Respuesta :
Answer:
[tex]V= 3.451 [/tex]
Step-by-step 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 [tex]f(x)=\frac{1}{\sqrt{x-1 }}[/tex]
We can find the volume of any solid by integrating its area
[tex]V=\int\limits^b_a {A} \, dx [/tex]
where [tex]A=\pi r^{2}[/tex]
[tex]r^{2} =f((x))^2=( \frac{1}{\sqrt{x-1} } )^2=\frac{1}{x-1 }[/tex]
Limits are [tex] a=2, b=4[/tex]
so,
[tex]V=\int\limits^b_a {\pi r^{2}\, dx[/tex]
[tex]V=\int\limits^4_2 {\pi (f(x) )^{2}\, dx[/tex]
[tex]V=\int\limits^4_2 {\pi (\frac{1}{x-1}) \, dx[/tex]
As we know the integral of [tex]\frac{1}{x-1}[/tex] is equal to [tex]ln(x-1)[/tex] therefore, integrating yields,
[tex]V=\pi ln(x-1)[/tex]
evaluating limits yields,
[tex]V=\pi (ln(4-1)-ln(2-1))[/tex]
[tex]V=\pi (ln(3)-ln(1))[/tex]
as we know ln(1) is equal to 0 therefore,
[tex]V=\pi ln(3)[/tex]
[tex]V=\pi (1.0986)[/tex]
[tex]V= 3.451 [/tex]
