What is the surface area produced by rotating #f(x)=1/(x^2+1), x in [0,3]# around the x-axis?

1 Answer
Truong-Son N.
Aug 3, 2017

Well, the surface area of a function #f(x)# is given by

#S = int 2pi f(x)ds#

#= int overbrace(2pi f(x))^"Circumference"overbrace(sqrt(1 + ((df)/(dx))^2))^("Arc Length")dx#

This surface would look like:

First we get the derivative.

#(df)/(dx) = -(2x)/(x^2 + 1)^2#

And square it to get:

#((df)/(dx))^2 = (2x)^2/(x^2 + 1)^4#

So, the surface area integral is:

#S = 2pi int_(0)^(3) 1/(x^2 + 1)sqrt(1 + ((2x)^2)/(x^2 + 1)^4)dx#

#= 2pi int_(0)^(3) sqrt(1/(x^2 + 1)^2 + ((2x)^2)/(x^2 + 1)^6)dx#

There is no elementary solution to this, so we can only get the numerical integration result, #color(blue)(2.72708pi)# #color(blue)("u"^2)#, or about #8.56737#.

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