2
$\begingroup$

I've come this far as to combine both functions in a graph:

f = Plot[4 - x^2, {x, -10, 10}]
g = Plot[-1 + 4 x, {x, -10, 10}]
Show[g, f, PlotRange -> All]

enter image description here

Solved the intersections:

sol = x /. NSolve[f == g, x, Reals]

{-5., 1.}

Next thing I would like to do is integrate and rotate the area between the intersects enclosed by the functions around the X axis and calculate the volume of the resulting rotational body. I'm fairly new to Mathematica and require some help to finish the plot.

EDIT: I have painted the area, so it is clear which area is meant to be revolved: enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ I'm not sure what you mean by "rotate the region". Do you want to create a solid of revolution using the region between the two curves? If yes, which axis are you rotating around? $\endgroup$
    – Szabolcs
    Sep 9, 2017 at 19:04
  • 3
    $\begingroup$ 1) I think you want 4 - x^2 rather than 4 - x. 2) What line in 3D space to want as the axis of revolution? $\endgroup$
    – m_goldberg
    Sep 9, 2017 at 19:13
  • $\begingroup$ @Szabolcs around the X axis. I mean the Area, the 2 functions are enclosing between the intersections. Do you have an idea how to plot this? $\endgroup$
    – BlkPengu
    Sep 9, 2017 at 22:03

2 Answers 2

5
$\begingroup$

This is not an answer, but an explanation of why the question, as currently posed, is not clear.

If the two curves are rotated about the x-axis, they do not enclose a simple closed region -- they produce as region that self-intersects and for which it is difficult to define volume. Here are two views of a half-revolution plot that show the difficulty of determining the volume.

plots

However, if the two curves are translated upward by 21, then revolving them about the x-axis produces a simple closed region for which has a volume that can be computed with reasonable effort.

simple_plot

Is volume enclosed by the translated curves the one you want?

Update

Code for producing the half-revolution plots

f[x_] := 4 - x^2
g[x_] := -1 + 4 x
fSurface = 
  RevolutionPlot3D[f[x], {x, -5, 1}, {u, 0, π}, 
    ColorFunction -> (White &), RevolutionAxis -> "X"];
gSurface = 
  RevolutionPlot3D[g[x], {x, -5, 1}, {u, 0, π}, 
    ColorFunction -> (White &), RevolutionAxis -> "X"];
Show[
  fSurface, gSurface,
  BoxRatios -> {1, 1, 1}, PlotRange -> All, Lighting -> "Neutral"]

This may be the answer you are looking for. I will solve the problem by translating the curves.

The functions after translating the curves upward by 21.

f[x_] := 25 - x^2
g[x_] := 20 + 4 x
Plot[{f[x], g[x]}, {x, -5, 1}]

2Dplot

The surfaces of revolution

fSurface = 
  RevolutionPlot3D[f[x], {x, -5, 1}, {u, 0, 2 π}, 
    ColorFunction -> (White &), RevolutionAxis -> "X"];
gSurface = 
  RevolutionPlot3D[g[x], {x, -5, 1}, {u, 0, 2 π}, 
    ColorFunction -> (White &), RevolutionAxis -> "X"];
Show[
  fSurface, gSurface, 
  BoxRatios -> {1, 1, 1}, PlotRange -> All, Lighting -> "Neutral"]

3Dplot

The inner surface of revolution and the volume it encloses

inner = 
  Volume @ 
    ImplicitRegion[y^2 + z^2 <= g[x]^2, {{x, -5, 1}, {y, -24, 24}, {z, -24, 24}}]

1152 π

The outer surface of revolution and the volume it encloses

outer = 
  Volume @ 
    ImplicitRegion[y^2 + z^2 <= f[x]^2, {{x, -5, 1}, {y, -25, 25}, {z, -25, 25}}]

(11376 π)/5

The volume enclosed between the two surfaces

outer - inner

(5616 π)/5

Getting the volume by integration (which for this problem is much faster)

Integrate[π (f[x]^2 - g[x]^2), {x, -5, 1}]

(5616 π)/5

$\endgroup$
3
  • $\begingroup$ I edited the post to depict the exact area I want to revolve around the X axis. I'm not sure which one of your plots that is, but I get what the problem is. Those look almost like Klein Bottles in some way. However the last plot looks promising. $\endgroup$
    – BlkPengu
    Sep 10, 2017 at 11:19
  • $\begingroup$ Could you add the syntax used to plot the objects pictured above? I could learn a great deal from it. $\endgroup$
    – BlkPengu
    Sep 10, 2017 at 11:23
  • $\begingroup$ @BlkPengu. I have updated my post to provide the code you asked for and some more that I think will answer your question. $\endgroup$
    – m_goldberg
    Sep 10, 2017 at 22:18
0
$\begingroup$

This is an approach (note volume of RegionDifference does not seem to work):

f[x_] := 4 - x^2
g[x_] := 4 x - 1
region = ImplicitRegion[
   0 < y^2 + z^2 <= Max[f[x]^2, g[x]^2], {{x, -5, 1}, y, z}];
mr = ImplicitRegion[
   0 < y^2 + z^2 <= Min[f[x]^2, g[x]^2], {{x, -5, 1}, y, z}];
ro = RegionDifference[region, mr];
rplot = RegionPlot3D[ro, PlotPoints -> 100, PlotStyle -> Opacity[0.6],
    Background -> Black];
rot[a_] := RotationMatrix[a, {1, 0, 0}]
ppl[a_] := 
 ParametricPlot3D[{rot[a].{x, 0, f[x]}, rot[a].{x, 0, g[x]}}, {x, -5, 
   1}, PlotStyle -> Directive[Red, Thick]]
Manipulate[Show[rplot, ppl[a]], {a, 0, 2 Pi}]

enter image description here

Volume:

Through[{Simplify, N}[Volume[region] - Volume[mr]]]

yields:

{-(8/5) (-394 + 49 Sqrt[7]) π, 1328.81}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.