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In cylindrical coordinates, the infinitesimal surface area is $dA=sd\theta dz$.

In order to find the surface area of the curved portion of a cone,with radius R and height h, I compute the integral:

$$A = \int_{\theta=0}^{2\pi}\int_{z=0}^{h}dA = \int_{\theta=0}^{2\pi}\int_{z=0}^{h} sd\theta dz$$

Using the straight line equation which gives $s = \frac{R}{h}(h-z)$, I obtain $A = \pi R h$.

This however does not give the literature solution, $ A = \pi R (R + \sqrt{R^2 + h^2})$. Have I gone wrong somewhere in my calculations?

Tian
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  • Infinitesimal surface area in cylindrical coordinates is $r\sqrt{1+(dr/dz)^2}d\theta dz$. In addition, you forgot to add the area of the base. – Intelligenti pauca Sep 15 '17 at 17:16
  • The lateral area of a sphere is $2\pi r l$ where $l$ is the "slant height" or $\sqrt {r^2 + h^2}.$ From this we can deduce that $s = \frac {h}{r}\sqrt{r^2 + h^2}$ – user317176 Jul 23 '21 at 04:25

2 Answers2

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Parameterize your surface.

$S = (r\cos\theta, r\sin\theta, r\frac {H}{R})$

Find $dS$ and $\|dS\|$

$dS = (\cos\theta, \sin\theta, \frac {H}{R})\times(-r\sin\theta, r\cos\theta, 0)\ dr\ d\theta\\ dS = (-\frac {H}{R}r\cos\theta, -\frac {H}{R}r\sin\theta, r)\\ \|dS\| = r\sqrt {\frac {H^2}{R^2} + 1}$

Giving us the integral.

$\int_0^{2\pi}\int_0^R r\sqrt {\frac {H^2}{R^2} + 1}\ dr\ d\theta$

And resolve the integral:

$2\pi (\frac 12R^2)(\sqrt {\frac {H^2}{R^2} + 1})\\ \pi R\sqrt{H^2 + R^2}$

Doug M
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What you tried to calculate is the surface of the lateral area, not including the bottom of the cone. So you did 2 mistakes, you didn't take the $S_{circle} =πR^2 $ area of the circle in mind and also you made a mistake calculating the lateral area.

Now for the lateral surface there is an older question already so you can go check it out: Setting Up an Integral to Find A Cone's Surface Area which gives you $S_{lateral} = πR \sqrt{ R^2 + h^2 } $ so the result is what you expect.

EDIT: the reason you are wrong is because the infinitesimal surface you used is that of a surface of constant radius (so you can use that in a cylinder for example). But in a cone the radius, the height and the azimuth all change.

Arbiter
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  • I have used a substitution for $s$, is that still insufficient to capture the change in the radius? – Tian Sep 15 '17 at 15:28
  • As you saw no, or else you'd have found the right result. By the way you made a mistake when integrating, your result should be $πh(h-2R)$ – Arbiter Sep 15 '17 at 15:46
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    If you want to find the area of a surface you parameterize the surface such as $ \vec{S} (x,t) $. Then the surface element is $ || \frac{\partial{ \vec{S}}}{\partial{x}} \times \frac{\partial{\vec{S}}}{\partial{t} }|| dx dt $ – Arbiter Sep 15 '17 at 15:53