Math IA
IB Math AA HL
Mehdi Khfifi
Approximating the volume of a pyramid using 3 dimensional Reimann sums
Introduction
The Reimann Sum is a calculus phenomenon developed by the German
Mathematician Bernhard Reimann. Reimann sums are a unique method for finding
the area under a line or a curve. By partitioning the area into rectangles, we can
approximate the area under the curve using the area of those rectangles. The more
rectangles we use, and the finer their widths, the better the approximation gets.
When first introduced to Reiman sums, our teacher had a computer model
that displayed the sums visually, and we could see that as the number of divisions
grew, the approximation became more accurate. I found this partitioning to be
fascinating, and inspired me to dive deeper into the topic. I went back home and
read through an advanced calculus textbook to further understand these sums. The
book featured the following image in figure 1 with a caption, “Volume
approximation.” This series of pictures was part of a more advanced calculus section,
concerning multivariable calculus and double integrals.
Figure 1a displays a 3D concave down surface. Colored in light blue, the
surface is continuous over the region labeled R. Beneath the graph is a square grid
superimposed on the x-y coordinate plane; the tiles of the grid are colored where the
surface fully encloses their area. The objective is to find the volume of the solid
region lying between the surface given by some function f and the x-y plane. If we
choose some point (𝑥! , 𝑦! ) which lies in one of the enclosed squares, as shown in
Figure 1b, we can approximate the volume between the surface and the square
region by multiplying its area by the value of f at (𝑥! , 𝑦! ), seen visually as box or
prism extruding out of the grid, retaining a height of 𝑓(𝑥! , 𝑦! ). If we repeat this
procedure for all squares lying in the region R and sum the volumes of their
corresponding boxes, we obtain an approximation of the volume present between the
surface and the x-y plane.
I thought of this process as a “3D Reimann sum;” much like the partitioning
of an area into rectangles of equivalent widths but varying lengths, this process
partitioned the volume under a surface into smaller boxes of equivalent areas but
varying heights. This image, despite it being related to a calculus concept beyond the
syllabus, captivated me, and I wanted to explore how Reimann sums can be applied
to approximating the volumes of 3D shapes, without the need for advanced
multivariable calculus.
Figure 1: Picture from Larson Calculus I, II, III textbook
Figure 1a
Figure 1b
Figure 1c
Plan
What do 3D Reimann sums look like
Instead of partitioning the areas under 2D curves into rectangles of equal
widths, as in the case for traditional Reimann sums, I will be partitioning the volume
under 3D graphs and surfaces into rectangular prisms or “boxes” of congruent square
bases. Much like the fact that there are right endpoint, left endpoint, and midpoint
Reimann Sums shown in Figure 2, the same case holds for 3d Reimann sums, as
shown in Figure 3. I researched online that the boxes used to approximate volumes
under curves can scan touch the function being approximated at either its endpoints
or center(citation).
Figure 2: Left, right and midpoint Reimann Sums
Left Reimann Sum
Right Reimann
Midpoint Reimann sum
Figure 3: Corner and center 3D Reimann Sums
Figure 3a Endpoint 3D Reimann Sum
Figure 3b Center 3D Reimann Sum
To approximate the volume of my pyramid, I decided to choose the center
method indicated by Figure 3b. Since midpoint Reimann sums are known to be
considerably more accurate than left or right Reimann sums, I deduced this would be
the same case for 3D Reimann Sums. I chose the center method assuming it would
result in minute overestimates compared to the great underestimates visible in
endpoint method in Figure 3a.
In addition, I decided that these boxes will have congruent square bases, for
which their widths will be decided. By having congruent bases, the task of finding
the volume of each individual box will be simplified; the main challenge of my
investigation would hence be to find the respective heights of each box.
Choice of shape
I needed a shape for which the height of the rectangles extruding out of its
base can be modelled using a single variable function f(x). This is better understood
in pictures than in words. Figure 2a displays a triangular. Figure 2b displays a
volumetric approximation of the shapes using 3D Riemann sums. Boxes of the same
row retained equivalent heights. As you can see in figure 2b, the bases of the shape
can be represented using a single variable function- an absolute value function. The
heights of the boxes had some respective relationship with their position of rows and
columns. The volume of the prism can be represented using the following equation:
"
𝑉𝑜𝑙𝑢𝑚𝑒 = lim 2 𝐴 ∙ 𝑓(𝑥! ) ∙ 𝑛
"→ %
!&'
Where n represents the number of subdivisions of the length of the prism, 𝐴
represents the area of the square bases of each box, and 𝑓(𝑥! ) represents some value
of 𝑓 in the ith interval, as shown in figure X. Note that the Area 𝐴 is dependent on
the number of subintervals and the length of the prism.
