The following problem is something I thought about myself, trying to improve my integration and multivariable calculus skills. I would be glad to hear your review about it (maybe I forgot to mention something relevant to the problem), and of course a way to solve it (since I couldn't do it myself).
Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ be a continuously differentiable scalar function, which is defined on a smooth (differentiable) and simple curve $\{\bar\gamma(t) \mid t\in[a,b]\}$ in $\mathbb{R}^3$, such that $f(x,y,z)>0$ for every $(x,y,z)\in\mathbb{R}^3$, with the exception of $f(\bar\gamma(a))=f(\bar\gamma(b))=0$ . The curve $\bar\gamma(t)$ is not closed
($\bar\gamma(a)\neq\bar\gamma(b)$).
For every point $t_0\in[a,b]$ on the curve, we will define and construct a circle $C_{t_0}$, such that the circle is laying on the plane that is perpendicular to the curve at that point (meaning that the normal of the plane is the tangent vector to the curve at that point). The radius of $C_{t_0}$ is $f(\bar\gamma(t_0))$, and its center is $P=\bar\gamma(t_0)$.
Now we will define a volume $V\subset\mathbb{R}^3$ such that:
$$\partial V=\bigcup_{t\in[a,b]}C_{t}$$
Prove (or maybe disprove?) that $V$ is compact, and suggest and construct a way to compute its volume.
Thanks!
P.S.: A multivariable calculus solution would be the best for me, but of course anything else would also be great.
HINT: If you assume your curve $\gamma$ is reasonably smooth, with $\gamma'(t)\ne 0$ and $\gamma''(t)\ne 0$ for all $t$, then there is a Frenet frame along $\gamma$. (You can certainly get around this assumption if you have to.) You can parametrize your region $V$ by using "cylindrical coordinates" — $t$ along $\gamma$, and $(r,\theta)$ in the normal planes, with $0\le r\le f(\gamma(t))$ for $t$ fixed.