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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,602

I just received this email from "Allan":

Can I ask you an urgent question please.

I am trying to find out the external square metres of a dome shaped roof.

Its 8.5 metres accross ( 4.25mts in Radius.

And the highest point in the centre is 5 metres high.

The roof is an exact half of eclipse shape.

Help!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,602

Dear Allan,

I am not sure if this is the exact answer for you ... but perhaps you can clarify it a bit more after reading

You mention a radius, so let us assume the dome is half a sphere.

The area of a sphere is 4 × π × r², and you want just half of that.

So, the pure half dome will have Surface Area = 2 × π × r² = 2 × π × (4.25)² = 113.5 m²

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,082

And the highest point in the centre is 5 metres high.

The roof is an exact half of eclipse shape.

Contradiction!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,602

True.

Perhaps it means that there is a circular "ring wall" under the dome that is (5-4.25)=0.75 m high? If so, then multiply the height (0.75) by the circumference of the circle (2 × π × r) to get the area of the wall.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,082

Had that condition not been given, the solution would have been simple, as you had posted, Sir. The dome is not a perfect hemishphere. It resembles an ellipsoid.

Maybe, this link would be of some help in determining the surface area.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,602

The "ring wall" plus hemisphere will get close, I think (considering relative sizes). Maybe Allan could tell us how critical it is to get an exact answer. I emailed him to let him know I started a thread on his problem.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,082

I meant a hemiellipsoid.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

I believe that this is the answer. If we model the roofline as a curve and then treat the roof as a surface of revolution, I believe that it produces the answer needed.

Using points where x = -4.25, 0, 4.25 we get the curve:

y = -32x^2/61 + 1092x/1037 + 5

y = (-544x^2 + 1092x + 5185) / 1037

Sorry for the large constants, but it was specified an exact answer was needed.

V = ∫2 pi x f(x) dx evaluated from zero to r

V = (pi / 1037) (-272r^4 + 728r^3 + 5185r^2)

I spared you the simplification process, but I am sure that you all know how to do this.

V/thickness = A , in our case V/r = A

A = (pi / 1037) (-272r^3 + 728r^2 + 5185r)

Since our radius is 4.25 our area is 1683pi / 122 (exact)

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