How can I plot a 3-variables function to identify the neutral surface and a region

I wanted to plot a 3-D solid region, in which my function f(x,y,z)=

x (1/(1 + 0.06))^3 + y (1/(1 + 0.06))^2 + (2 z^2)/15 (Sin[10 Degree])^2 -z/3 Cos[10 Degree]

is non-positive,where x, y and z are 3 variables. As can be seen in the following fig.

I think I should first find out the neutral surface, on which f(x,y,z)=0, it is denoted as the red surface in the fig. I have tried Plot3D and ParametricPlot3D:

Plot3D[x (1/(1 + 0.06))^3 + y (1/(1 + 0.06))^2 + (2 z^2)/15 (Sin[10Degree])^2 – z/3 Cos[10 Degree], {x, 0, 8}, {y, 0, 8}, {z, 0, 82}]

and

ParametricPlot3D[x (1/(1 + 0.06))^3 + y (1/(1 + 0.06))^2 + (2 z^2)/15 (Sin[10 Degree])^2 – z/3 Cos[10 Degree], {x, 0, 8}, {y, 0, 8}, {z, 0, 82}]

But MMA reports nonopt: Options expected (instead of {z,0,82}) beyond position 3

I do know the plot of 2 variables define the 3D surface. And the plot of 3 variables define 3D space in 4D space. So, how can I to visualize the plot of 3 variables. Is there some way to use ParametricPlot3D, Plot3D or any other Mathematica functions to perform this task?

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2 Answers
2

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sf[x_, y_, z_] := x (1/(1 + 0.06))^3 + y (1/(1 + 0.06))^2 +
(2 z^2)/15 (Sin[10 Degree])^2 – z/3 Cos[10 Degree];

ContourPlot3D[sf[x, y, z] == 0, {x, 0, 8}, {y, 0, 8}, {z, 0, 82}]

sz[x_, y_] := z /. Solve[sf[x, y, z] == 0., {z}];

Plot3D[Evaluate@Quiet@sz[x, y], {x, 0, 8}, {y, 0, 8}, BoundaryStyle -> None,
Mesh -> None, BoxRatios -> 1]

To plot a “solid region” as you request, you can use RegionPlot3D

f[x_, y_, z_] :=
x (1/(1 + 0.06))^3 +
y (1/(1 + 0.06))^2 + (2 z^2)/15 (Sin[10 Degree])^2 –
z/3 Cos[10 Degree]

RegionPlot3D[
f[x, y, z] < 0, {x, -2000, 2000}, {y, -2000, 2000}, {z, -2000, 2000}]      Thanks, @kguler! I have learned the ContourPlot3D from you. – can Nov 2 '14 at 13:04      Thank you,@rhermans! I have learned the use of RegionPlot3D from your answer! Although I have accepted the reply offered by @kguler, I do think your answer is very useful as well! But stack only allows to accept one answer, sorry about that! – can Nov 2 '14 at 13:06