Question

Find the volume of the solid of revolution formed by rotating the bounded region about the x-axis.

f(x)=1/√x-1, y=0, x=2, x=4.

Answer 1

`V= 3.451`

Explanation:

In this problem we want to find the volume of the solid pf revolution formed by rotating the bounded region about x-axis on the interval [a,b]=[2,4] for the following function
`f(x)=(1)/(√(x-1 ))`

We can find the volume of any solid by integrating its area

`V=\int\limits^b_a {A} \, dx`

where
`A=\pi r^(2)`

`r^(2) =f((x))^2=( (1)/(√(x-1) ) )^2=(1)/(x-1 )`

Limits are
`a=2, b=4`

so,

`V=\int\limits^b_a {\pi  r^(2)\, dx`

`V=\int\limits^4_2 {\pi (f(x) )^(2)\, dx`

`V=\int\limits^4_2 {\pi ((1)/(x-1)) \, dx`

As we know the integral of
`(1)/(x-1)` is equal to
`ln(x-1)` therefore, integrating yields,

`V=\pi ln(x-1)`

evaluating limits yields,

`V=\pi (ln(4-1)-ln(2-1))`

`V=\pi (ln(3)-ln(1))`

as we know ln(1) is equal to 0 therefore,

`V=\pi ln(3)`

`V=\pi (1.0986)`

`V= 3.451`