Ta có: $S(x)=x\sqrt{3-x}$$\\$ $V_\Phi=\displaystyle\int_{0}^{3}S(x)dx=\displaystyle\int_{0}^{3}x\sqrt{3-x}dx$$\\$ Đặt $t=\sqrt{3-x}$$\\$ $\Rightarrow t^2=3-x\\\Rightarrow x=3-t^2$$\\$ Ta có:$\\$ $\dfrac{d}{dt}(x)=\dfrac{d}{dt}(3-t^2)\\\Rightarrow dx=(3-t^2)'dt\\\Rightarrow dx=(0-2t)dt\\\Rightarrow dx=-2tdt$$\\$ Đổi cận:$\\$ $\begin{array}{\|c\|c\|c\|} \hline x & 0 & 3 \\ \hline t & \sqrt{3} & 0 \\ \hline \end{array}$$\\$ Khi đó: $\displaystyle\int_{0}^{3}x\sqrt{3-x}dx\\=\displaystyle\int_{\sqrt{3}}^{0}(3-t^2).t.(-2tdt)\\=\displaystyle\int_{0}^{\sqrt{3}}(3-t^2).2t^2dt\\=\displaystyle\int_{0}^{\sqrt{3}}(6t^2-2t^4)dt\\=\left(\dfrac{6t^3}{3}-\dfrac{2t^5}{5}\right)\Bigg|_0^{\sqrt{3}}\\=\left(2t^3-\dfrac{2}{5}t^5\right)\Bigg|_0^{\sqrt{3}}\\=\left[2(\sqrt{3})^3-\dfrac{2}{5}(\sqrt{3})^5\right]-0\\=6\sqrt{3}-\dfrac{18\sqrt{3}}{5}\\=\dfrac{12\sqrt{3}}{5}$$\\$ $\Rightarrow\rm C$