Solid of Revolution: Infinite yet Finite

Calculus Level 4

Consider the section of graph of the function $f(x) = \frac{1}{x^n},\ \ \ \ x \in \langle 0, 1],\ n > 0.$ This section of graph is revolved around the $x$ -axis, resulting in a solid of revolution.

For which values of $n$ does this solid of revolution have a finite volume?

Clarification :

If you are uncomfortable revolving a graph involving a vertical asymptote: consider the volume of the solid of revolution for $x \in \langle \varepsilon, 1]$ , and take the limit for $\varepsilon\to 0$ .

All

n > 0

n < 1/e

n < 2

This is never true

n < \tfrac12

1 solution

Arjen Vreugdenhil
Feb 18, 2016

The volume is equal to $\pi\int_0^1 (f(x))^2\:dx = \pi\int_0^1 x^{-2n}\:dx \\ = -\frac\pi{1-2n} \left[x^{1-2n}\right]_0^1 = \begin{cases} -\frac\pi{1-2n} + 0 & \text{if}\ 1-2n > 0; \\ -\frac{\pi}{1-2n} + \infty & \text{if}\ 1-2n < 0 \end{cases}$ This value is finite the exponent $1 - 2n > 0$ . It follows that $\boxed{n < \tfrac12}$ .

Solid of Revolution: Infinite yet Finite

1 solution

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