What is the surface area of the solid created by revolving #f(x)=-sqrtx/e^x# over #x in [0,1]# around the x-axis?

1 Answer
Gió
Jan 22, 2018

I got as far as the definite integral but then I got stuck at the solution of the integral....probably there is a mistake or there is an easier way...

Explanation:

I considered the function and the rotation:

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then I used the standard technique:

enter image source here

giving the expression of the surface area #S# as:

#S=int_0^1(2pi)(-sqrt(x)/e^x)*sqrt(1+(4x^2-4x+1)/(4e^(2x)*x))dx#

which I am not able to solve!!!

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