The surface you integrate over in the divergence theorem has to be closed. The two surfaces you mention, the "hemiellipse" and the disk, are not individually closed, but together form a closed surface. Hence the integral over both is zero, by the divergence theorem, so the integral over the hemiellipse is the negative of the integral over the disk.
The divergence (Gauss-Green) theorem can be used to define the improper integral of the divergence of (weakly) singular vector fields $\mathbf{F}$ with isolated singular points $\mathbf{p}_o=(x_0,y_o,z_0)\in V$. Customarily, the definition goes as follows
$$
\begin{split}
\int\limits_{V}\nabla\cdot\mathbf{F}(x,y,z)\,\mathrm{d}V& \triangleq \lim_{R\to 0} \Bigg[\,\int\limits_{V\setminus B(\mathbf{p}_o,R)} \nabla\cdot\mathbf{F}(x,y,z)\, \mathrm{d}V - \int\limits_{\partial B(\mathbf{p}_o,R)} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS\Bigg]\\
\\
&\triangleq \int\limits_{\partial V} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS,
\end{split}\label{1}\tag{1}
$$
where, for the small volume $\delta$, a small ball $B(\mathbf{p}_o,R)$ with radius $R>0$ centered on the singular point of $\mathbf{F}$ is customarily chosen. The definition is clearly consistent if and only if the limits of the two integrals in formula \eqref{1} exist and are finite.
An example.
The most famous example of use of \eqref{1} as a definition is perhaps the calculation of the integral of the divergence of the following field:
$$
\begin{split}
\mathbf{F}(x,y,z)&=\nabla{\bigg[\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 }\,\bigg]^{-1}}\\
&=\nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|}
\end{split}
$$
where $\mathbf{p}=(x,y,z)\in V$. This vector field is, apart from a multiplicative constant, the gradient of the fundamental solution of the laplacian: therefore, the integral of the divergence of this vector field is zero in every domain $V\subset\Bbb R^3\setminus\mathbf{p}_o$ since $\nabla\cdot\mathbf{F}$ is zero in such domains. However, applying \eqref{1} we have
$$
\begin{split}
\int\limits_{V}\nabla\cdot\mathbf{F}(\mathbf{p})\,\mathrm{d}V&=
-\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \cdot\hat{n}\, \mathrm{d}S \\
&= -\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial \hat{n}} \frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \mathrm{d}S\\
&=-\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial r} \frac{1}{r}\mathrm{d}S\\
&=\lim_{R\to 0} \frac{1}{R^2} \int\limits_{ \partial B(\mathbf{p},R)} \mathrm{d}S = 4\pi,
\end{split}
$$
and thus we can also define the flux of $\mathbf{F}$ throug $\partial V$.
Final notes
- All the above development is done under the hypothesis $\delta=B(\mathbf{p},R)$: however, formula \eqref{1} is valid for more general classes of small volumes $\delta$, provided that the singularity of $\mathbf{F}$ is sufficiently "weak".
- Let's precise the exact meaning of the locution "weak singularity" in the context of fields with a single isolated singularity. Being the field $\mathbf{F}$ singular, we can say that, near the singular point $\mathbf{p}_o\in V$,
$$
|\mathbf{F}(\mathbf{p})|\le K{|\;\mathbf{p}-\mathbf{p}_o|^{-\alpha(|\mathbf{p}-\mathbf{p}_o)|}}\label{2}\tag{2}
$$
where $\alpha:\Bbb R_+\to \Bbb R_+$ is a non negative function ($\alpha\ge0$). Then we have that
$$
\lim_{R\to 0}\bigg|\int\limits_{ \partial B(\mathbf{p},R)}\mathbf{F}(\mathbf{p})\cdot\hat{n}\, \mathrm{d}S \Bigg|<\infty\iff \lim_{R\to 0}R^{-\alpha(R)+2}<\infty\iff \lim_{R\to 0} \alpha(R)\le 2
$$
Be it noted that this condition is stronger than the simple condition of local integrability of the field $\mathbf{F}$: this one implies that
$$
\lim_{R\to 0} \alpha(R)<3
$$
in estimate \eqref{2}, thus there are locally integrable vector fields for which formula \eqref{1} is not applicable.
- As I stated clearly at the beginning of my answer, formula \eqref{1} is not really a divergence (Gauss-Green) theorem for singular vector fields: it is a definition which uses the standard theorem applied to non-singular regions of $F$ (by eventually cutting small volume pieces $\delta$) to extend its range of applicability to a class of singular vector fields. Therefore you will not find it stated as a theorem: however, in books on partial differential equations which do not use the theory of distributions, \eqref{1} is silently used in the proof of Green's formula. See for example Tikhonov and Samarskii [1], chapter IV, ยง2.1, pp. 316-318.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
Best Answer
By Reynolds transport theorem you have that $\frac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}\,dV\right) = \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} + \mathbf{f}\,\boldsymbol{\nabla} \cdot \mathbf{v}\right)\,dV$, so using $f=1, v=\dfrac{d}{dt}\Phi_t(X)=\mathbf F(\Phi_t(X)),\Omega(t)=D_t$ you get what you want.
$$\dfrac{d}{dt}\text{Vol}(D_t)=\frac{d}{dt}\left( \iiint_{D_T}dV\right) = \iiint_{D_T} \left( 0 + 0 +\boldsymbol{\nabla} \cdot \mathbf{\dfrac{d}{dt}\Phi_t(X)}\right)\,dV=\iiint_{D_t}\nabla\cdot\mathbf FdV$$
By Jacobi's formula we have $\frac{d}{dt} \det A =\mathrm{tr} ( \mathrm{adj}\ A \; \frac{dA}{dt})$ or in the equivalent form $\frac{d}{dt} \det A = \det A \; \mathrm{tr} \left(A^{-1} \frac{dA}{dt}\right)$ and for $\mathbf F$ we have that + $\frac{\mathbf dF}{dt}=F\nabla\varphi$ so putting everything together $$\frac{d}{dt} \det \mathbf F =\mathrm{tr}( \mathrm{adj}\ \mathbf F \; \frac{d\mathbf F}{dt})=\det \mathbf F \; \mathrm{tr}(\mathbf F^{-1} \frac{d\mathbf F}{dt})=\det \mathbf F \; \mathrm{tr}(\nabla\varphi)=\det \mathbf F \; \nabla\varphi$$
and we recover $\frac{d}{dt} \det \mathbf F =\det \mathbf F \; \nabla\varphi$ that is how it is used in the first link.
Hope it helps