       Re: 3D graph with assumptions

• To: mathgroup at smc.vnet.net
• Subject: [mg50889] Re: [mg50814] 3D graph with assumptions
• From: Garry Helzer <gah at math.umd.edu>
• Date: Sun, 26 Sep 2004 05:31:52 -0400 (EDT)
• References: <200409220411.AAA18719@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Mukhtar,

Some years ago I wrote a package to address this question using
Mathematica version 2. A description finally appeared in The
Mathematica Journal volume 8, number 4 (Title: Clipping Polygons). A
syntax change in version 5 broke the package but it can be repaired by
replacing  FindRoot[f[p+t(q-p)]==0,{t,{0,1}}] with
FindRoot[f[p+t(q-p)]==0,{t,0,1}] throughout (one less pair of curly
braces). The package handles general situations. Here is a rough and
ready approach for your case.

The idea is to chop your set of xy values into small polygons
(triangles or quadrilaterals), map the polygons to space, and display
them.

Mapping the polygons is easy. Suppose that your function is f[x,y] and
poly2D is a list of polygons in the plane. Define (note the braces)
F[{x_,y_}]:={x,y,f[x,y]}. The list of transformed polygons is displayed
by

poly3D=poly2D/.Polygon[a_]:>Polygon[F/@a]
Show[Graphics3D[poly3D], . . . your choice of options here. . .]

Note that this code is indifferent to the nesting levels at which the
polygons appear in the lists.

The annoying part is producing the list poly2D. Since you are using the
area under the curve h[x], start with, say, the left endpoint Riemann
sum.

d=(b-a)/n
poly2D=Table[Polygon[{{a+k d,0},{a+(k+1)d,0},{a+(k+1)d,h[a+k
d},{a+(k+1)d,h[a+k d}}],{k,0,n}]

These rectangles need to be chopped up into shorter rectangles. A
function that will chop each rectangle in a list into two rectangles
is(we will need the segment ab horizontal)

halve[L_]:=L/.Polygon[{a_,b_,c_,d_}]:>{{Polygon[{a,b,(b+c)/2,(a+d)/
2}],Polygon[{(a+d)/2,(b+c)/2,c,d}]}

To split each Riemann rectangle horizontally into 2^m rectangles use

poly2D=Nest[halve,poly2D,m]

The rectangles in poly2D can be made as small as you like by choosing m
and n large enough.

I have come to believe that, except in the simplest cases, it is better
to ignore Mathematica's built in Plot functions and go directly to the
primitives.

On Sep 22, 2004, at 12:11 AM, Mukhtar wrote:

> Suppose I have f(x,y) that I need to Plot in 3D, however, the range of
> x is (0,1) while the range of y is (0,g(x)), where g(x) is some
> explicit function of x, say x^1/2.  Is there a way to do it since
> specifying it directly the way I outlined above in Plot3D gives me the
> error message that "Plot3D :: plln: Limiting value g(x) in {y,0,g(x)}
> is not a machine-size real number". Perhaps I can do this indirectly
> by plotting two surfaces and then consider the relevant part of
> surface f(x,y), however, it would look ugly. Thanks.
>
>
Garry Helzer
gah at math.umd.edu

```

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