[pstricks] Volumes/Solids of Revolution Examples?
Dougherty, Michael
michael.dougherty at swosu.edu
Sun Nov 20 00:48:05 CET 2016
Thank you all for the responses. Those are all useful.
To answer Hubert's question: While I wouldn't mind having "smooth" solids, with gradient effects and all of that, I do like all of the suggestions I've seen so far. I've been so far keeping my graphics simple, and trying to do everything on PSTricks instead of importing pictures generated elsewhere. I should also mention that I've been assuming my book would be in black and white (and gray). I want it to be inexpensive and easily LaTeX-able.
I'm happy to see ANY examples, and appreciate them all.
Herbert, yours has appeal, and I appreciate you pointing out \psVolume, which I now see in your book. For limited applications (rotations around the x-axis) it looks really useful and I probably will use it in an example or two.
Hubert, I like yours very much too. If you want to share some code I'd appreciate it. Trying to get those discs into good perspective has been a challenge. I thought I'd try ellipses but those aren't obviously ellipses.
Randy, I might bug you for some of your code. I especially like the big curly braces, and should investigate those!
It's funny that the volume of a cone came up a couple times already. Without making it a "solid of revolution" problem I already did just that for my book project, and recently (like two days ago):
\documentclass{article}
\usepackage{pst-func,multicol,fouriernc,amsmath}
\begin{document}
\begin{multicols}{2}
\underline{Solution}:
Our illustration is on the right.
We note that at each height $y\in[0,h]$,
the cross sectional area is
$\pi(\text{rad})^2$, where the
radius ``rad'' is a function of $y$. If we give such a circle a ``thickness,''
or ``height'' $dy$, then it forms an infinitesimally thin cylinder
of volume $dV=\pi(\text{rad})^2\,dy$.
What is left to do is to find the radius, labeled ``rad,''
as a function of $y$, and
then set up and compute the resulting integral. From the diagram we
can see that by a similar triangles argument,
$(h-y)/h=(\text{rad})/r$---or equivalently $\text{rad}/(h-y)=r/h$.
>From this we get
$\text{rad}=(h-y)\frac{r}h$, and so
\begin{center}
\begin{pspicture}(-2,-.4)(2,5.5)
\psset{xunit=1.2cm}
\parametricplot[linestyle=dashed]{0}{180}{t cos 2 mul t sin .5 mul}
\parametricplot{180}{360}{t cos 2 mul t sin .5 mul}
\psline(-2,0)(0,5)(2,0)
\psline{*-}(0,0)(2,0)
\rput(1,0.2){$r$}
\psline(0,0)(0,5)
%\parametricplot{0}{180}%
% {t cos 2 mul 3 div 5 2 3 div mul t sin 3 div sub}
%\parametricplot{180}{360}{t cos 2 mul t sin .5 mul}
\parametricplot[linestyle=dashed]{0}{180}%
{t cos 3 mul 5 div 2 mul t sin .5 mul 3 mul 5 div 2 add}
\parametricplot{180}{360}%
{t cos 3 mul 5 div 2 mul t sin .5 mul 3 mul 5 div 2 add}
\parametricplot[linestyle=dashed]{0}{180}%
{t cos 3 mul 5 div 2 mul t sin .5 mul 3 mul 5 div 2.1 add}
\parametricplot{180}{360}%
{t cos 3 mul 5 div 2 mul t sin .5 mul 3 mul 5 div 2.1 add}
\psline[linewidth=1.5pt]{*-}(0,2)(0,5)
\rput[l]{270}(.2,3.5){$h-y$}
\psline[linewidth=1.5pt]{*-}(0,2)(1.2,2)
\rput(.625,2.2){rad}
\psline{<->}(2.2,0)(2.2,5)
\rput[l](2.4,2.5){$h$}
\rput[l](.2,1){$y$}
\psline(-2.3,2)(-1.7,2)
\psline(-2.3,2.1)(-1.7,2.1)
\psline{->}(-2,2.4)(-2,2.1)
\psline{<-}(-2,2.0)(-2,1.7)
\rput(-2,2.6){$dy$}
\psline(0,0)(.2,0)(.2,.2)(0,.2)
\end{pspicture}
\end{center}
\end{multicols}
\begin{align*}V&=\int_0^h
\pi\left[(h-y)\frac{r}h\right]^2\,dy
=\pi\cdot\frac{r^2}{h^2}\int_0^h
(h^2-2hy+y^2)\,dy
=\pi\cdot\frac{r^2}{h^2}\left.\left[h^2y-hy^2+\frac13y^3\right]\right|_0^h\\
&=\pi\cdot\frac{r^2}{h^2}\left[h^3-h^3+\frac13h^3\right]-0
=\pi\cdot\frac{r^2}{h^2}\cdot\frac13h^3=\frac13\pi r^2h.
\end{align*}
\end{document}
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