曲線：y=x^2-2x を45°時計回りに回転させると、xを(x-y)/√2、yを(x+y)/√2 と置き換えて、
(x+y)/√2=(x-y)^2/2-2(x-y)/√2、2(x+y)=√2(x-y)^2-4(x-y)、√2x^2-2√2xy+√2y^2-6x+2y=0
√2y^2-2(√2x-1)y+√2x^2-6x=0、y={(√2x-1)-√(1+4√2x)}/√2
V=π∫[x=0〜3√2] y^2 dx = (π/2)∫[x=0〜3√2] {(√2x-1)-√(1+4√2x)}^2 dx
= π∫[x=0〜3√2] x^2+√2x-x√(2+8√2x)+√(1+4√2x) dx、