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# Volume of an Object by Using Calculus

12/16/2015 2:43 PM

On this picture we see a octagonal dome. I am trying to calculate the volume of this object by integral calculus but I can't find a way. How would you calculate this? https://dl.dropboxusercontent.com/u/17974596/Sk%C3%A6rmbillede%202015-12-17%20kl.%2002.14.48.png

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#1

### Re: Volume of an Object by Using Calculus

12/16/2015 3:01 PM

It is the sum of eight slices. Each slice has a triangle as basis limited on both sides by planes and the 3rd surface is to be defined if it is a cylinder or generated by any other 2nd degree curve translated along the triangle side.

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#2

### Re: Volume of an Object by Using Calculus

12/16/2015 4:11 PM

Calculate it as a hemisphere then deduct ~4%...

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#3

### Re: Volume of an Object by Using Calculus

12/16/2015 4:34 PM

I think you can do this by simply calculating for only one slice of the eight sections. Then multiply that result by eight for the answer.

The base of one slice will form a triangle subtended at 45° and the opposite angles will be 67.5° and 67.5° respectively. Calculating the area is now simple.

Now think of each subsequent horizontal triangular slice stacked in the vertical direction. Volume is simply the sum of the area of each stacked triangle.

Since the top is a hemisphere you should be able to simply calculate the lengths of each side of each subsequent stacked triangle based on the profile of that hemisphere, but the curve will not be a pure curve, rather it will be half of an eclipse due to the rotation of the two vertical edges by 22.5° (as viewed from a god's eye view).

Again, you can calculate the eclipse created and therefore the subsequent sides of each triangle with a little geometry.

You should be able to figure out the geometry. Just don't forget to multiply the result by eight!

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#9

### Re: Volume of an Object by Using Calculus

12/17/2015 4:41 AM

Why must be the top a hemisphere ?

As I mentioned the sides can be also a cylinder surface. Why not ?

And the cylinder must not have its axis at the center !

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#4

### Re: Volume of an Object by Using Calculus

12/16/2015 5:46 PM

It doesn't say it's symmetrical, but let's assume it is. I.e., the 'radius' is the same as the height of the dome.

Let r be the radius of the dome. This will also the be length, r1, of the hypotenuse of the bottom-most octagonal piece.

The area of this triangle is A1 = r12 *cosθsinθ, where θ is 22.5 degrees.

The volume of the first slice will be this area times the thickness t of the slice. (You should make t as thin as possible). V1 = A1*t

The area of the next triangle is A2 = r22 *cosθsinθ.

Here, r2 is given by r2 = √(r2 - t2), and then V2 = A2*t

Continue calculating the slices n number of times, where n is r/t.

Then Sum the triangles to get the volume of one chord of the dome.

V(total of 1 chord) = ∑ (V1 + V2 +... Vn)

V(dome) = 8*V(total of 1 chord)

The sketch is a bit cluttered where I show r1, r2, and r3 along the curve of the dome. I also drew-in, but didn't label r4. Nevertheless, this should be enough to illustrate my approach.

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#5

### Re: Volume of an Object by Using Calculus

12/16/2015 6:04 PM

The angle subtended is 45°, not 22.5° for the triangle.

The base is an octagon, so it will be 45°.

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#6

### Re: Volume of an Object by Using Calculus

12/16/2015 6:57 PM

I should have labeled the angle I'm calling theta.

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#7

### Re: Volume of an Object by Using Calculus

12/16/2015 9:39 PM

Yes, that helps!

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#8

### Re: Volume of an Object by Using Calculus

12/16/2015 9:55 PM

Details are missing on the shape of the red curves, and how high the dome is compared to its width.

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#10

### Re: Volume of an Object by Using Calculus

12/17/2015 5:27 AM

Fill it up completely with water, and measure the volume needed to fill it using a flowmeter on the fill pipe.

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#11

### Re: Volume of an Object by Using Calculus

12/17/2015 5:42 AM

The dome looks like Filippo Brunelleschi's work at the cathedral of Santa Maria del Fiore in Florence (Firenze).

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#12

### Re: Volume of an Object by Using Calculus

12/17/2015 6:01 AM

Using Calculus: Get a formula for circumference C as a function of height h. Then integrate C with respect to h, i.e. ∫C(h)dh.

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#13

### Re: Volume of an Object by Using Calculus

12/17/2015 6:53 AM

You need a maths expression for the shape of the curve of the dome to start with.

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#14

### Re: Volume of an Object by Using Calculus

12/17/2015 7:21 AM
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#15

### Re: Volume of an Object by Using Calculus

12/17/2015 12:27 PM

The figure shows a dome with an octagonal basis. Most probably because of the building constrains the surfaces are generated along a at least second degree guide. The equation for this guide line can be: l= a-P(z) with P(z) = a0*z^2+a1*z+a2

The coefficients can be determined with the boundary conditions:

At z= 0 l=a ; dP/dz= c → a2=0 and c= a1

at z= H i=b → b= a0*H²+c*H → (b - c*H)/ H²= a0

thus l(z)=a-[(b-c*H)/H²*z²+c*z]

If the guide line is normal to the base plane at z=0 then c=0.

For this particular case the red triangle area will be S(z)= l²*tan(22.5°) in a general case with a different number of sides S(z)= l²*tan(α) where α= π/n n being the number of sides the polygon has.

The elementary volume is dV= S(z)*dz = l²*tan(α)*dz= {a-[(b-c*H)/H²*z²+c*z]}²*tan(α)*dz.

The integral (from z=0 to z=H) is easy to solve since it is the integral of a polynomial.

Result should to be multiplied by "n".

To validate the result follow the given counsel: fill it with water and measure how much was in.

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#16

### Re: Volume of an Object by Using Calculus

12/17/2015 1:00 PM

(deleted comment)

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#17

### Re: Volume of an Object by Using Calculus

12/18/2015 5:48 AM

In previous comment I proposed a general solution.

A particular case is the dome enclosed in a spherical envelope.

In this case a= R cos(α)

and the guide line is an elliptical arc with the equation:

(l(z)/a)^2+(z/R)^2=1 → l(z)^2= a^2*(1-(z/R)^2) = (R*cos(α))^2*(1-(z/R)^2)

The triangle area is S(z)= tan(α)*l(z)^2 = tan(α)*(cos(α))^2*(R^2-z^2) = 0.5*sin(2*α)*(R^2-z^2)

The slice volume will thus be

H

V1=0.5*sin(2*α )*∫(R^2-z^2)*dz =

0

= 0.5*sin(2*α )*(R^2*H-H^3/3)= 0.5*sin(2*π/n)*R^3*(χ-χ^3/3)

V= n*V1 χ=H/R

The solution I suggested was for cases when the guide line equation is not as simple as in this particular case.

The solution is correct since if n→∞ V= 2/3*π*R^3 which is half of the sphere volume.

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