Let $[a, b] \subset \Bbb R, a, b \in \Bbb R, a < b$, and let $f: [a, b] \rightarrow [0, \infty)$ be a continuous function. Define:
$K := \{(x, y, z) \in [a, b] \times \Bbb R^2 : y^2 + z^2 \le (f(x))^2 \} \subset \Bbb R^3$.
Prove that $\lambda^3(K) = \pi \int_a^b (f(x))^2 dx$.
After receiving an answer with examples on how $K$ looks like (https://math.stackexchange.com/a/2087291/3445309), I tried to solve the problem like this:
First, we note that $f$ is continuous on a compact interval, thus, it is bounded, thus, $K$ is compact itself. Therefore, we are allowed to use the principle of Cavalieri here. In order to do so, we need to define a cut surface and work in the $2$-dimensional space, so we need to handle circle surfaces right now. We define:
$K_t := K_2(\sqrt{(f(x))^2 - t^2})$ for $|t| \le (f(x))^2, \emptyset$ otherwise.
I define it like this because the radius is obviously given by $(f(x))^2$. But how do I need to apply Cavalieri now? I started with something like this:
$\lambda^3(K) = \int_a^b \lambda^2(K_t) dt = \theta_2 \int_a^b (f(x))^2 - t^2) dt$.
Note that $\theta_n$ is identical with the Lebesgue-measure of the $n$-dimensional ball with radius $1$, so $\theta_2$ is just $\pi$, which is part of the solution. But the integral there doesn't work out correctly. It would work out correctly if I would choose $t = 0$, and I think that this even makes sense when one regards that $(f(x))^2$ already depends on $x \in [a, b]$, so I would automatically get different radians. Is it true though?
This might be relevant: