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Area of a dome

by DaveC426913
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DaveC426913
#1
Dec25-04, 03:11 PM
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Trying to figure out the answer to another thread.

What is formula for the surface area of a dome?

Googling got me [tex]2\pi r^2 h[/tex]
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Integral
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Dec25-04, 03:36 PM
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Dome, is a pretty general term. You will need to be more specific if you want a specific formula.

A dome may be simply a hemisphere, if it is geodesic then you can get close with a hemisphere but to get accurate you would need to know the size and shape of the tiles.

The surface area of a portion of a sphere is 2 pi R h, R = Radius of the sphere, h is the height of the portion. Your formula cannot work for surface area, it has units of volume.
dextercioby
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Dec25-04, 03:48 PM
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Quote Quote by DaveC426913
Trying to figure out the answer to another thread.
What is formula for the surface area of a dome?
Googling got me [tex]2\pi r^2 h[/tex]
1.It's wrong.The area has dimensions of m^{2},while your formula has dimensions of volume:m^{3}.
2.You'll find the area in terms of the diameter and the focal distance here (hope u have a good connection,the page is HTML and is pretty big).
3.If you're interested in the volume,though,and the formula u found was for the volume,check out this site:wolfram

Daniel.

DaveC426913
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Dec26-04, 12:48 AM
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Area of a dome

OK. I'm going to try to post this one more time without it screwing up on me.

It didn't occur to me how many types of domes there might be. I'm talking about a dome that is a section of a sphere.

Take a sphere, radius r.
Slice it with a plane, at some distance d from the centrepoint of the sphere.
Throw away the larger piece, leaving a dome-shape with a "height" equal to (r-d).

Is there a formula for calculating the surface area of this dome?

Examples:
1] If d = zero, you have cut the sphere exactly through the centre; the dome's surface area is equal to half the sphere's area (4pi*r^2) / 2.

2] If d is equal to r, you have placed the plane tangent to the sphere, throwing away everything, leaving a surface area of 0.

3] What if you set d = some intermediate value, say, 1/2 r? What is the general case i.e. What is the surface area of the dome as a function of d?

(Fingers crossed. Please post without messing up!)
Justin Lazear
#5
Dec26-04, 01:23 AM
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Just evaluate (2) in http://mathworld.wolfram.com/SurfaceofRevolution.html for the equation of a circle with the proper bounds on the integral.

--J
Integral
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Dec26-04, 02:36 AM
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The best I can interpret what you are saying the formula I gave you on my first post is the one you want.

[tex] S = 2 \pi R h [/tex]

Where R is the Radius of the sphere
h would be the distance from the ground to the top of the dome.
HallsofIvy
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Dec26-04, 07:21 AM
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Quote Quote by Integral
The best I can interpret what you are saying the formula I gave you on my first post is the one you want.

[tex] S = 2 \pi R h [/tex]

Where R is the Radius of the sphere
h would be the distance from the ground to the top of the dome.
No, Integral, that's the lateral area of a cylinder of radius R and height h.

Here's how I would do it:

Set up a coordinate system have origin at the center of the sphere and positive z axis perpendicular to the base of the dome.
The equation of the sphere is F(x,y,z)= x2+ y2+ z2= R2 and we think of that as a "level surface" for a function F(x,y,z). Then grad F= 2xi+ 2yj+ 2zk is normal to the sphere at each point. We can "normalize" that, using the xy-plane as reference by dividing by the z component: dS= ((x/z)i+ (y/z)j+ k). The "differential of surface area" is The length of that, &sqrt;((x2/z2)+ (y2)/z2)+ 1)dA. That is the same as (&sqrt(x2+ y2+ z2)/z)dA= (R/z)dA.

z= √(R2- x2- y2) so it is simplest to do the integral in polar coordinates: the differential of surface area is (R/√(R2- r2))rdrdθ

Now, if the height of the dome is h< R, the z= R-h and the equation of the sphere becomes x2+ y2+ (R-h)2=
r2+ R2- 2hR+ h2= R2 so
that r2= 2hR- h2 and the integration over r will be from 0 to &radic;(2hr- h2).

Since &theta; does not appear explicitely in the integral the &theta; integral is just 2&pi;.

The surface area is 2&pi; integral of (R/&radic;(R2- r2)rdr with limits of integration r=0 to &radic;(2hr- h2).

That can be integrated by letting u= R2- r2 so that
du= -r dr and the integration is from u= R to R2- 2hr+ h2= (R-h)2.

In terms of u, the surface area is 2&pi;Rintegral of u-1/2du=
4&pi;Ru[sup]1/2[sup] evaluated between (R-h)2 and R2: that is, of course, 4&pi;R(R- (R-h))= 4&pi;Rh, exactly twice what Integral said.
dextercioby
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Dec26-04, 07:56 AM
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Quote Quote by HallsofIvy
No, Integral, that's the lateral area of a cylinder of radius R and height h.Here's how I would do it:Set up a coordinate system have origin at the center of the sphere and positive z axis perpendicular to the base of the dome.The equation of the sphere is F(x,y,z)= x2+ y2+ z2= R2 and we think of that as a "level surface" for a function F(x,y,z). Then grad F= 2xi+ 2yj+ 2zk is normal to the sphere at each point. We can "normalize" that, using the xy-plane as reference by dividing by the z component: dS= ((x/z)i+ (y/z)j+ k). The "differential of surface area" is The length of that, &sqrt;((x2/z2)+ (y2)/z2)+ 1)dA. That is the same as (&sqrt(x2+ y2+ z2)/z)dA= (R/z)dA z= √(R2- x2- y2) so it is simplest to do the integral in polar coordinates: the differential of surface area is (R/√(R2- r2))rdrdθNow, if the height of the dome is h< R, the z= R-h and the equation of the sphere becomes x2+ y2+ (R-h)2=2+ R2- 2hR+ h2= R2 so
that r2= 2hR- h2 and the integration over r will be from 0 to √(2hr- h2).Since θ does not appear explicitely in the integral the θ integral is just 2π.
The surface area is 2π integral of (R/√(R2- r2)rdr with limits of integration r=0 to √(2hr- h2).
That can be integrated by letting u= R2- r2 so that
du= -r dr and the integration is from u= R to R2- 2hr+ h2= (R-h)2.In terms of u, the surface area is 2πRintegral of u-1/2du= 4πRu[sup]1/2[sup] evaluated between (R-h)2 and R2: that is, of course, 4πR(R- (R-h))= 4πRh, exactly twice what Integral said.

Interesting,HallsofIvy,you mind taking a look at these 2 pages on wolfram site
Integral was right
Make [itex] b\rightarrow 0 [/itex] in the formula for the spherical zone area,to get the result that Integral predicted,which is found
here
HallsofIvy
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Dec26-04, 08:11 AM
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Ouch! I see what I did. Because I used symmetry, (I used z2 and not just z) I got the area of the section above AND below the xy-plane- exactly twice what I wanted.

In particular, if h= R, your formula gives A= 2&pi; R2, the area of a hemisphere, while mine would give A= 4&pi;R2, the area of the full sphere.

It is very surprising to me that the area is precisely the same as the lateral area of a cylinder!

Thanks.
DaveC426913
#10
Dec26-04, 11:54 AM
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Quote Quote by Integral
The best I can interpret what you are saying the formula I gave you on my first post is the one you want.

[tex] S = 2 \pi R h [/tex]

Where R is the Radius of the sphere
h would be the distance from the ground to the top of the dome.
So... the surface area of the dome (which curves in 2 dimensions) varies linearly with a change in its third dimension.

So... a dome that has a height of .5r will have a surface area equal to .5 of the hemisphere.
And a dome with height .1r will have a surface area equal to .1 of the hemisphere.

I'm thinkin' there's something wrong here.

2 pi r h That's the formula for the surface area of an open-ended cylinder.
dextercioby
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Dec26-04, 12:00 PM
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There's nothing wrong with that formula.Why didn't u check the 2 webapges on wolfram.com that i indicated.They containt the calculations and therefore prove that the area of a dome,seen as spherical cup,is given by:
[tex] S_{sperical\cup}=2\pi Rh[/tex]
,where 'R' is the radius of the sphere and 'h' is the height of the cup.
Daniel.

PS.Aren't u familiar with integrals??
DaveC426913
#12
Dec27-04, 12:53 AM
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Quote Quote by dextercioby
Why didn't u check the 2 webapges on wolfram.com ...
PS.Aren't u familiar with integrals??
Because the math is beyond my ability. My Calculus education is 20 years old.
(which is why I'm asking here.)

I'm taking your word on faith, and trying to visualize it.

