Chọn phương án C.

Từ phương trình của \((E)\) ta suy ra $$y=\pm b\sqrt{1-\dfrac{x^2}{a}}.$$
Khi đó, \(S=2\displaystyle\int\limits_{-a}^{a}b\sqrt{1-\dfrac{x^2}{a}}\mathrm{\,d}x\).

Đặt \(x=a\sin t\) \(\left(-\dfrac{\pi}{2}\leq t\leq\dfrac{\pi}{2}\right)\).

• \(\mathrm{d}x=a\cos t\mathrm{\,d}t\)
• \(x=a\Rightarrow t=\dfrac{\pi}{2}\)
• \(x=-a\Rightarrow t=-\dfrac{\pi}{2}\)

Khi đó: $$\begin{eqnarray*}
S&=&2\displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}b\sqrt{1-\dfrac{a^2\sin^2t}{a^2}}\cdot a\cos t\mathrm{\,d}t\\
&=&2ab\displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\sqrt{1-\sin^2t}\cdot\cos t\mathrm{\,d}t\\
&=&2ab\displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\cos^2t\mathrm{\,d}t\\
&=&2ab\displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\dfrac{1+\cos2t}{2}\mathrm{\,d}t\\
&=&ab\left(t+\dfrac{\sin2t}{2}\right)\bigg|_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\\
&=&\pi ab.
\end{eqnarray*}$$