Show that: $ \lambda_3(K) = \pi \int_{a}^{b} f(x)^2 dx$. Why is $\lambda_2(K_x)$ = $\pi \cdot f(x)^2 $? $K_x$ circle with radius of $f(x)$

Question

Let $[a,b] \subset \mathbb{R}$ be an interval, $ f: [a,b] \rightarrow \mathbb{R_+} $ a continuous function.

We define: $ K$ := { $(x,y,z) \in [a,b] x \mathbb{R^2}$ | $ y^2 + z^2 \le f(x)^2 $ }

Show that: $$ \lambda_3(K) = \pi \int_{a}^{b} f(x)^2 dx.$$

Attempt : I know that I have to use Cavalieri's principle.

So for $ x \in \mathbb{R} $ define $ K_x$ := { $(y,z) \in \mathbb{R^2}$ | $(x,y,z) \in K$ }. So far so good, I think. Now $K_x$ is empty if $x \notin [a,b] $.

if $ x \in [a,b] $ then $K_x$ = { $(y,z) \in \mathbb{R^2}$ | $y^2 + z^2 \le f(x)^2$ }.

Now I can use Cavalieri: $ \lambda_3(K) =\int_{a}^{b}\lambda_2(K_x) d\lambda_1x $.

Now I have a few questions: I know intuitively that $\lambda_2(K_x)$ = $\pi \cdot f(x)^2 $ because $K_x$ is a circle with radius of f(x). But how can I show this? Can you help me here? Second question: Do I have I mistake in my proof? Third and last question: Why I need that $f$ is contiuous? Only to say that f is integrable?

Thank you for your help! :)

Answer 1

First question: For $x \in [a,b]$ we have

$K_x=\{(y,z) \in \mathbb R^2: y^2+z^2 \le f(x)^2\}$ as you noted. Write $r=f(x)$, then

$K_x=\{(y,z) \in \mathbb R^2: y^2+z^2 \le r^2\}$. This is a circle with radius $r$.

Second question: no there are no mistakes.

Last question: it suffices to assume that $f: [a,b] \rightarrow \mathbb{R_+}$ is measurable. This ensures that $K$ is (Borel-) measurable (why ?).