What is the volume of a square-based dome with a flexible cover?

[Huskerseth]

i am trying to figure out how to find the volume of a "dome" with a square base. think a swimming pool with a flexable cover attached to all sides. if the length of the base is 10 and the width is 20 and the max height the cover can move up above the base is 15 how do i figure out the volume that is under the cover since the max height is only acheived in the very center?

 

[berkeman]

i am trying to figure out how to find the volume of a "dome" with a square base. think a swimming pool with a flexable cover attached to all sides. if the length of the base is 10 and the width is 20 and the max height the cover can move up above the base is 15 how do i figure out the volume that is under the cover since the max height is only acheived in the very center?

Welcome to the PF.

What is the context of the question? What is the application?

Do you know how to calculate a double integral? Do you have a formulat for the height of the dome as a function of x and y?

 

[Huskerseth]

Welcome to the PF.

What is the context of the question? What is the application?

Do you know how to calculate a double integral? Do you have a formulat for the height of the dome as a function of x and y?

what we are doing is colecting methane gas under the "dome" and using it to run boiler to produce steam. what i am trying to figure out is how much methane is under the "dome" to determin how long a boiler can run off the colected gas. i don't know how to calculate a double integral. and i don't have a formulat for the height of the dome as a function of x and y? is this information i will need before i can procede?

 

[HallsofIvy]

There are a number of different ways of constructing such a dome that will give different volumes. I think you are referring to the case in which any horizontal cross section of the dome is a rectangle. Set up a coordinate system so that the corners of the rectangular bottom are at (10, 5), (-10, 5), (-10, -5), and (10, -5). A diagonal will have length \sqrt{20^2+ 10^2}= \sqrt{500}= 10\sqrt{5}. In order to have height, in the center, of 15, I am going to assume that the line of the dome cover from, say, (-10, 5) to (10, -5) is an ellipse with "semi-major axis" of length 10\sqrt{5} and "semi-minor axis" of length 15. That will have equation 225s^2+ 500z^2= 112500 (225 is 15^2 and 112500 is (225)(500).) That is, for a given z,
s= \frac{112500- 500z^2}{15}
where s is the diagonal length of the rectangle at height h. We also know that all the rectangles, at every height z, are similar to the base, which is 10 by 20 meters. That is, each rectangle is 10k by 20k meters so, by the pythagorean theorem 100k^2+ 400k^2= 500k^2= 112500- 500z^2 so that k^2= 225- z^2 and k= \sqrt{225- z^2}. That is, a cross section of the dome at height z is a rectangle with length 20\sqrt{225- z^2} and width 10\sqrt{225- z^2} and so area 200(225- z^2).

The volume of such a rectangular layer, of thickness \Delta z would be 200(225- z^2)\Delta z so the total volume of such layers would be
\sum 200(225- z^2)\Delta z
and, in the limit, that becomes the integral
\int_{z=0}^{15} 200(225- z^2)dz

 

[Huskerseth]

ok so it looks like i am totally out of my leage with this problem. thanks for all the help

 

[berkeman]

ok so it looks like i am totally out of my leage with this problem. thanks for all the help

Even without calculus, you should be able to estimate the volume to within about 10% or so.

Use Halls' initial forumulation, where you divide up the volume into a bunch of horizontal slabs. As you make each slab thinner (and therefore break the volume up into more slabs), the approximation gets more accurate. But you can start with slabs of thickness 1, say, and still get a pretty close answer.

Do your best to sketch what you think the length and width is for each horizontal slab of thickness 1 to stack them up to approximate the volume of the dome, and see what you get for an answer. The volume of each slab is just length X width X height.

To get an even better approximation, you could do one slab stack that is slightly larger than the dome extent, and one stack that is slightly smaller than the dome extent, and take the average of the two answers.

Let us know what you get!

 

[Huskerseth]

so berkeman what you are saying is to slice this dome in layers starting with 10 x 20 x 1 and slowly reduce each number till i hit my desired height? so i would have a ton of equations that looked like 10 x 20 x 1, 9 x 20 x 1, 9 x 19 x 1 ... 1 x 1 x 1 and then add all the answers to get a good estimation of volume?

 

[berkeman]

so berkeman what you are saying is to slice this dome in layers starting with 10 x 20 x 1 and slowly reduce each number till i hit my desired height? so i would have a ton of equations that looked like 10 x 20 x 1, 9 x 20 x 1, 9 x 19 x 1 ... 1 x 1 x 1 and then add all the answers to get a good estimation of volume?

Exactly. You could even take a couple pictures of the structure from the sides, and use those to help you estimate the base area for each slice as you go up.

 

[Huskerseth]

i am going to ask 1 last stupid question. in halls explanation can i input that final fourmula into excell where i then could change my height to get the vloume or would i have to do it by hand every time if i ever figure out what all the symbols mean and where all the different leters came from?

 

[berkeman]

i am going to ask 1 last stupid question. in halls explanation can i input that final fourmula into excell where i then could change my height to get the vloume or would i have to do it by hand every time if i ever figure out what all the symbols mean and where all the different leters came from?

You could use Excel to do the summation. You just vary the slab thickness delta z, and correspondingly have more slabs to sum up.

You should probably do it both ways (by hand estimation and with Excel), to see how close the answers are that you get.