Integrals and volume (calculus)

Having trouble figuring these two out:

1. Consider the solid obtained by rotating the region bounded by the given curves about the y-axis.
y^2=2x, x=y
Find the volume of this solid

2. Consider the solid obtained by rotating the region bounded be the given curves about the line y=-1.
y=(2/x), y=0, x=1, x=3
Find the volume of this solid

I can integrate once I get it set up, Im just having trouble getting them set up.

Thanks in advance!

1.- The curves cut when y^2 = 2y , ie, at y=0 and y=2 , since y=x , they cut at x=0 and x= 2
Shell method , A= 2pi x H , where H (x) = sqrt2x - x .- ( H is Y parabola - Y line at x)

dV = Adx
dV= 2pi x( sqrt 2x-x) dx
V= 2pi INT x( sqrt2x -x) dx
0
V= 2pi INT sqrt2 x^( 3/2) -x^2) dx
V= 2pi ( sqrt2 ( 2/5) x^(5/2) -x^3/3)
V= 2pi ( sqrt2 (2/5) (4sqrt2 -(8/3)
V= 2pi( 16/5 -8/3)
V= 2pi ( 8/15)
V=16pi /15

2.- Disk washers , Radius of revolution R= (2/x) -(-1) = (2/x)+1
V= INT piR^2 dx
V= pi INT (2/x)+1)^2 dx
1 V= pi ( INT (4/x^2 +4/x +1) dx
V= pi ( ( -4)(1/x) + 4 lnx +x)
V= pi (( -4/3+ 4ln3+3) - (-4+4ln1+1))
V=pi( 5/3+4ln3 +3-0)
V=pi ( 4Ln3 +14/3)