Need help with triple integral plz

Find the volume of the first octant solid bounded by the graph z=1-y^2, y=2x, and x=3.

So the way I set up the integrand is

Triple integral of dzdydx where
0 0 0
when I evaluate the integral I keep getting a negative volume of -45. Could someone help me out please?

Projecting z = 1 - y^2 with x in [0, 3] onto the xy-plane leaves the region x in [0, 3], y in [0, 1],
since 1 - y^2 = 0 ==> y = 1 (with y > 0).

Now, we intersect this region with y = 2x.
This is most conveniently realized as x = y/2 to x = 3 for y in [0, 1].

So, the volume equals
∫(y = 0 to 1) ∫(x = y/2 to 3) ∫(z = 0 to 1 - y^2) 1 dz dx dy
= ∫(y = 0 to 1) ∫(x = y/2 to 3) (1 - y^2) dx dy
= ∫(y = 0 to 1) (3 - y/2) (1 - y^2) dy
= (1/2) ∫(y = 0 to 1) (6 - y) (1 - y^2) dy
= (1/2) ∫(y = 0 to 1) (y^3 - 6y^2 - y + 6) dy
= (1/2)(1/4 - 2 - 1/2 + 6)
= 15/8.

I hope this helps!

Yes, that result is correct
http://www.wolframalpha.com/input/?i=%E2…
but there is something wrong about the limits:

you see, in the first octant, when z = 0 then y = 1 ;
so 0 <= y <= 1 ;
and 0 <= x <= y/2 ;
then x = 3 does not belong here.

The new integral here :
http://www.wolframalpha.com/input/?i=%E2…