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

Well, the surface area of a function

#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,

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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