Bài tập 3.33 trang 178 SBT Toán 12

a) Ta có: \(2 - {x^2} = 1 \Leftrightarrow {x^2} = 1\)

Khi đó 

\(\begin{array}{l}
V = \pi \int \limits_{ - 1}^1 \left| {{{\left( {2 - {x^2}} \right)}^2} - 1} \right|dx\\
 = \pi \int \limits_{ - 1}^1 \left| {{x^4} - 4{x^2} + 3} \right|dx\\
 = \pi \left| {\int \limits_{ - 1}^1 \left( {{x^4} - 4{x^2} + 3} \right)dx} \right|\\
 = \pi \left| {\left. {\left( {\frac{{{x^5}}}{5} - \frac{4}{3}{x^3} + 3x} \right)} \right|_{ - 1}^1} \right|\\
 = \pi \left| {\frac{1}{5} - \frac{4}{3} + 3 + \frac{1}{5} - \frac{4}{3} + 3} \right| = \frac{{56\pi }}{{15}}
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
2x - {x^2} = x \Leftrightarrow {x^2} - x = 0\\
 \Leftrightarrow x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x = 1}
\end{array}} \right.
\end{array}\)

Khi đó 

\(\begin{array}{l}
V = \pi \int \limits_0^1 \left| {{{\left( {2x - {x^2}} \right)}^2} - {x^2}} \right|dx\\
 = \pi \int \limits_0^1 \left| {4{x^2} - 4{x^3} + {x^4} - {x^2}} \right|dx\\
 = \pi \left| {\int \limits_0^1 \left( {{x^4} - 4{x^3} + 3{x^2}} \right)dx} \right|\\
 = \pi \left| {\left. {\left( {\frac{{{x^5}}}{5} - {x^4} + {x^3}} \right)} \right|_0^1} \right|\\
 = \pi \left| {\frac{1}{5} - 1 + 1} \right| = \frac{\pi }{5}
\end{array}\)