Monday
September 15, 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.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

calculus - Find the volume of the solid that is obtained when the region under ...
CALCULUS - The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/(sqrt(...
CALCULUS HELP PLZ - The region R is bounded by the x-axis, y-axis, x = 3 and y...
calculus - let R be the region of the first quadrant bounded by the x-axis and ...
AP Calculus - Let R be the region bounded by the x-axis and the graph of y=6x-x^...
Calculus - solid is formed by rotating the region bounded by the curve y=e−...
Calculus - solid is formed by rotating the region bounded by the curve y=e&#...
calculus - Consider the graphs of y = 3x + c and y^2 = 6x, where c is a real ...
Calculus - A solid is formed by rotating the region bounded by the curve y=e&#...
Calculus - Let R be the region bounded by the x-axis, x = 5 and the curve y = x...

Search
Members