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So I am supposed to find the volume of a solid that I get from rotating $$f(x)= \frac{1}{\sqrt{1+x^2}}$$ around the $x$-axis when the absolute value of $x$ is greater than or equal to $1$. I know the equation for area when rotating around the $x$-axis is $\pi f(x)^2$ and I know the interval is on a certain interval, but I only know the left endpoint is one. What is the right endpoint of the interval?

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There isn’t one: you’ll have an improper integral of the form $$\int_1^\infty\text{something}~dx\;.$$ – Brian M. Scott Dec 12 '12 at 2:42
up vote 3 down vote accepted

As pointed out in the comment by @BrianMScott , you don't have an upper endpoint. So the integral that you want to find is $$ \begin{align} \int_1^{\infty} \pi f(x)^2 \; dx &= \pi\int_1^\infty \frac{1}{1+x^2}\; dx.\\ &= \pi\lim_{k\to \infty}\int_1^k \frac{1}{1+x^2}\;dx \end{align} $$ If you know an antiderivative of $\frac{1}{1+x^2}$, then this is not too hard to find. In fact you get $$ \pi\lim_{k\to \infty} \left[\arctan(k) - \arctan(1)\right] $$

The above is the area of the solid that is rotated around the $x$-axis for $x\geq 1$. You then have a similar piece for $x \leq -1$. So the total area is twice what I wrote. (Thanks to Andre for pointing this out)

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