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April 17, 2014

Homework Help: Calculus

Posted by mathstudent on Sunday, January 13, 2008 at 4:36pm.

Suppose that the region between the x-axis and the curve y=e^-x for x>=0 has been revolved around the x-axis. Find the surface area of the solid.

I got 3*pi
The book shows an answer of pi * [sqrt(2) + ln(1 + sqrt(2))]

Where do I go wrong? For the sides of the surface, I integrated 2*pi*e^-x over [0,inf] and got 2*pi. Then for the base of the surface, I pi*(e^0)^2 = 2*pi.

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