# How to create 3d region of f[x,y] over triangular region on xy plane?

I need to create a VOLUME under f[x,y]= x (y^3 + 1)^(1/2) that's above the triangular region bounded by y=x/3, y=2, and x=0. I know how to make the plot of the function over an area, such as Plot3D[f[x, y], {x, 0, 6}, {y, 0, 2}]. However, I do not know how to make it exclusively over the triangle.

f[x_, y_] = x (y^3 + 1)^(1/2);

Plot3D[f[x, y], {x, 0, 6}, {y, x/3, 2}, Filling -> 0]


Plot3D[f[x, y], {x, 0, 6}, {y, 0, 2},
RegionFunction -> Function[{x, y}, x/3 <= y <= 2 && x >= 0],
Filling -> 0]


RegionPlot3D[0 <= z <= f[x, y] && x/3 <= y <= 2 && 0 <= x <= 6,
{x, 0, 6}, {y, 0, 2}, {z, 0, 18}, PlotPoints -> 51]


Maximize[{f[x, y], x >= 0, x/3 <= y <= 2}, {x, y}]


{18, {x -> 6, y -> 2}}

Several methods. one of them is:

Plot3D[x (y^3 + 1)^(1/2), {x, 0, 6}, {y, 0, 2},
RegionFunction -> Function[{x, y}, y <= 2 && y >= 3 x && x >= 0]]

• Great catch. thanks :) Nov 24, 2014 at 4:01
• You're welcome! (+1 already.) Nov 24, 2014 at 4:01

In V10+:

Plot3D[x (y^3 + 1)^(1/2),
{x, y} ∈ Polygon[{{0, 0}, {6, 2}, {0, 2}}],
AxesLabel -> Automatic, Filling -> 0]


Also

DiscretizeRegion[
ImplicitRegion[0 <= z <= x (y^3 + 1)^(1/2), {{x, 0, 6}, {y, 0, 2}, {z, 0, 18}}]
]

• I tried this and realized I wrote the problem incorrectly. I don't need the slice of f[x,y] over the triangle. Instead I need the volume between the triangle on the xy plane and f[x,y]. It should be a solid with a triangular base. Thank you for your quick response, and I apologize for my poor wording. This is my first time using the program.
– Nick
Nov 24, 2014 at 4:06

NIntegrate x (y^3 + 1)^(1/2) over Polygon[{{0, 0}, {6, 2}, {0, 2}}]:

NIntegrate [x (y^3 + 1)^(1/2), {x, y} \[Element] Polygon[{{0, 0}, {6, 2}, {0, 2}}]]