A1  ^?? Derive an expression for the elementary work $\delta A$ that the external field has to do to change the polarization of a unit volume of ferroelectric by $\mathrm dP$.

When the polarization changes, the field does charge separation work equal to:

A2  ^?? Derive an equation relating the heat $\delta Q$ released inside the ferroelectric, the work $\delta A$ done by external field and the change $\mathrm dU$ in ferroelectric internal energy.

Work $\delta A$, heat $\delta Q$, and change in internal energy $\mathrm dU$ are related to each other according to the first law of thermodynamics:

A3  ^?? Express as some proper integral the total work $A$ that the external field has to do to a unit volume of the material during the whole cycle of field's increase and decrease. By how much $\Delta U$ does the internal energy density changes during the cycle? How much heat $Q$ is released in a unit volume?

The work is expressed as an integral:\[A=\oint E\,\mathrm dP\]over the whole cycle. In order to calculate the integral explicitly (since only the dependences $P_{\pm}(E)$ are known), the integral must be transformed by integration by parts:\[\begin{aligned}A&=\oint E\,\mathrm dP=EP\Bigg|_{(E_1,P_1)}^{(E_1,P_1)}-\oint P\,\mathrm dE=\\&=-\left[\int\limits_{E_1}^{E_2}P_-\,\mathrm dE+\int\limits_{E_2}^{E_1}P_+\,\mathrm dE\right]=\\&=\int\limits_{E_1}^{E_2}(P_+-P_-)\,\mathrm dE.\end{aligned}\]Since the internal energy of a ferroelectric is a state function and thus can depend only on its polarization $P$ and external field $E$, its change in the cycle must be equal to $\Delta U=0$. Then $Q=A$.

B1  ^?? Calculate what amount of heat $Q$ was released in the film.

Since $|V_{\min,\max}| > 300~V$, then:
\[|E_{\min,\max}| > \dfrac{300~V}{10~\text{µm}} = 300~\dfrac{kV}{cm},\]the voltage range completely covers the BTO hysteresis loop, and the dissipated heat will be equal to:
\[\begin{aligned}Q&=V\int\limits_{E_{\min}}^{E_{\max}}(P_+-P_-)\,\mathrm dE\approx \\&\approx V\int\limits_{-300~kV/cm}^{300~kV/cm}(P_+-P_-)\,\mathrm dE=\\&=d \cdot S\cdot S_{\text{graph.}}=10^{-3}~cm^3\cdot S_{\text{graph.}}, \end{aligned}\]where $S_{\text{graph.}}$ is the area under the graph. It can be found by counting cells:
\[S_{\text{graph.}}\approx6.40\cdot10^3~\frac{kV}{cm}\cdot\frac{\text{µC}}{cm^2}=6.40~\frac{J}{cm^3}\implies\]

B2  ^?? Calculate the average thermal power $\dot Q$ released in the film.

Effective voltage is the root mean square voltage for sinusoidal oscillations, related to the amplitude of oscillations $V_\mathrm m$ as:
\[V_{\mathrm{eff}}=\frac{V_\mathrm m}{\sqrt2}\implies V_{\mathrm m}=\sqrt2 V_{\mathrm {eff}}\approx 311~V.\]This once again covers the entire hysteresis loop of BTO, so the dissipated power is calculated as:
\[\dot Q=\nu\cdot Q = 0.32~W,\]where $Q$ is the heat found in the previous task.