As you can see, I simplified the task of finding a volume by modelling the
height of each prism as a function. Since boxes of the same were arranged in the
same row, I was able to find the volume of one of the boxes and multiply by the
number of boxes in its respective row. Repeating this step for all the subinterval
rows hence resulted in a volumetric approximation. As the number of rows goes to
infinity, the approximation gets more and more accurate, as shown with the limit.
I wanted to approximate the volume of shapes that we were not as intuitive
as the volume of prisms. The shape I would investigate would also need to retain
some form of height pattern as in the case of the triangular prism, for which I could
simplify the height of the boxes into a simple function.
Initially, I’d chosen to approximate the volume of a paraboloid (parabola
revolved around a central axis) as shown in Figure 3a. I initially chose this shape as
it resembled the graph presented in the Larson textbook. I made a 3D model using a
pre-existing algorithm that applies 3D Reimann sums onto the volume under the
surface. As you can see in Figure 3a, the heights of each of the boxes, regardless of
the number of partitions, don’t follow a predictable pattern. In addition, it’s difficult
to decide whether boxes near the edge of the shape should be included into the
approximation, as the surface does not enclose their square base on the grid. I tried
looking for any patterns within the shape, for which you could refer to Appendix A,
but couldn’t find any, and I hence discarded using it.
After much research and talks with my teacher, I came upon choosing a right
square pyramid. This shape retains a square base, for which fixes the issue of empty
edges as in the case for the paraboloid. More importantly, however, boxes of the
same height followed an interesting pattern, as shown in Figure 4, for which I
wanted to investigate.
Reflection 1
The task of finding the volume of a square pyramid seemed a little bit of large
and tedious task, and I’d been thinking critically over how I could simplify my
problem.
Pondering back about the symmetric nature of square based pyramids, I
graphed a pyramid on a 3D coordinate plane centered at the origin (0,0,0) (Figure
4a) using GeoGebra. I found that it had it had two convenient planes of symmetry
that lined up with the x and y axes. I modeled these planes using GeoGebra as
shown in Figure 4b, with the blue plane lining up with the y-axis and the green
plane lining up with the x-axis. This was interesting, as the pyramid seemed to be
split equivalently into the four different quadrants of the x-y coordinate plane.
Finding out that each of these “quarters” sliced from the pyramid had an
equivalent volume, I concluded that finding the volume of the pyramid in one
quadrant and multiplying it by 4 would equate to a volumetric approximation of the
entire pyramid. I pondered a bit longer, and concluded that finding the volume of
the pyramid in the first quadrant, instead considering the pyramid as a whole, would
be a much more convenient and easier approximation method.
Figure 4 Partitioning of a pyramid into four “quarters”
Figure 4a
Figure 4b
Exploration
Exploring the nature of the shape
Having decided that I will approximate the volume of a “quarter pyramid,” I
modeled the shape I’d envisioned using the 3D software, as shown in Figure 5a and
Figure 5b. To get an even better understanding of the shape, I asked the school’s
design teacher to teach me how to use the 3D printer, for which then I printed my
model as shown in Figure 5c.
Figure 5: Process of 3D printing the quarter pyramid shape.
Figure 5a: Using hole boxes to cut out a quarter of
the pyramid
Figure 5b: 3D model of cut quarter pyramid
Figure 5c: 3D printed model of
quarter pyramid
I re-implemented the box approximations, trying to investigate the pattern
present in the heights of the boxes discovered earlier. I realized boxes of the same
height were aligned in this interesting “L” arrangement. Retaining the same height,
these boxes hence also retained the exact same volume. This caught my attention
greatly. I figured, I could find the volume of one of the boxes and multiply it by the
number of boxes in the L region. If I repeat this procedure for all of the L regions,
then I end up with a volumetric approximation of the quarter pyramid, as shown in
Figure 8.
Figure 7 Isometric(7a) and bird’s eye(7b) view of “L” shape pattern
discovered
Figure 7a Boxes of the
same height followed this
interesting pattern
Figure 7c
Figure 7b From a bird’s eye view,
it looks like an L. As the height
increased, the number of squares
in these “L” regions decreased.
