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