by Charles Gonzales Gagui - CleVer Vibration
If you are currently studying solid geometry, finding the volume is just one formula away. This formula can save you time, and as I like to call it, save some brain space too.
Have you ever wondered, why do we need to memorize all these formulas and equations? There is one equation that you can use to solve any three-dimensional euclidean space's volumes. Solid figures like trapezoid, cubes, pyramids, prisms, spheres, cylinders, cones, and truncated cones.
You will only need to have somewhat acceptable ability level of manipulation skills and basic trigonometry.
Introducing the Trapezoid Volume Formula
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where;
V = Volume
Atop = Top Surface Area of the Solid Figure
Amiddle = Middle Area of the Solid Figure
Abottom = Bottom Area of the Solid Figure
h = height
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This is the only formula you will need to remember during exams and computing any geometrical objects you know in three-dimensional space.
Here are some examples:
Here are some examples:
Example 1. Truncated Square Pyramid/ Pyramidal Frustrum
A 50 mm thick blue square bowl is shaped as a pyramidal frustum. The open top of the bowl has an outer edge length of 14.2 cm having an angle of 45 deg. The height of the bowl is 5.5 cm. Approximate the liquid it can carry in cubic centimeter.
Given:
Thickness, t = 0.5 cm
Outside Edge Length, OLtop= 14.2 cm
Slope Angle at the Side of the Bowl, α = 45o
Overall Height of the Bowl, H = 5.5 cm
Required to Find:
Volume of fluid it can carry in cubic centimeters.
Solution:
Isolating just half of the bowl, we can figure out the values we need to formulate the area.
From this example, you can see how it can be applied to
Example 2. Spheres
Charles' revolutionizes a juice drink by obtaining juice in spheres with thin soluble membrane that explodes immediately when it comes in contact with your tongue. These spheres are to fill a spherical shape container. How many spherical shaped juice are needed to fill the spherical container, if each spheres is 1/480 of a fluid ounce and each sphere container will have 4 oz of juice? Also, what is the minimum inside radius of the bottle?
From here, we can estimate and recommend minimum radius above 3.4 cm, say like 4 to 5 cm or 2 in.
From this example, you can see how it can be applied to
V = (Atop + 4Amiddle
+ Abottom) h/6
Isolating just half of the bowl, we can figure out the values we need to formulate the area.
where:
Atop = ILtop2
where:
ILtop
= Inside top length
of the bowl
ILtop
= OLtop – 2 ta
where:
ta = thickness adjacent to
the bowl
ta
= t * √2 or
ta = 2 * t cos 45
ta = 0.7
ILtop
= 14.2 – 2 * 0.7
ILtop
= 12.8 cm
Atop
= (12.8) 2
Atop
= 163.8 cm2
Amiddle = ILmiddle2
where:
ILmiddle
= Inside middle
length of the bowl
ILmiddle
= ILtop – 2 xmiddle
where:
xmiddle = Half Difference of Length Top and Middle
Length
xmiddle
= (H – t)/2
tan 45
xmiddle
= 2.5 cm
ILmiddle = 7.8 cm
Amiddle
= 60.8 cm2
Abottom = ILbottom2
where:
ILbottom
= Inside bottom
length of the bowl
ILbottom
= ILtop – 2 xbottom
where:
xbottom = Half Difference of Length Top and Bottom
Length
xbottom
= (H – t)
tan 45
xbottom
= 5 cm
ILbottom = 2.8 cm
Abottom
= 7.8 cm2
h = H – t
h = 5 cm
Therefore:
V = 349.4 cu. cm
We can safely say that
the bowl can hold approximately 350 cu. cm of fluid. As other dimensions was not provide, this solution assumes that the bowl has straight corners. In reality, the actual volume could be more as you need to consider such variables like surface tension of the fluid and the fillets of the bowl. As this is only an example we will not go deeper into that. This solution is good enough basis, if someone is to use this bowl as a measurement for baking.
Example 2. Spheres
Charles' revolutionizes a juice drink by obtaining juice in spheres with thin soluble membrane that explodes immediately when it comes in contact with your tongue. These spheres are to fill a spherical shape container. How many spherical shaped juice are needed to fill the spherical container, if each spheres is 1/480 of a fluid ounce and each sphere container will have 4 oz of juice? Also, what is the minimum inside radius of the bottle?
Given:
Volume
Juice per Sphere, Vs = 1/480
oz.
Total
Volume of Juice in the Container, VJ
= 4 oz.
Required
to Find:
No.
of Sphere Juice in each Sphere Container, Ns
Inside
Radius of Sphere Container, rC
Solution:
Assuming that spheres with juice and sphere container are perfect spheres and each sphere juice are equal.
Ns = VJ / Vs
Ns = 4 / (1/480)
Ns = 1,920 spheres in each container
equating,
rC
from,
AC = π rC2
where:
Ac = Area at the middle
of the container.
from Volume
Formula:
VC = (Atop + 4Amiddle + Abottom) h/6
where:
VC = Volume of container
Atop = 0, as there is no area at the top of the sphere
Atop = 0, also as there is no area at the bottom
of the sphere
h = 2 rC
VC = (0+ 4Amiddle
+ 0) (2 rC / 6)
then,
Amiddle = (3 VC) / (4 rC)
Ac = Amiddle
then,
π rC2 = (3 VC) / (4 rC)
therefore,
rC = 3√[(3 VC) / (4 π)]
As the problem is asking the minimum radius, we will get an estimation, if the spheres are in a close-packed structure inside the container, using average density of π / √18 or 0.74048.
Using 0.74048 = VJ /VC
then,
VC = VJ /0.74048
rC = 3√[(3 (VJ
/0.74048)) / (4 π)]
rC = 3√[(3 *
((4oz. (375 cu.cm / 12oz.)) / 0.74048) / (4 π)]
rC = 3.4 cm
From here, we can estimate and recommend minimum radius above 3.4 cm, say like 4 to 5 cm or 2 in.
From this example, you can see how it can be applied to
