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December 22, 2014

December 22, 2014

Posted by **Morgan** on Monday, January 23, 2012 at 8:14pm.

- Calculus II -
**Reiny**, Monday, January 23, 2012 at 10:08pmintersection 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.

- Calculus II -
**Morgan**, Monday, January 23, 2012 at 10:43pmCould you show how to do this problem using integrals?

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