How do I solve for this 2 π ∫ from 1 to 3 (y (1-(y-2^2)dy) the answer is 16π/3

Question

This is what I have try but couldn't get to the answer
2 π ∫ (y (1-((y-2)(y-2))dy )
2 π ∫ (y (1-(y^2-4y+4))dy )
2 π ∫ (y (1-y^2+4y-4)dy )
2 π ∫ (y (-y^2+4y-3)dy )
2 π ∫ (-y^3+4y^2-3y)dy )
2 π [ (-y^4)/4+(4y^3)/3-(3y^2)/2]from 1 to 3
2 π [ ((-3^4)/4+(4*3^3)/3-(3*3^2)/2)-(-1^4)/4+(4*1^3)/3-(3*1^2)/2)]
2 π [ 81/4+108/3-27/2+1/4-4/3+3/2]
2 π [ 82/4+104/3-24/2]
2 π [ 246/12+416/12-144/12]
2 π [ 518/12]
1036 π/12 or 259π/3
Foes anybody know where are my mistakes that cause my answer to not match the correct answer?

Answer 1

-3^4 = -(3^4) = -81
So, between that and the missing parenthesis, you should have had
2π[ ((-3^4)/4+(4*3^3)/3-(3*3^2)/2)-((-1^4)/4+(4*1^3)/3-(3*1^2)/2)]
2π(-81/4 + 108/3 - 27/2 + 1/4 - 4/3 + 3/2)
16π/3