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could anyone explain how to calculate the surface area of a curved plane? I am trying to calculate the surface area of a "vaulted" ceiling that is 24' long, 7' wide, and the height of the curve is 4' at the mid-point so that I can figure out how much paint I need to buy.

Here is an illustration:

Illustration of surface

If it were a simple flat ceiling, I would just multiply length x width, but that's not what I have here and a google search is leaving me empty-handed. Any ideas?

Thank you very much! -Neal

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It depends on the precise shape of the surface. Is this particular one supposed to be half a cylinder? – Javier Jun 4 '13 at 20:11
If you take it as an approximate cylinder then a conservative estimate is $\pi r L$. easy part $L=24$. Hard part what is $r$, you can take $r=\max\{7/2,4\}$ that is $r=4$ and have a quick answer $\pi * 4 *24$ – Maesumi Jun 4 '13 at 20:54
up vote 0 down vote accepted

If I understand your description correctly, we can figure out the area as follows:

A rough estimate (taking the width as 8 instead of 7, which gives a half-circle) gives $\pi r l = 96 \pi \approx 300$.

$\tan \theta = \frac{4}{3.5} $, $r \sin \theta = \frac{\sqrt{4^2+(3.5)^2}}{2}$. Since $\sin \theta = \frac{4}{\sqrt{4^2+(3.5)^2}} $, we have $r = \frac{4^2+(3.5)^2}{8} = \frac{113}{32}$.

$\theta = \arctan \frac{4}{3.5} \approx 0.85917$.

Hence the length of the arc is $r (4 \theta) = (\frac{113}{32}) 4 \arctan \frac{4}{3.5} \approx 12.034$. Hence the required area is $\approx 24 \cdot 12.034 = 288.82$. This is consistent with the rough estimate.

enter image description here

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Fascinating... It's been a long time since college for me but I remember now why I enjoyed math. Thanks so much for this! – Neal L Jun 4 '13 at 21:18

use the formulae you derived in part 1 to calculate the curved surface area and the total surface area of the cylinder given below

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Your answer is correct, Can you please add more details? – Iuʇǝƃɹɐʇoɹ Dec 21 '14 at 16:30

Just calculate the cross-section length and multiply it with 24". This is the simplest solution, without integration.

In order to deal with this more precisely, we must know the exact shape of the cross-section i.e. if it's a part of a circle or not

EDIT: I just did some thinking, and this cannot be a part of a circle obviously. It is illogical to me that what we are looking at are actually two arcs glued together. This is a parabola - once we find out its equation, we will find the surface - I'm working on that.

Here - the equation of parabola is $$h(1-\frac{x^2}{a^2})=y$$

The length of parabola arc is $$\sqrt{(a^2+4*h)}+\frac{2a^2}{2h}sinh^{-1}(\frac{2h}{a})$$

Here, a and h are the half-width which is 3.5 and height which is 4.

When you enter the numbers, the arc length is 7.418', and the surface is $$7.418*24=178.032$$

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