I investigated this pattern by drawing a graphic model of the base of the
quarter pyramid using Adobe Illustrator, as shown in Figure 9. Splitting the base of
the quarter pyramid into a 1x1 grid resulted in only one square, which I still
considered an “L” region and I labeled it green. Splitting the base into a 2x2 grid
resulted in 2 “L” regions, both the previous green one and an additional red one,
containing three squares. Splitting the base into a 3x3 grid resulted in 3 “L” regions:
the original green “L” region containing one square, the second red “L” containing
three squares, and finally the third “L” region-which I labeled blue-containing 5
squares. It turns out, the number of prisms in each L region follows the sequence
1,3,5,7,9, as the number of partitions grew. I realized that the number of squares in
each “L” region followed an odd number sequence, starting from 1, and continuing
depending on the number of partitions.
Figure 9: Investigating “L” pattern
I will label each “L” region 𝐿" , I’ll consider the green L region to be the first L
region 𝐿' , the red being the second 𝐿( , the blue being the third 𝐿) etc.
Figure X Assigning an order to the “L” regions
It turns out the number of squares in each “L region” is odd. I learnt in class
various useful sequences that can model specific number patterns, one of which were
odd numbers. Hence, the relationship between the order of the L region and the
number of squares inside it is given by:
𝑛 ∈ 𝑅*
𝑆" = 2𝑛 − 1
Equation 1
Where Sn represents the number of squares in 𝐿" ,, and n is an element of positive real
numbers, since the order of the L regions begins at 1.
To confirm this works, I inputted 4 for n into Equation 1to find out the
number of squares in 𝐿+ , as shown below:
𝑆" = 2𝑛 − 1
𝑆+ = 2(4) − 1
𝑆+ = 7
From the work above, we see there are exactly 7 squares in the fourth L
region, which correlates with the graphic I had made in Figure 9, as there are 7
orange squares regardless of how partitioned the grid is.
In addition to the discovery of the L regions, I also found that the height of
each of these “L” regions can be modeled into a linear function. I designed a 3D of
the volumetric approximation of the quarter pyramid, whilst respecting the color
codes used earlier for each “L” region. I switched my view to looking at the faces
perpendicular to the pyramid’s base, the slant height of the pyramid as well as the
heights of the boxes in their respective “L” regions almost seemed 2 dimensional. I
took a screenshot of my view and graphed it onto a x-y plane using GeoGebra. I
realized I could model the slant height using a linear function. Evidently, this
function passed through the boxes at their midpoints, due to my implementation of
the center method.
With all the boxes retaining a fixed area for their square bases, it came
to my mind that finding the height of one of the boxes from each “L” region is
suffice to approximate the volume of the quarter pyramid. Retaining the same
height, these boxes hence also retained the exact same volume. I figured, I could find
the volume of one of the boxes and multiply it by the number of boxes in the L
region. Repeating this procedure for all of the L regions, and taking the sum of their
volumes, would result in a volumetric approximation of the quarter pyramid.
Using real data
Hence, finalizing my approach, I decided to try to approximate the volume of a
square pyramid with a width of 8 units and a height of 4 units, as shown in Figure
11a. Slicing the pyramid into four equivalent parts, as done previously on the
coordinate plane, the width and length of the quarter pyramid are 4 units long, as
shown in Figure 11b. I decided that I will partition the base into a 4 by 4 grid, as
shown in Figure 11c, which hence means there will be four "L" regions.
Figure 11a
Figure 11b
Figure 11c
With a height and width of 4 units, the function that models the slant height of the
pyramid vertically and horizontally aligned with the origin will hence be
𝑓(𝑥) = −𝑥 + 4
Equation 2
The function f is graphed on the x-y coordinate plane as shown in Figure 12.
Figure 12 Slant height modeled as a linear equation
Partitioning the base into a 4x4 grid means there will be exactly four “L”
regions. I hence mapped the four boxes present on the face of the quarter pyramid
onto my graph, as shown in Figure 13.
As you can see, the heights of each of the respective boxes is given by the
value 𝑓 at the midpoint of bottom side of each box as shown in Figure 14.
I found the midpoints of the bottom side lengths of each of the boxes in their
respective “L” regions simply using my graph, in addition to the fact that midpoints
are 1 unit apart since the side lengths of all of the boxes are 1 unit. I’ll consider the
sum of the volume of the boxes situated in 𝐿" to be 𝑉" .
Finding 𝑉' .
