# Area of volume of revolution

I want to calculate the exact area of this volume:

RevolutionPlot3D[
f[x_] = Sqrt[28 x]
, {x, 0, 21}
, Axes -> True
]


Additionally, I wish to see $x$ and $y$ axis.

Before you vote down my question because I don't know how mathematica works and dare asking, I did try to solve it in the documentation. But that page does not reply to my question.

Area[f[x]]


gives output

Area::reg: 20.105968871208372 is not a correctly specified region. Area[20.106]

• Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that all users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. Jun 27, 2018 at 8:19
• Ok, I understand. Do you wish me to unaccept it? Jun 27, 2018 at 8:23
• It's up to you, you could un-check now and come back tomorrow and accept the best answer then, or you could just take the idea and apply it in the future. Jun 27, 2018 at 8:26
• I take the idea then. Your code works beautifully so seems unfair to uncheck. Jun 27, 2018 at 8:33

It seems that f[x] is evaluating to a number. Probably you Set an OwnValues for x? Check by evaluating ?x.

Area is only defined for regions, parametric surfaces and sets of coordinates, not single numbers nor plots.

Therefore, my the strategy is to first, start with a fresh kernel. Then, create a Cartesian parametric expression for the surface and use Area on that.

To create the parametric expression in Cartesian coordinates the easiest way is to start with the cylindrical coordinantes {r, θ, f[r]} and then use CoordinateTransform

CoordinateTransform[
"Cylindrical" -> "Cartesian"
, {r, θ, Sqrt[28 r]}
]


See also this other answer using ParametricRegion.

Here my take on it.

With[
{
parmcartsn = CoordinateTransform[
"Cylindrical" -> "Cartesian"
, {r, θ, Sqrt[28 r]}
]
},
Column[
{
ParametricPlot3D[
parmcartsn
, {r, 0, 21}
, {θ, 0, 2 π}
],
Area[
parmcartsn
, {r, 0, 21}
, {θ, 0, 2 π}
]
}
]
] • Thank you. I would have never come to that answer. I don't know how to parametrize with cos and sin, how can I do it with an easy f(x) function? Jun 27, 2018 at 8:15
• @Dovendyr see my edit. Jun 27, 2018 at 8:26
• Let me try that thing ''cartesian to parametric'' with another math problem. I'll post a new question if I'm lost... Jun 27, 2018 at 8:43
r = ParametricRegion[{r Cos[ϕ], Sin[ϕ] r,
Sqrt[28 r]}, {{r, 0, 21}, {ϕ, 0, 2 π}}] Area[r]
`

49/2 π (14 Sqrt - ArcSinh[Sqrt])

• Thank you, Rom38! That was also short and neat. Jun 27, 2018 at 11:46

Maybe I'm missing something but shouldn't $f(x) = \sqrt{28 x}$ yield a concave function as opposed to the convex examples shown in the anwers above? My approach would be to use the formula for Solids of Revolution : $$V=\pi \cdot \int_{a}^{b}f(x)^{2}dx$$ analogous to $V = \pi r^2\cdot h$ for a cylindrical body where $r$ is the radius and $h$ is the cylinder height. This, evaluated by Mathematica or otherwise (I did it quick'n dirty by hand, so I hope I didn't make an error) should give: $$\pi \cdot 12348 \approx 38792.39$$

After misunderstanding the question at first, mistakenly stopping at giving the Volume of the Solid of revolution, the addendum for the surface area follows here: Further reading in the wikipedia article linked above also gives a pretty simple equation for the surface area with respect to rotation around the x-axis: $$S = 2 \pi \int_{a}^{b} f(x) dx$$ in analogy to $S = 2\pi r \cdot h$ for a cylindrical body. Because an exact result is required, I'll stop short of giving an value, as square roots are involved. But it should be easy enough from here.

• The exercise is to calculate an area so I don't know if it applies.... Jun 27, 2018 at 9:51
• yeah, 'area of volume' in the OP sounds a bit confusing. I think the above approaches somehow include an axis swap from x to y. This may be perfectly reasonable, though I..don't really get it. Jun 27, 2018 at 10:10
• Your formula gives a volume; the question is the area of the volume, like its skin (?). Don't know what the correct English expression would be. Jun 27, 2018 at 11:43
• oh, the surface area? Yeah, in that case, my solution won't help 😐 Jun 27, 2018 at 12:56
• thanks for trying to help in any case! Jun 27, 2018 at 18:03