Question

How do you find the volume bounded by `y = e^x` , y=0 , x = 0, x = ln2 revolved about the x=axis?

Answer 1

`(3pi)/2~~4.71`

Explanation:

`V=piint_a^by^2dx`

`y^2=(e^x)^2=e^(2x)`

`int_0^ln2e^(2x)dx=[e^(2x)/2]_0^ln2`

`pi[e^(2x)/2]_0^ln2=pi(e^(2ln2)/2-e^(2(0))/2)=pi(4/2-1/2)=(3pi)/2~~4.71`