integral by hand, to find surface area

[studentMCCS]

The problem is,

SA = 4pi*integral((4-y)^(1/2)*(1-(1/(4-y)))^(1/2), y)

I tried a few things, but haven't been able to figure out how to do it by hand.

bringing everything into a single square root,

= 4pi*integral(y^2-7y+12,y)

Then I brought in the 4,

= pi*integral(16y^2-112y+192,y)

which is close, if the it were 196 instead of 192 it would be a perfect square.

or I could have just brought in a 4

= 2pi*integral(4y^2-28y+48,y)

= 2pi*integral((2-7)^2-1,y)

Can anyone give me a hint.

 

[studentMCCS]

Very sorry, I just realized my mistake.

Should have been,

SA = 4pi*integral((4-y)^(1/2)*(1+(1/(4-y)))^(1/2), y)

which I think boils down to

4pi*integral((5-y)^(1/2),y)

then use u-sub...