f(x，y)=y？ +2y+φ(x)

∴f(y,y)=y？ +2y+φ(y)

∫f(y，y)=(y+ 1)？ -(2-y)lny

∴φ(y)= 1-(2-y)lny

∴φ(x)= 1-(2-x)lnx

∴f(x,y)=y？ +2y+ 1-(2-x)lnx

Therefore, f(x, y)=0 is

(y+ 1)？ =(2-x)lnx

The two intersections of f(x, y)=0 and straight line y=- 1 are respectively

( 1,- 1),(2,- 1)

Therefore, the volume of the rotating body is

V=∫( 1~2)π(y+ 1)？ Advanced (short for deluxe)

=π∫( 1~2)(2-x)lnx dx

=π∫( 1~2)lnx d(2x-x？ /2)

=π(2x-x？ /2)lnx |( 1~2)

-π∫( 1~2)(2x-x？ 1/x dx

=2π ln2-π∫( 1~2)(2-x/2) dx

=2π ln2-π(2x-x？ /4) |( 1~2)

=2π ln2-5π/4