Chọn phương án D.

\(\begin{aligned}
V&=V_1+V_2\\
&=\pi\displaystyle\int\limits_{0}^{1}\left(\sqrt{x}\right)^2\mathrm{\,d}x+\pi\displaystyle\int\limits_{1}^{2}(2-x)^2\mathrm{\,d}x\\
&=\pi\displaystyle\int\limits_{0}^{1}x\mathrm{\,d}x+\pi\displaystyle\int\limits_{1}^{2}\left(4-4x+x^2\right)\mathrm{\,d}x\\
&=\dfrac{\pi x^2}{2}\bigg|_0^1+\left(4x-2x^2+\dfrac{x^3}{3}\right)\bigg|_1^2\\
&=\dfrac{5\pi}{6}.
\end{aligned}\)