I have a function $f(x) = \frac{1}{\sqrt{x}}$ on $[0, \pi]$ and want to find the area of the function, then the volume of the function were it revolved around the x-axis.
For the area:
$$\begin{align} A &= \int_0^\pi \frac{1}{\sqrt{x}} \, dx \\ &= \int_0^\pi \frac{1}{x^{\frac{1}{2}}} \, dx \\ &= \int_0^\pi x^{\frac{-1}{2}} \, dx \\ &= \left[ \frac{x^{\left(\frac{-1}{2}+1\right)}}{\left(\frac{-1}{2}+1\right)} \right]_0^\pi \\ &= \left[\frac{x^{\frac{1}{2}}}{\left( \frac{1}{2}\right)}\right]_0^\pi \\ &= \left[2x^{\frac{1}{2}}\right]_0^\pi \\ &= \left[2\sqrt{x}\right]_0^\pi \\ &= 2\sqrt{\pi} - 2\sqrt{0} \\ &= 2\sqrt{\pi} \end{align}$$
The area $A$ of $f(x)$ on $[0, \pi]$ is finite. Now revolving around the x-axis:
$$\begin{align} V &= \int_0^\pi \pi\left(\frac{1}{\sqrt{x}}\right)^2 \, dx \\ &= \int_0^\pi \pi\left(\frac{1^2}{\left(\sqrt{x}\right)^2}\right) \, dx \\ &= \int_0^\pi \pi\left(\frac{1}{x}\right) \, dx \\ &= \int_0^\pi \frac{\pi}{x} \, dx \\ &= \left[\pi \ln{x}\right]_0^\pi \\ &= \pi \left[\ln{x}\right]_0^\pi \\ &= \pi (\ln{\pi} - \ln{0}) \end{align}$$
Note $\ln{x}$ is undefined for $x = 0$ and complex for $x \lt 0$, thus the volume $V$ of $f(x)$ over $[0, \pi]$ is not finite.
How can an area of a function be finite, yet the volume of the same function rotated around an axis be infinite/undefined? Is this a quirk of the mathematics or I haven't yet learned how to find the volume of shapes like this? I don't know if the mathematics here is definitive or it's undefined as a consequence of my (lack of) knowledge.
When you integrate to compute the volume of the solid of revolution, you are integrating the square of the function. It turns out that $\frac 1{\sqrt x}$ is not square integrable, and the integral diverges to $+\infty$. Any function that is integrable but not square integrable will give finite area but infinite volume.