Differentiating under the integral sign in Kelvin's theorem I know well that this has already been asked here:
Differentiation under the integral sign - line integral?
..but the answer given there already assumes that the length of the contour does not depend on time.. which is the actual point of the question. So I'll rewrite here the problem.
Suppose you have an ideal isoentropic fluid, and consider the circulation on a contour $C(t)$ evolving in time
$$\Gamma(t) =  \oint_{C(t)}\textbf{v}(\textbf{x}(s,t),t)\cdot\frac{\partial}{\partial s}\textbf{x}(s,t)ds.$$
Kelvin's Theorem proves $\Gamma$ to be constant in time. But its time derivative is
$$ \frac{d\Gamma}{dt} = \oint_{C(t)}\frac{d\textbf{v}}{dt}\cdot\frac{\partial}{\partial s}\textbf{x}(s,t)ds + \oint_{C(t)}\textbf{v}\cdot\frac{d}{dt}\frac{\partial}{\partial s}\textbf{x}ds + BTs,$$
where I denoted with $BTs$ the boundary terms.
Now, the issue is that, according to my textbook, and to the aforementioned answer (and to the wikipedia page https://en.wikipedia.org/wiki/Kelvin%27s_circulation_theorem as well), these boundary terms are vanishing.
So..why is that?
Shouldn't the length of the line on which we take the circulation depend on time?
 A: To see that the approach in Differentiation under the integral sign - line integral? is valid, we can consider the one-dimensional case.
If $f$ is sufficiently smooth, then using the Leibniz integral rule we get
$$\tag{1}\frac{d}{dt} \int_0^{C(t)}f(x,t ) \, dx =  \int_0^{C(t)}\frac{\partial f}{\partial t}(x,t ) \, dx + f(C(t),t) C'(t)$$
Using the approach in question, we make the change of variables $x = C(t)s$ where $0 \leqslant s \leqslant 1$ and the integral becomes
$$\tag{2}\int_0^{C(t)}f(x,t ) \, dx = C(t)\int_0^1f(C(t)s,t) \, ds$$
Differentiating with respect to $t$, we get
$$\tag{3}\frac{d}{dt} \int_0^{C(t)}f(x,t ) \, dx= \underbrace{C(t) \frac{d}{dt} \int_0^1 f(C(t)s,t) \, ds}_{A(t)}+ \underbrace{C'(t)  \int_0^1 f(C(t)s,t) \, ds}_{B(t)}  $$
Using (2), the second term $B(t)$ is
$$\tag{4} B(t) = \frac{C'(t)}{C(t)}  C(t)\int_0^1 f(C(t)s,t) \, ds= \frac{C'(t)}{C(t)}  \int_0^{C(t)} f(x,t) \, dx$$
In the first term $A(t)$, we can pass the derivative under the integral to obtain
$$\tag{5}A(t) = C(t) \int_0^1 \frac{\partial f}{\partial t}(C(t)s,t) \, ds + C(t) \int_0^1 \frac{\partial f}{\partial x} (C(t)s,t)C'(t) s \, ds \\= \int_0^{C(t)} \frac{\partial f}{\partial t}(x,t) \, dx + \frac{C'(t)}{C(t)}\int_0^{C(t)}\frac{\partial f}{\partial x} (x,t) x \, dx$$
Integrating the second integral by parts, we get
$$\int_0^{C(t)}\frac{\partial f}{\partial x} f(x,t) x \, dx = \left. xf(x,t)\right|_0^{C(t)}- \int_0^{C(t)} f(x,t) \, dx = C(t)f(C(t),t) - \int_0^{C(t)} f(x,t) \, dx$$
and substitution into (5) yields
$$\tag{6}A(t) = \int_0^{C(t)} \frac{\partial f}{\partial t}(x,t) \, dx + f(C(t),t) C'(t) - \frac{C'(t)}{C(t)} \int_0^{C(t)}f(x,t) \, dx$$
Finally, substituting into (3) with (4) an (6), we obtain
$$\frac{d}{dt} \int_0^{C(t)}f(x,t ) \, dx =  \int_0^{C(t)}\frac{\partial f}{\partial t}(x,t ) \, dx + f(C(t),t) C'(t)$$
This matches the result (1) obtained with the Leibniz integral rule and demonstrates the validity of the change of variables approach.
