# Volume of a region of revolution

I'm given the region bounded by $x=0$ and $x=4y^2-y^3$.

Show[
ContourPlot[{x == 4 y^2 - y^3, x == 0}, {x, -1, 10}, {y, 0, 4}],
RegionPlot[0 <= x <= 4 y^2 - y^3, {x, 0, 256/27}, {y, 0, 4}],
Axes -> True
]


Show[
RevolutionPlot3D[{4 y^2 - y^3, y, 0}, {y, 0, 4},
RevolutionAxis -> {1, 0, 0}, AxesLabel -> {"x", "y"}],
RevolutionPlot3D[{0, y, 0}, {y, 0, 4}, RevolutionAxis -> {1, 0, 0}],
BaseStyle -> Opacity[0.4]
]


Now, this is similar to a question I posted at Area of surface of revolution. I've found the volume using the cylindrical shell method.

Integrate[2 \[Pi] y (4 y^2 - y^3), {y, 0, 4}]


The answer is $512\pi/5$. I'd like to use Mathematica's Volume command to find the volume, but I've been unsuccessful. Tough problem. Does anyone have a suggestion?

Update: This can help show that it is a rotation about the x-axis.

Manipulate[
ParametricPlot3D[{4 y^2 - y^3, y Cos[t], y Sin[t]}, {y, 0, 4}, {t, 0,
tau}, AxesLabel -> {"x", "y"}, PerformanceGoal -> "Quality",
PlotStyle -> Opacity[0.8], PlotRange -> {{0, 10}, {-4, 4}, {-4, 4}}],
{{tau, 0.1}, 0, 2 \[Pi]}]

• @C.E. Yes it is. The option RevolutionAxis -> {1, 0, 0}] causes the revolution across the x-axis. I've also updated my original post, adding some axes labels and a manipulate activity to show the rotation. Jan 31, 2017 at 18:26
• My bad, I misread the plot. Jan 31, 2017 at 18:31
• reg = ImplicitRegion[ 0 <= Sqrt[y^2 + z^2] <= 4 && 0 <= x <= 4 (y^2 + z^2) - (y^2 + z^2)^(3/2), {x, y, z}] // Region; {reg, Volume@reg} only get the numeric result , I don't know why. Apr 5 at 7:54

ir = ImplicitRegion[

yields: $512\pi/5$