# surface area of a dome with a square base

• Mar 1st 2013, 07:27 AM
ak1990
surface area of a dome with a square base
I am looking to find out how to calculate the surface area of a domed roof which rests on a quadrangle (ie a dome with squared edges) and I was wondering if anybody knew how to do this? I have attached a sketch of the kind of shape I am thinking of; the roof is a solid surface.
• Mar 1st 2013, 03:03 PM
Prove It
Re: surface area of a dome with a square base
Is the dome a perfect hemisphere, or is it just smoothed out so that the edge touches the edges of the square base?
• Mar 1st 2013, 03:46 PM
ak1990
Re: surface area of a dome with a square base
Thank you for the reply. I am a civil engineering student looking to gain some approximate sizes based on a roof shape of a specified design (the domed shape on square base). As a result of this at this stage either a domed smoothed out or a perfect hemisphere would be suitable I am just looking for a way of working out values to help me proceed with my design. I you have any information regarding either it would help me greatly.
• Mar 1st 2013, 04:27 PM
Prove It
Re: surface area of a dome with a square base
A perfect hemisphere would be easiest. The curved surface area of a hemisphere is \displaystyle \begin{align*} A = 2\pi\, r^2 \end{align*}.

Notice that if you drew a diagonal on your square base, it gives the diameter of the hemisphere, so half of it is the radius.

To evaluate the length of the diagonal, use Pythagoras' Theorem.
• Mar 1st 2013, 07:41 PM
ibdutt
Re: surface area of a dome with a square base
It may not be as simple as it appears. If we consider hemisphere then it will not coincide with the base and if it coincides with the base and is tapered top wards then it would not be a hemisphere. Visualize one side of square base and the shape tapering towards top. We need to look in for different techniques. Think on these lines and i too will consider it.
• Mar 2nd 2013, 07:43 AM
ak1990
Re: surface area of a dome with a square base
After further inspection I now believe the perfect hemisphere calculation to be too approximate. My sturcture involves the peak of the dome at a height of 2.14m above the 4 quadrangle walls, with each of these walls being 15m in length.