]`
`color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[-pi/2+1+pi/4]`
`color(white)(int_(pi/4)^(pi/2)cot^2xdx)=0.674`