a) Đ, b) Đ, c) S, d) S

a) \(V = \pi \int\limits_0^1 {{{\left( {2 - x} \right)}^2}dx} = \frac{{7\pi }}{3}\).

b) Ta có \({S_1} = \int\limits_0^1 {\sqrt x dx} + \int\limits_1^2 {\left| {2 - x} \right|dx} \)\( = \left. {\frac{2}{3}{x^{\frac{3}{2}}}} \right|_0^1 + \left. {\left( {2x - \frac{{{x^2}}}{2}} \right)} \right|_1^2\)\( = \frac{2}{3} + 4 - 2 - \left( {2 - \frac{1}{2}} \right) = \frac{7}{6}\).

c) Ta có \({S_2} = {S_{\Delta OAB}} - {S_1} = \frac{1}{2}.2.2 - \frac{7}{6} = \frac{5}{6}\). (với \(A\left( {0;2} \right),B\left( {2;0} \right)\))

d) \(V = \pi \int\limits_0^1 {{{\left( {\sqrt x } \right)}^2}dx} + \pi \int\limits_1^2 {{{\left( {2 - x} \right)}^2}dx} \)\( = \left. {\pi \frac{{{x^2}}}{2}} \right|_0^1 - \left. {\pi \frac{{{{\left( {2 - x} \right)}^3}}}{3}} \right|_1^2\)\[ = \frac{\pi }{2} + \frac{\pi }{3} = \frac{{5\pi }}{6}\].