Volume benda putar yang terbentuk apabila daerah yang dibatasi oleh garis [tex] {x= \sqrt{2y} + 1 } [/tex], garis [tex] {y=2} [/tex], garis [tex] {x=0} [/tex],

batas :
1) x = √(2y) + 1
=> x - 1 = √(2y)
=> 2y = (x - 1)^2
=> y = 1/2 (x - 1)^2
2) y = 2
3) x = 0 => sumbu y
4) y = 0 => sumbu x

Perpotongan y = 2 dengan y = 1/2 (x - 1)^2
1/2 (x - 1)^2 = 2
(x - 1)^2 = 4
(x - 1) = ± 2
x = 3 atau x = -1
diputar mengelilingi y = 3
agar mempermudah kita geser y = 3 ke y = 0 sehingga batasnya menjadi
1) y = 1/2 (x - 1)^2 - 3
2) y = 2 - 3 => y = -1
3) x = 0
4) y = 0 - 3 => y = -3

Volume = π (batas 0 - 3) Int (1/2 (x - 1)^2 - 3)^2 - (-1)^2) dx
= π (batas 0 - 3) int (1/4 (x - 1)^4 - 3(x - 1)^2 + 9 - 1) dx
= π (batas 0 - 3) int (1/4 (x - 1)^4 - 3(x - 1)^2 + 8) dx
= π [1/20 (x - 1)^5 - 3/3 (x - 1)^3 + 8x] batas 0 - 3
= π [(1/20 (3 - 1)^5 - (3 - 1)^3 + 8(3)) - (1/20 (0 - 1)^5 - (0 - 1)^3 + 8(0))]
= π [32/20 - 8 + 24 - (-1/20 + 1 + 0)]
= [33/20 + 15] π
= (333π)/20