Take out the constant `2Pi` , and rewrite the integral.

`int_0^1(2 pi (3-y)(1-y^2)dy`

= 2`pi` `int_0^1(3-y)(1-y^2) dy`

`=2pi int_0^1 (3(1-y^2) -y(1-y^2)dy= 2pi int_0^1(3-3y^2-y+y^3)dy`

`=2pi(3y-3y^3/3-y^2/2+y^4/4)|(0,1)`

`2pi(3y-y^3-y^2/2+y^4/4)|(0,1)`

`=2pi(3-1-1/2+1/4 - 0)`

`=2pi(2-1/2+1/4)`

`=2pi((2*4 -2+1)) / 4`

`=2pi(8-1)/4 = 2pi(7/4) = 7pi/2`

Thus the volume of the required solid is 7`pi/2`

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