Thể tích cần tìm: $V=\pi \int\limits_{0}^{1}{\frac{dx}{{{\left( 1+\sqrt{4-3x} \right)}^{2}}}}$

Đặt $t=\sqrt{4-3x}\Rightarrow dt=-\frac{3}{2\sqrt{4-3x}}dx\Leftrightarrow dx=-\frac{2}{3}tdt\left( x=0\Rightarrow t=2;x=1\Rightarrow t=1 \right)$

Khi đó: $V=\frac{2\pi }{3}\int\limits_{1}^{2}{\frac{t}{{{\left( 1+t \right)}^{2}}}dt}=\frac{2\pi }{3}\int\limits_{1}^{2}{\left( \frac{1}{1+t}-\frac{1}{{{\left( 1+t \right)}^{2}}} \right)dt}=\left. \frac{2\pi }{3}\left( \ln \left| 1+t \right|+\frac{1}{1+t} \right) \right|_{1}^{2}=\frac{\pi }{9}\left( 6\ln \frac{3}{2}-1 \right)$