Find the volume of the solid generated by revolving R about the x-axis where R is the region enclosed by the larger curve y=(e^2x)/3, the smaller curve y=1 and the line x=ln(3)

Answer 1

intersection of y = (1/3)e^(2x) and y = 1
e^2x = 3
2x = ln3
x = (1/2)ln 3
so let's take the volume of the whole region below (1/3)e^(2x) from x = (1/2)ln3 to ln3 and subtract the small cylinder

Vol = π∫y^2 dx - inside small cylinder
= π∫(1/9)e^(4x) dx - i.s.c.
=π[(1/36)e^(4x) from (1/2)ln3 to ln3 - i.s.c.
= π[( (1/9)(81) - (1/9)(9) ) - i.s.c.
= π(9-1) - i.s.c.
= 8π - inside small cylinder

the inside small cylinder has a radius of 1 (from y=1) and a height of ln3 - (1/2)ln3 = (1/2)ln3
its volume is π(1^2)(1/2)ln3
= πln3 /2

whole volume = 8π - (1/2)(π)(ln3) or appr 23.4

I am pretty sure of my method, but you better check my arithmetic and calculations.

Answer 2

Could you show how to do this problem using integrals?