1

I'm trying to fill the region under the surface z = x^2 + y^2, over the rectangle 0<=x<=1, 0<=y<=1. How may I do so? Also, how do I orient the 3-d plot to as to have the x-axis coming towards me, the y-axis going to the right, and the z-axis going straight up. Here is what I have:

\documentclass[10pt]{article}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
    \begin{axis}[xlabel=$x$,ylabel=$y$,zlabel=$z$]
        \addplot3
            [surf,faceted color=blue,
             samples=20,
             domain=-2:2,y domain=-2:2]
            {x^2 + y^2};
    \end{axis}
\end{tikzpicture}
\end{document}

1 Answer 1

1

This sets the view which you seem to be asking for and fills the area as indicated (in this view) but is not really spectacular IMHO.

\documentclass[10pt]{article}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\begin{document}
\begin{tikzpicture}[declare function={f(\x,\y)=\x*\x+\y*\y;}]
    \begin{axis}[xlabel=$x$,ylabel=$y$,zlabel=$z$,view={00}{00}]
        \addplot3
            [surf,faceted color=blue,
             samples=21,
             domain=-2:2,y domain=-2:2]
            {f(x,y)};
    \path[fill=blue]    (1,1,0) -- plot[samples=5,domain=1:0,variable=\x]
    (\x,1,{f(\x,1)}) --(0,1,0)--cycle;
    \end{axis}
\end{tikzpicture}
\end{document}

enter image description here

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .