Chọn phương án C.

Phương trình hoành độ giao điểm:

• \(x^2=\dfrac{x^2}{8}\Leftrightarrow\dfrac{7}{8}x^2=0\Leftrightarrow x=0\).
• \(x^2=\dfrac{27}{x}\Leftrightarrow\begin{cases}x\neq0\\ x^3=27\end{cases}\Leftrightarrow x=3\).
• \(\dfrac{x^2}{8}=\dfrac{27}{x}\Leftrightarrow\begin{cases}x\neq0\\ x^3=216\end{cases}\Leftrightarrow x=6\).

Ta có: $$\begin{eqnarray*}
S&=&S_1+S_2\\
&=&\displaystyle\int\limits_0^3\left|x^2-\dfrac{x^2}{8}\right|\mathrm{\,d}x+\displaystyle\int\limits_3^6\left|\dfrac{27}{x}-\dfrac{x^2}{8}\right|\mathrm{\,d}x\\
&=&\displaystyle\int\limits_0^3\left(x^2-\dfrac{x^2}{8}\right)\mathrm{\,d}x+\displaystyle\int\limits_3^6\left(\dfrac{27}{x}-\dfrac{x^2}{8}\right)\mathrm{\,d}x\\
&=&\dfrac{7x^3}{24}\bigg|_0^3+\left(27\ln|x|-\dfrac{x^3}{24}\right)\bigg|_3^6\\
&=&\dfrac{63}{8}+\left(27\ln2-\dfrac{63}{8}\right)\\
&=&27\ln2.\end{eqnarray*}$$