Does bounded area of a curve imply bounded volume of associated solid of rotation?

Question

Say you have that an improper integral (say in the x-z plane) has finite area, i.e., $ \int_{0}^{\infty} f(x) dx = A$ where A is some positive real number. Suppose this curve is rotated around one of its axes (WLOG, say, around the z axis axis) to create a surface in 3D. Is the region bounded by this surface and the x-y plane guaranteed to have finite volume?

Answer 1

No.

Consider the function $$f(x) = \frac{1}{(x+1)^2}.$$ Then $$\int_0^\infty f(x) dx = -\frac{1}{(x+1)} \Bigg|_0^\infty = 1.$$ However, rotating it around the $z$-axis, we can use the shell method to compute the volume. We get the volume is $$ \begin{aligned} 2 \pi \int_0^\infty xf(x) dx &= 2 \pi \int_0^\infty \frac{x}{(x+1)^2} dx \\ &= \lim\limits_{t \to \infty} \left(2 \pi \int_0^t \frac{x+1}{(x+1)^2} dx - 2 \pi \int_0^t \frac{1}{(x+1)^2} dx \right)\\ &= \lim\limits_{t \to \infty} \left(2 \pi\int_0^t \frac{1}{(x+1)} dx - 2 \pi \int_0^t \frac{1}{(x+1)^2} dx \right) \\ &= \lim\limits_{t \to \infty} \left(2 \pi\ \log (x + 1)\Bigg|_0^t + 2 \pi \frac{1}{(x+1)} \Bigg|_0^t \right) \\ &= \infty. \end{aligned} $$