To find the volume of all the boxes in 𝐿' , I need to find the volume of one box
situated in 𝐿' , and hence multiply it by the number of boxes in the region. The
height boxes in 𝐿' are given by 𝑓 (0.5), shown in Figure 14. Since the volume of
rectangular prism is given by the product of its length and width and height, the
volume of any box lying 𝐿' is given by:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿' = (1)(1)𝑓(0.5)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿' = −(0.5) + 4
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿' = 3.5 𝑢𝑛𝑖𝑡 )
To find the number of boxes in 𝐿' we refer back to equation 1:
𝑆" = 2𝑛 − 1
𝑆' = 1
Hence, sum of all the volumes of boxes in 𝐿' is:
𝑉' = (1)(3.5 𝑢𝑛𝑖𝑡 ) )
𝑉' = 3.5 𝑢𝑛𝑖𝑡 )
This process is represented visually in Figure 15
Finding 𝑉( .
We can repeat this process again to find the sum of volumes of boxes situated
in 𝐿( . Again, from figure 14, the height boxes in 𝐿( is given by 𝑓 (1.5).
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿( = (1)(1)𝑓(0.5)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿( = −(1.5) + 4
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿( = 2.5 𝑢𝑛𝑖𝑡 )
To find the number of boxes in 𝐿( we refer back to equation 1:
𝑆" = 2𝑛 − 1
𝑆( = 3
Hence, sum of all the volumes of boxes in 𝐿( is:
𝑉( = (3)(2.5 𝑢𝑛𝑖𝑡 ) )
𝑉( = 7.5 𝑢𝑛𝑖𝑡 )
This process is represented visually in Figure 16
Finding 𝑉) .
We can repeat this process again to find the sum of volumes of boxes situated in 𝐿) .
Again, from figure 14, the height boxes in 𝐿) is given by 𝑓 (2.5).
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿) = (1)(1)𝑓(2.5)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿) = 1.5 𝑢𝑛𝑖𝑡 )
To find the number of boxes in 𝐿( we refer back to equation 1:
𝑆" = 2𝑛 − 1
𝑆) = 5
Hence, sum of all the volumes of boxes in 𝐿( is:
𝑉) = (5)(1.5 𝑢𝑛𝑖𝑡 ) )
𝑉) = 7.5 𝑢𝑛𝑖𝑡 )
This process is represented visually in Figure 17
Finding 𝑉+ .
We can repeat this process again to find the sum of volumes of boxes situated in 𝐿) .
Again, from figure 14, the height boxes in 𝐿+ is given by 𝑓 (2.5).
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿+ = (1)(1)𝑓(3.5)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑏𝑜𝑥 𝑠𝑖𝑡𝑢𝑎𝑡𝑒𝑑 𝑖𝑛 𝐿+ = 0.5 𝑢𝑛𝑖𝑡 )
To find the number of boxes in 𝐿( we refer back to equation 1:
𝑆+ = 2𝑛 − 1
𝑆+ = 7
Hence, sum of all the volumes of boxes in 𝐿+ is:
𝑉+ = (7)(0.5 𝑢𝑛𝑖𝑡 ) )
𝑉+ = 3.5 𝑢𝑛𝑖𝑡 )
This process is represented visually in Figure 18
Approximation of the volume of the quarter pyramid:
The sum of 𝑉' , 𝑉( , 𝑉) , 𝑉+ would reflect an approximation of the volume of the
quarter pyramid, as shown in Figure 19.
Multiplying this sum by 4 would then produce an approximation of the volume of
the entire pyramid, as shown in Figure 20.
Hence, an approximation of the volume of the pyramid is:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑦𝑟𝑎𝑚𝑖𝑑 ≈ 4 ∙ (𝑉' + 𝑉( + 𝑉) + 𝑉+ )
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑦𝑟𝑎𝑚𝑖𝑑 ≈ 4 ∙ (3.5 + 7.5 + 7.5 + 3.5)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑦𝑟𝑎𝑚𝑖𝑑 ≈ 88 𝑢𝑛𝑖𝑡 )
To make sure my approximation made relative sense, I needed to find actual
volume of the pyramid. Using the formula for the volume of right-pyramids:
1
∙𝐴∙ℎ
3
Where 𝐴 is the area of the base, and ℎ is the height
𝑉,!-./ 01,23!4 =
The volume of my pyramid is found in the workings below:
𝑉,!-./ 01,23!4 =
1
∙𝐴∙ℎ
3
𝑉,!-./ 01,23!4 =
1
∙ (8 ∗ 8) ∙ (4)
3
𝑉,!-./ 01,23!4 =
1
∙ (8 ∗ 8) ∙ (4)
3
𝑉,!-./ 01,23!4 = 85.33P
Since the two slant heights of the function have the same function, I was able
to map the prisms representing their respective L regions onto a 2-dimensional
graph, as shown in Figure 13. As you can see, by investigating the L region present
on the quarter pyramid’s base, as well as how the slant height can be modeled using
a function, I was able to simplify my 3D problem into a 2d problem.
