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

Question

Christopher Allers · Accepted Answer

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#.

Christopher Allers · Answer

undefined To find the surface area produced by rotating $ f(x) = \frac{1}{x^2+1} $, where $ x $ is in the interval $[0,3]$, around the x-axis, you can use the formula for the surface area of revolution:

$$ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx $$

First, calculate $ f'(x) $:

$$ f'(x) = -\frac{2x}{(x^2+1)^2} $$

Now, substitute the values and integrate:

$$ S = 2\pi \int_{0}^{3} \frac{1}{x^2+1} \sqrt{1 + \left( -\frac{2x}{(x^2+1)^2} \right)^2} \, dx $$

After integrating, you'll get the surface area.