Kristy Kemp
Dr. Churchill
Calc 2
April 11th 2023
1) Archimedes bounded π by 3 1/7 and 3 10/71 by inscribing and circumscribing regular
polygons with 96 sides around a circle. What are the bounds on π that you get if you
inscribe and circumscribe squares?
Answer:
Archimedes bounded π by 3 1/7 and 3 10/71 since he used a polygon with 96 sides
inscribed and circumscribed in the circle. The more sides on the polygon circumscribed in the
circle the more accurate and the smaller the bounds will be for the value of π. Since I am only
using squares to be circumscribed and inscribed in the circle the bounds are going to be a bit
larger and the values not as accurate.
Figure 1: Circumscribed Square
Figure 2: Inscribed Square
Since Archimedes told us that the side of the circumscribed square is equal to the diameter of the
circle we know that each side of the inscribed triangle is ½ of that. Using his inscribed and
circumscribed polygons method we can find the upper and lower bounds for π. In this instance
The two sides of the triangle become radii and the equation for a radius of a circle is ½ of the
diameter. We can then use the pythagorean theorem to find the missing side length for AB. When
we calculate this we get about 0.707 which is shown in Figure 3 below this.
Figure 3:
Since 0.707 is equal to side AB but we know that π is equal to all four sides of the inscribed
square. So 4 x 0.707 is how we get the first bound for π at 2.828. To find the upper bound for pi
we now need to use the circumscribed square in order to do so. In Figure 2 is our circumscribed
square along with the diameter labeled MN. Through geometry we can prove that MN is a line
segment since point m is a tangent line we know that the angle for FMC is 90 degrees then we
can prove that line CM is parallel to FG therefore MN is a straight line segment. Since we now
know MN is a line segment and it passes through the center of the circle we can say that it is
equal to the diameter of the circle, which means the length of MN=1. If the length of MN=1 and
we know that FGMN is a rectangle we know that the parallel line of MN; FG is also equal to 1.
Hence, all the sides for the square FGJH are equal to 1. Now the upper bound of π is equal to the
perimeter of the circumscribed circle so we know that 4 x 1=4. Hence we know that the upper
bound for π is 4.
2) “The volume of a sphere is two-thirds of that of its circumscribed cylinder.” Draw a
sphere with radius r, circumscribe a cylinder around it, and translate the words of
Archimedes into an equation. Is it right?
Figure 4:
Answer: Archimedes considered three volumes when brainstorming this theory. He considered
the volumes of a cylinder, a cone and a sphere all having the same base area. Knowing this we
know that the cone and sphere had a diameter of D=2r and height of H=2r. Archimedes was able
to prove that if the volumes of the cone and sphere are added the result is the volume of a
cylinder.
Figure 5:
Figure 6:
After Archimedes proved this, he introduced a method later rediscovered and known as
Cavalieri’s principle. This principle states that if 2 figures have the same height and same cross
sectional area at every point along that height, they shall have the same volume. Using this
principle Archimedes proves that a cone has the same volume as a pyramid of the same base area
and height. He was then able to prove that these three volumes are in the ratio 1:2:3. Thus he
proved that the volume of the sphere is ⅔ the volume of the cylinder.
Figure 7:
Figure 8:
Although when you directly compare the normal V of a sphere equation with the volume
of a sphere equation Archimedes came up with it does not directly seem like his equation holds
any weight. Then you look closer.. In the proof Archimedes claims that the inscribed cone is ⅓
of the volume of the circumscribed cylinder while the volume of the sphere is ⅔ of the
circumscribed cylinder. When looking at any drawing The tip of the cone is intersecting with a
part of the sphere. Once you find the volume of the cone, that part of the sphere is now missing
as it can not be accounted for twice. Using this figure for the volume of the sphere we now can
prove Archimedes was correct when he said that the volume of an inscribed sphere is ⅔ that of
its circumscribed cylinder.
3) “[A sphere’s] surface area is equal to the area of four great circles”. Translate that into an
equation.
Answer:
Archimedes started with a sphere, and similarly to the inscribed sphere and its
circumscribed cylinder Archimedes created an inscribed sphere and circumscribed it with four
circles. So it resembled Figure 9. Now Archimedes realized if he slowly connected the four
circles he circumscribed the sphere with, they would slowly overlap with each other and with the
inscribed sphere as shown in Figure 10.
Figure 9:
Figure 10:
2
Since we know that the area of a circle is π𝑟 and the four “great” circles when hundreds
of them are overlapped become a sphere. Surface area is only used for 3D objects and the 2D
equivalence would be the standard area equation. Since we realize all this we can figure that 4
times the area of a circle should equal the surface area of the sphere. Therefore, the surface area
2
of the sphere is equal to 4π𝑟 .
4) Show that Archimedes was correct when he said “the area of a segment of a parabola is
four-thirds that of its inscribed triangle of greatest area.” Although Archimedes did not
have analytic geometry, we do, so you can use Figure 4.
Answer:
Before Archimedes' time there was not much knowledge about parabolas and no such
thing as analytical geometry. All that was known was that the tangent line at P is parallel to AB,
The straight line through P is parallel to the axis of the parabola that intersects AB at its midpoint
2
M and the equation of the parabola starts with 𝑎𝑥 . Archimedes has not much to go on but he
manages to create a method by which to find the area of a parabolic segment. He created
something called the Method of Exhaustion which is where a triangle or polygon is placed inside
of a parabolic segment until one is found that exhausts or is the closest to the area of the
segment.