Finding boundary to a triple integral

Questions
1

Hi,
What would be the limits used in a triple integral that is bounded by the function 4x^2 + y^2 +z = 128, x=0, x=4, y=0,y=4?
Limits for x and y would be 0 to 4. Would the limits for z be 48 to (128-4x^2-y^2) or maybe 48 to 128?
Please let me know what limits to use, and the method of going about it.
Thank you for the help.

2

The way you state the problem, there are two regions that are bound by those surfaces, one above and one below the parabolic cylinder 4x^2 + y^2 +z = 128, as you can see from the picture:

ContourPlot3D[{4 x^2 + y^2 + z == 128, x == 0, x == 4, y == 0, y == 4},
 {x, -1, 5}, {y, -1, 5}, {z, 50, 150},
 ContourStyle -> Opacity[0.8], AxesLabel -> Automatic]
3

Perhaps like this?

Integrate[x y z, {y, 0, 4}, {x, 0, 4}, {z, 0, 128 - y^2 - 4 x^2}]