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calculus and analysis - Finding the volume enclosed by two surfaces of revolution


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




Answer



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



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