calculus

can somebody show I how to get to the answer 256π/15.
This is how I solve for it basing on what the last tutor show I but got the wrong answer, so I don't know what I did wrong. The question is 2π∫ from 0 to 8 [y(√(y/2)-(y/4)]dy.

These are how I do but got the wrong answer
2π∫ from 0 to 8 [y(y^(1/2)/√2)-(y/4))]dy
2π∫ from 0 to 8 [(y^(3/2)/√2)-(y^2)/4]dy
2π [(2y^(5/2))/5√2)-(y^3)/12]from 0 to 8
2π [((2*8^(5/2))/5√2)-(8^3)/12)-((2*0^(5/2))/5√2)-(0^3)/12]
2π [((2*81.02)/5√2)-512/12)-(0-0)]
2π [(162.04/5√2)-(512/12)]
2π [(12*162.04)/(12*(5√2))-((5√2)*512)/((5√2)*12)]
2π [(1944.48/60√2-(2560√2)/60√2]
2π [-615.52/60√2]
1231.04π/60√2

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1. (2*8^(5/2))/(5√2) = 256/5
fix that and the rest follows
You're carrying a lot of extra √2 stuff around

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2. can you show the steps from (2*8^(5/2))/(5√2) to = 256/5 steve?

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3. Oh come on.
2π [(2*8^(5/2)/(5√2)-(8^3)/12)-((2*0^(5/2))/5√2)-(0^3)/12]
2π [256/5 - 512/12]
2π [3072/60 - 2560/60]
2π [512/60]
2π [128/15]
256π/15

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4. (2*8^(5/2))/(5√2)
(2*2^(15/2))/(5√2)
(2*2^7*√2)/(5√2)
(2*128√2)/(5√2)
256/5

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