Figure 13: How the height of the rectangular prisms could be found using
the slant height function
Since the length of the quarter pyramid is 4 units, and the base will be
partitioned into a 4x4 grid, each prism will have a square base length of 1 unit. I
noticed as well that there were four L regions, so hence, to approximate the volume
of the quarter pyramid, I will need to find the volume of each of the L regions, and
sum them together.
Starting off by finding the volume of the first "L" region, I made sure that I
used my 3D modeling software to help me out. Since I am using the midpoint
method, the respective height of this prism will be given by:
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 "𝐿" 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝑓(𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑏𝑎𝑠𝑒𝑠 𝑖𝑛 1𝑠𝑡 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛)
The midpoint of the first prism(green) is given by a distance of 0.5 units from
the origin, or x=0.5. This is better seen visually in Figure 14. As you can see, by
mapping the prisms on a graph, I was able to map the intersection of the center of
the prisms to the slant height.
As you can see, by mapping the prisms on a graph, I was able to map the
intersection of the center of the prisms to the slant height. Since the slant height is
the same on both faces of the quarter pyramid, the figure above could represent the
viewpoint from either perspective, as shown in Figure XX.
Hence, the height of the prism in the first “L” region is hence given by:
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 "𝐿" 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝑓(0.5)
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 L 𝑟𝑒𝑔𝑖𝑜𝑛 = −(0.5) + 4
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 L 𝑟𝑒𝑔𝑖𝑜𝑛 = 3.5
From the work above, I concluded that the prisms situated in the first L
region in our quarter pyramid, displayed by the color green, will all have a height of
3.5 units.
It is known that the volume of a rectangular prism is given by the product of
its width and its length and height. Hence, I applied that formula using the height I
had just found to find the volume of one of the prisms located in the first L region
(we don’t how many there are yet).
𝑉,56/2"-782, 0,!93 = (𝑙 )(𝑤)(ℎ)
𝑉,56/2"-782, 0,!939 !" :!,9/ ; ,5-!<" = (1 𝑢𝑛𝑖𝑡 )(1 𝑢𝑛𝑖𝑡)(3.5 𝑢𝑛𝑖𝑡)
𝑉,56/2"-782, 0,!93 !" :!,9/ ; ,5-!<" = 3.5 𝑢𝑛𝑖𝑡 )
Commented [MK1]: I really need to fix this part. Trying to
communicate that by mapping the prisms 2 dimensionally
on a grid, their respective widths have coordinates, which
can be inputted to find the height of the function at those
centers.
From the workings shown above, we’ve calculated that all of the rectangular
prisms present in the first L region have a volume of 3.5 unit3.
To find the number of prisms in the 1st L region, we refer back to equation 1,
substituting 1 for n, since n represents the order of the “L” region:
𝑆" = 2𝑛 − 1
𝑆' = 2(1) − 1
𝑆' = 1
From the workings above, I calculated there is exactly 1 prism in the first L
region, which has a height of 3.5 units.
Now to find the entire volume of the first “L” region, I would simply multiply
the volume of one of the rectangular prisms located in the first L region by the
number prisms in the first L region, as shown in the following formula:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 =
(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑝𝑟𝑖𝑠𝑚 𝑖𝑛 𝑡ℎ𝑎𝑡 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛) (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑡ℎ𝑎𝑡 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛)
Equation 3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 = (3.5 𝑢𝑛𝑖𝑡 ) ) (1)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 = 3.5 𝑢𝑛𝑖𝑡 )
The volume of the first L region is 3.5 unit3. The calculation process can be
better understood by viewing the 3D representation of the volume just found.
We repeat the exact same process all over again for the second "L" region to
find its volume.
As you can see in Figure X, the midpoint of the prisms in the second L region
can be 2D dimensional represented by the x coordinate x=1.5. This made sense to
me, as since all the prisms had bases of 1 units in width, their midpoints must also
be 1 unit in width. The midpoint of the 1st L region x=0.5 is in fact unit away from
the midpoint of the second "L" region x=1.5.
Figure X
Hence, now that we know the x coordinate of the center of the second "L" region, we
input that x value into Equation 2 to find the height of the prisms in the second "L"
region.
