Calculus II
posted by Morgan .
Find the volume of the solid generated by revolving R about the xaxis 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)

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(91)  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. 
Could you show how to do this problem using integrals?
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