Prove that $\iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int _a ^b tF'(t)dt$

Question

Suppose $f\in C^1(\mathbb{R}^3)$, for every $t\in \mathbb{R}$, $f^{-1}(t)$ is a simple(1) closed surface, and let the volume of the 3-D object surrounded by $f^{-1}(t)$ be $F(t)$. If $F:[0,+\infty] \to \mathbb{R}$ is continuously differentialble, prove that $$\iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int _a ^b tF'(t)dt.$$ (1):To be clear, here is the definition of a simple domain:

My first thought was to use the coarea formula. Then $$ \iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int_a^b dt \int _{f^{-1}(t)} \frac{f}{|\nabla f|} d\sigma = \int _a ^b tdt \int _{f^{-1}(t)} \frac{1}{|\nabla f|} d\sigma. $$ Well, as you have seen the RHS is pretty close to $\int _a^b tF'(t)dt$. But from here I met 2 major difficulties:(1)the conditions required by the coarea formula were not all satisfied: we need $f\in C^2(\mathbb{R})$.(2) even if the coarea formula is allowed, I still cannot proceed. By definition $F(t)=\iiint _D 1dxdydz$ where $\partial D=f^{-1}(t)$. But how should I get $F'(t)=\int _{f^{-1}(t)} \frac{1}{|\nabla f|} d\sigma$?
Thank you in advance.
Example: let $f(x,y,z)=x^2+y^2+z^2,[a,b]=[0,1]. F(t)=\frac{4\pi t^{3/2}}{3},F'(t)=2\pi \sqrt {t}.$ Hence, $$ \iiint _{f^{-1}[0,1]}f(x,y,z)dxdydz=\iiint _{x^2+y^2+z^2 \leq 1}(x^2+y^2+z^2)dxdydz= \\ \int _{[0,2\pi] \times [0,\pi] \times [0,1]} r^2 \cdot r^2 \sin \phi d\theta d\phi dr = 4\pi/5. $$ $$ \int _0 ^1 tF'(t)dt = \int _0 ^ 1 t \cdot 2\pi \sqrt {t} dt = 4\pi/5. $$

Answer 1

I won't go through the epsilontics, but I think your formula can be understood in terms of simple Riemann integrals. I write just ${\bf x}$ for your $(x,y,z)$.

The set $D:=f^{-1}\bigl([a,b]\bigr)$ can be viewed as an onion, foliated by the surfaces $f^{-1}\bigl(\{t\}\bigr)$, $a\leq t\leq b$.

Choose an $N\gg1$, and put $$t_k:=a+{k\over N}(b-a)\qquad (0\leq k\leq N)\ ,$$ $$D_k:=\bigl\{{\bf x}\in D\bigm| t_{k-1}\leq f({\bf x})\leq t_k\bigr\}\qquad(1\leq k\leq N)\ .$$ We then have $\bigcup_{k=1}^N D_k=D$, and therefore $$\int_Df({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^N \int_{D_k}f({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^Nf(\xi_k)\>{\rm vol}(D_k)\ .$$ By definition of $F$ we therefore have $$\int_Df({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^Nf(\xi_k)\bigl(F(t_k)-F(t_{k-1})\bigr)= \sum_{k=1}^Nf(\xi_k) F'(\tau_k)(t_k-t_{k-1})\ .$$ Since $\xi_k\in D_k$ we have $\tau_k^* :=f(\xi_k)\in[t_{k-1},t_k]$. The last sum therefore is a Riemann sum for the integral $$\int_a^b t\>F'(t)\>dt\ .$$