Ta có: \(\displaystyle \sin x = \frac{{2x}}{\pi } \Rightarrow \left[ \begin{array}{l}x = 0\\x = \frac{\pi }{2}\\x = - \frac{\pi }{2}\end{array} \right.\)
Khi đó \(\displaystyle V = \pi \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|dx} \)
Dễ thấy \(\displaystyle f\left( x \right) = \left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|\) là hàm số chẵn nên:
\(\displaystyle V = 2\pi \int\limits_0^{\frac{\pi }{2}} {\left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|dx} \)\(\displaystyle = 2\pi \int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right)dx} \) \(\displaystyle = 2\pi \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} - \frac{8}{\pi }\int\limits_0^{\frac{\pi }{2}} {{x^2}dx} \)
\(\displaystyle = \pi \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 2x} \right)dx} - \frac{8}{\pi }\int\limits_0^{\frac{\pi }{2}} {{x^2}dx} \) \(\displaystyle = \pi \left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^{\frac{\pi }{2}} - \frac{8}{\pi }.\left. {\frac{{{x^3}}}{3}} \right|_0^{\frac{\pi }{2}}\) \(\displaystyle = \pi \left( {\frac{\pi }{2} - 0} \right) - \frac{8}{\pi }.\frac{1}{3}.{\left( {\frac{\pi }{2}} \right)^3}\)
\(\displaystyle = \frac{{{\pi ^2}}}{2} - \frac{{{\pi ^2}}}{3} = \frac{{{\pi ^2}}}{6}\)
Chọn D.
