Phương trình hoành độ giao điểm của hai đường:$\\$ $x^2=-\dfrac{1}{3}x+\dfrac{4}{3}\\\Leftrightarrow3x^2+x-4=0\\\Leftrightarrow\left[\begin{aligned}x=1\\x=-\dfrac{4}{3}\ (l)\end{aligned}\right.$ Thể tích khối tròn xoay:$\\$$V=\pi\displaystyle\int_{0}^{1}(x^2)^2dx+\pi\displaystyle\int_{1}^{4}\left(-\dfrac{1}{3}x+\dfrac{4}{3}\right)^2dx\\\Rightarrow V=\pi\displaystyle\int_{0}^{1}x^4dx+\pi\displaystyle\int_{1}^{4}\dfrac{1}{9}(4-x)^2dx\\\Rightarrow V=\pi\left(\dfrac{x^5}{5}\right)\Bigg|_{0}^{1}-\pi\left[\dfrac{(4-x)^3}{27}\right]\Bigg|_{1}^{4}$ $\Rightarrow V=\dfrac{\pi}{5}-\pi\left(0-\dfrac{27}{27}\right)\\\Rightarrow V=\dfrac{\pi}{5}+\pi\\\Rightarrow V=\dfrac{6\pi}{5}$$\\$ $\Rightarrow\rm A$