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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,558

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: 7,558

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: 15,837

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

The roof is an exact half of eclipse shape.

Contradiction!

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,558

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: 15,837

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.

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,558

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: 15,837

I meant a hemiellipsoid.

Character is who you are when no one is looking.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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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