Respuesta :
Answer:
The statement is true
Step-by-step explanation:
Volume of a Solid of revolution:
when a region in the plane is revolved about a given line that is called axis of revolution, then we get a solid of revolution.
In this problem we want to find the volume of a solid formed by revolving the function [tex]f(x)=\sqrt{x^{2}+1 }[/tex] about the x-axis on the interval [1,2]
We can find the volume of any solid by integrating its area
[tex]V=\int\limits^b_a {A} \, dx [/tex] eq. 1
where [tex]A=\pi r^{2}[/tex]
and [tex]r^{2} =(f(x))^{2} =(\sqrt{x^{2}+1 })^{2}[/tex]
Limits are [tex] a=1 , b=2[/tex]
eq. 1 becomes
[tex]V=\int\limits^b_a {\pi r^{2}\, dx[/tex]
[tex]V=\int\limits^2_1 {\pi (f(x) )^{2}\, dx[/tex]
[tex]V=\int\limits^2_1 {\pi (\sqrt{x^{2}+1 } )^{2}\, dx[/tex]
Hence proved.
The volume of the solid formed by revolving the function [tex]f(x)=\sqrt{x^{2}+1 }[/tex] about the x-axis on the interval [1, 2] is given by [tex]V=\int\limits^2_1 {\pi (\sqrt{x^{2}+1 } )^{2}\, dx[/tex]
