I would like to get the area formed by two equations and the volume of the solid formed by this area around the Y axis.
Solve[x^2==6x-2x^2,x]
{{x->0},{x->2}}
Plot[{x^2,6x-2x^2},{x,0,2}]
I tried to use this idea but did not succeed;
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Sign up to join this communityI would like to get the area formed by two equations and the volume of the solid formed by this area around the Y axis.
Solve[x^2==6x-2x^2,x]
{{x->0},{x->2}}
Plot[{x^2,6x-2x^2},{x,0,2}]
I tried to use this idea but did not succeed;
I propose another way. You can use DiscretizeRegion
to create a region from the ParametricRegion
. Then, calculate the volume:
\[ScriptCapitalR]1 =
DiscretizeRegion[
ParametricRegion[{{x, y, z},6 Sqrt[x^2 + y^2] - 2 (x^2 + y^2) > z && x^2 + y^2 <= 4 &&x^2 + y^2 <= z && z >= 0}, {{x, -4, 4}, {y, -4, 4}, {z, 0, 5}}],
MaxCellMeasure -> #, ImageSize -> Small] & /@ {Automatic,0.1, 0.01,0.0005};
vols1 = Volume[\[ScriptCapitalR]1[[#]]] & /@ Range[4];
Row@(Labeled[\[ScriptCapitalR]1[[#]],Style["Volume = " <> ToString[vols1[[#]]], 12, FontFamily -> "Times New Roman"], Top]&/@ Range[4])
or use BoundaryDiscretizeRegion
:
\[ScriptCapitalR]2 =
BoundaryDiscretizeRegion[
ParametricRegion[{{x, y, z},6 Sqrt[x^2 + y^2] - 2 (x^2 + y^2) > z && x^2 + y^2 <= 4 &&x^2 + y^2 <= z && z >= 0}, {{x, -4, 4}, {y, -4, 4}, {z, 0, 5}}],
MaxCellMeasure -> #, ImageSize -> Small] & /@ {Automatic,0.1, 0.01,0.0005};
vols2 = Volume[\[ScriptCapitalR]2[[#]]] & /@ Range[4];
Row@(Labeled[\[ScriptCapitalR]2[[#]],Style["Volume = " <> ToString[vols2[[#]]], 12, FontFamily -> "Times New Roman"], Top]&/@ Range[4])
In both cases, the value of the calculated volume, and obviously the appearance, depends on the value of the option MaxCellMeasure
, as can be deduced from the plots.
It seems that BoundaryDiscretizeRegion
has a greater accuracy.
The volume can be calculated directly from ParametricRegion
, which is more exact, and compared to previous values:
NestList[N,Volume[ParametricRegion[{{x, y, z},
6 Sqrt[x^2 + y^2] - 2 (x^2 + y^2) > z && x^2 + y^2 <= 4 && x^2 + y^2 <= z && z >= 0},
{{x, -4, 4}, {y, -4, 4}, {z, 0, 5}}]],1]
(*{8 \[Pi], 25.1327}*)
The area is simply given by
Integrate[ 6x - 2x^2 - x^2, {x, 0, 2}]
(* 4 *)
Using the cylinder method, the volume is given by
2 Pi*Integrate[x*Abs[6 x - 2 x^2 - x^2], {x, 0, 2}]
(* 8 Pi *)
You can also get the area with Area[ImplicitRegion[x^2 <= y <= (6 x - 2 x^2), {x, y}]]
. For the volume, I would start by plotting the region with RevolutionPlot3D[6 x - 2 x^2 - x^2, {x, 0, 2}]
. There a many posts about this on the website.