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑 "𝐿" 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝑓(𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑏𝑎𝑠𝑒𝑠 𝑖𝑛 2"4 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛)
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑 "𝐿" 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝑓(0.5 + 1)
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑 L 𝑟𝑒𝑔𝑖𝑜𝑛 = −(1.5) + 4
ℎ𝑒𝑖𝑔ℎ𝑡 𝑝𝑟𝑖𝑠𝑚𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑 L 𝑟𝑒𝑔𝑖𝑜𝑛 = 2.5 𝑢𝑛𝑖𝑡
With a height of 2.5 units, the volume of one of the prisms located in the
second "L" region can hence be found using the known volume formula.
𝑉,56/2"-782, 0,!93 = (𝑙 )(𝑤)(ℎ)
𝑉,56/2"-782, 0,!939 !" :!,9/ ; ,5-!<" = (1 𝑢𝑛𝑖𝑡 )(1 𝑢𝑛𝑖𝑡)(2.5 𝑢𝑛𝑖𝑡)
𝑉,56/2"-782, 0,!93 !" :!,9/ ; ,5-!<" = 2.5 𝑢𝑛𝑖𝑡 )
Finding the volume of one of the prisms located in the second "L" region, I
next had to find the number of prisms in the second "L" region. I used equation 1
again, as shown in the workings below:
𝑆" = 2𝑛 − 1
𝑆( = 2(2) − 1
𝑆( = 3
Finally, to find the volume of the entire second "L" region, we refer back to
equation 3, inputting the values I calculated from the workings above:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 =
𝑉𝑜𝑙𝑢𝑚𝑒
𝑜𝑓
𝑜𝑛𝑒
𝑝𝑟𝑖𝑠𝑚
𝑖𝑛
𝑡ℎ𝑎𝑡
𝐿
𝑟𝑒𝑔𝑖𝑜𝑛
(𝑁𝑢𝑚𝑏𝑒𝑟
𝑜𝑓
𝑝𝑟𝑖𝑠𝑚𝑠
𝑖𝑛 𝑡ℎ𝑎𝑡 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛)
(
)
Equation 3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 = (2.5 𝑢𝑛𝑖𝑡 ) ) (3)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐿 𝑟𝑒𝑔𝑖𝑜𝑛 = 7.5 𝑢𝑛𝑖𝑡 )
The volume of the second "L" region is 7.5 unit3. I modeled the second "L"
region using the 3D software, correctly color coding it red for it to be distinguishable
from the other “L” regions, as shown in Figure X. Clearly, you can see that the
respective heights of the prisms in the second "L" region are shorter than the prism
in the first “L” region.
Figure X
Knowing that midpoints are 1 unit distances appart, the x coordinate of the
midpoint of the third "l" region can hence be deduced to occur at x = 2.5, and for
the fourth "L" region to be x=3.5. repeating all of the steps again for both
To approximate the volume of a shape using single variable calculus, the
height of the function needs to be modeled by a single variable function. The general
equation for the volume under a surface can be expressed using the following
equation:
"
𝑉𝑜𝑙𝑢𝑚𝑒 𝑢𝑛𝑑𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 2(𝐴)𝑓(𝑥! )
!&'
Choice of Shape to approximate
To begin investigating with these 3D Reimann Sums, I needed a particular
shape to explore. I decided that I will attempt to approximate the volume of a
square pyramid, for a couple of reasons. It’s symmetric, it has a square base which I
assumed would allow for convenient calculations and possibly generalizations. A
square pyramid, unlike a normal pyramid, also has rotational symmetry. In addition,
the formula for a square pyramid is widely known to be:
2
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒 𝑃𝑦𝑟𝑎𝑚𝑖𝑑 = ℎ𝑙 (
3
Where h represents the height of the square pyramid, and l represents the
width of the pyramid. Having a known formula, I could compare my approximation
of the volume of pyramid of a given dimensions to the “real” correct value obtained
by using the equation above, and determine the effectiveness of my method.
In addition, calculating the volume of a square pyramid, much like calculating
the volume of cone or a sphere, is not as intuitive as calculating the volume of a cube
or of a rectangular prism, which hence prompted me to investigate and understand
exactly how the formula for its volume works.
Commented [MK2]: This is the weakest part of the entire
IA. No justification as to why I chose a pyramid. The
reasoning is that other shapes, such as the paraboloid, or a
cone, or a hemisphere, have not height pattern. In that the
height of each individual box does not have some form of
correlation, as in the case for the pyramid. This part needs
some serious work, and is where I will lose the most marks
for reflection.