So, that also means if I slice a hemisphere with a meat slicer so that each slice is the same thickness, they will all have the same area. In this image, each slice has the same surface area, including the very top slice, and the very bottom slice.
Integral
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Dec27-04, 03:50 AM
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Frankly, I cheated. I picked up my copy of CRC Standard Math Tables (27th edition) on page 128 under Geometry, Mensuration formulas under Zone and segment of one base, the picture shows a sphere as you seem to be describing the formula I provided is the one they list for Surface area.

The next set of formulas is under the title of Zone and Segment of Two Bases, the picture indicates a section of sphere similar to one of your slices the Surface area formula given is the the same. [itex] S = 2 \pi R h [/itex] where h is the thickness of the slice.

Now if you tell us that you are not able to derive the results from fundamentals (do the integral) how can we convince you that the formula given is correct?

According to my book it is, what more do you want?

Look at the link provided by Justin Lazear, near the bottom of that page there is a list of Surface area formulas, look at the one for zone, be sure to look carefully at the picture with it. That is similar to the 2 base formula in my book.

Sorry, guys I know that I should have worked out the integral, its my NAME!, but what can I say, it was easier just to pick the book off the shelf. It's within an arms reach of my computer, I didn't even have to get out of my chair!
TenaliRaman
#14
Dec27-04, 03:54 AM
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Dave,
Take a string of length equal to that of the diameter.

Fix two poles(north pole and south pole) on the sphere and try to connect these poles with your string, hmm is it possible?
If yes? try again and if no? why wouldnt it connect? The curvature has anything to do with it? Hmm will it affect the surface areas as well then?? But i see the curvature of a sphere is uniform ?? Equal Surface areas??

-- AI
DaveC426913
#15
Dec27-04, 04:43 PM
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"Now if you tell us that you are not able to derive the results from fundamentals (do the integral) how can we convince you that the formula given is correct?

According to my book it is, what more do you want?"

Well, empirically is merely one way it can be shown.


"Look at the link provided by Justin Lazear,"

Yeah, that seems to be pretty clear all right.
DaveC426913
#16
Dec27-04, 04:47 PM
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TenaliRaman:
I think I see what you're getting at, but aren't you missing the fact that a string that could horizontally circumscribe the slice at the pole will be waaaay too short to horizontally circumscribe the slice near the equator?
Justin Lazear
#17
Dec28-04, 03:49 AM
P: 290
Suppose we take a slice near the center. This slice has a large radius, but it is pretty flat at the sides, so the fact that the slice is curving toward the center of the slice is not very significant. This slice is similar to a cylinder. Now, consider a slice from near the edge of the sphere. The radius is small, but you'll notice that the slice is no longer like a cylinder. The sides are now curving inward quite significantly, so there is a lot of extra surface area. The slice is much more like a plane than a cylinder.

It's certainly not a very quantitative argument, but it does show that there's a balancing act between how much the slice is curving toward the center and the radius of the slice, which might be enough to convince you that what the nice fellows here have ben saying isn't complete bull.

Even if your calculus is a little rusty, perhaps you still remember the integral of 1? The integral reduces to a bunch of constants multiplied by the integral of one, so it's more an exercise in algebra than anything. Just remember that dy/dx = -x/y for a sphere and do a bit of algebra and you can prove it to yourself.

--J
TenaliRaman
#18
Dec28-04, 09:38 AM
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Quote Quote by DaveC426913
TenaliRaman:
I think I see what you're getting at, but aren't you missing the fact that a string that could horizontally circumscribe the slice at the pole will be waaaay too short to horizontally circumscribe the slice near the equator?
I think you are getting what i am trying to say but not completely.

Lets take another view of the same thing. Consider a circle. Rotate it abt its axis, what u get finally is the surface area of a sphere. Now consider 1 line of height h. If you rotate this lines about the axis which goes through the a point which is at a distance R from the line,then we will get the surface area of a cylinder. Do u agree with me thus far?

If yes, then lets continue.

Now take the initial circle again and break it into pieces just like you have done in your image. Take each of these pieces and straighten them, will they be all of different sizes? I am sure you would agree that they definitely are of different sizes. Now if rotate each of these straightened lines about a point at Radius R, i will get many cylinders of different sizes and different surface areas (since i am keeping the base radius R constant for all of them).

Now if i want to make their surface areas equal, then i will have to change the base radius for each of them. Now u seem to have no problems with the formula of surface area of cylinder. You might notice that surface area of cylinder is linear with the base radius. That means if i linearly decrease the radius of each of the larger cylinders , their surface areas diminishes linearly.

Thereby finally after decreasing their radii's i get all cylinders with equal surface areas.

-- AI


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