Area of a Parabolic Segment
Archimedes of Syracuse lived from roughly 287 BC to 212 BC and spent much of his life in Alexandria and Syracuse. He devoted his life almost entirely to mathematics. Although the true nature of his death is not known, three different stories each share a common theme as to what Archimedes was doing at his time of death. One story explains that during the capture of Syracuse a soldier slew Archimedes after he refused to leave with him until he had completed his meditation on a mathematical problem. Another version of this story is that a soldier drew his sword to kill Archimedes when Archimedes begged him to wait until he solved the current problem he was working on. However, while he was begging, another soldier killed him anyway. Another story suggests Archimedes left with that soldier but found it necessary to take his mathematical instruments with him which another soldier mistook for gold and thus killed him for it. The final and most probable story explains that during the capture of Syracuse Archimedes was slain by an invading soldier who merely did not know who Archimedes was and how important he was to the world. Whatever his death, it is not unlikely that Archimedes died doing what he devoted his life to, mathematics.

Archimedes utilized the method of exhaustion in proving many of his propositions or theorems such as those which led up to his geometrical proof about the value of the area of a parabolic segment. In this proof, Archimedes exhausts the parabola by drawing an infinite number of triangles within the parabola. The area of the parabola is then the sum of the area of each triangle. This is obviously an infinite sum. However, Greek mathematicians never liked to mention things that were infinitely small or infinitely large. Thus, Archimedes expressed this value in a more appropriate way that would be accepted. However, his method of finding the area of a parabola was a prelude to his anticipation of integral calculus.

Actually, Archimedes proved what the area under a parabola was two different ways. In Propositions 1 - 17 of Quadrature of the Parabola, Archimedes used the following method of finding the area of a parabola:


First, draw a line from the maximum height of the parabola to the midpoint of its base.

How do you determine P? (I know it's the point on the parabolic arc that farthest away from the base qQ, but how would you get that point?)

These pictures should be generated with Maple. Here are some sample pictures.

> f := x -> x^2: a:=-1: b := 2:

> parab := plot(f(x),x=a..b,thickness=2,color=blue):

> seg := plot([[a,f(a)],[b,f(b)]],thickness=2,color=magenta):

> secslope := (f(b)-f(a))/(b-a):

> triang := plots[polygonplot]([[a,f(a)],[(a+b)/2,f((a+b)/2)],
[b,f(b)]],color=red):

> plots[display]([parab,seg,triang],scaling=constrained,axes=none);

> drawit := proc(a,b,clr)
local f,parab,seg,triang;
f := x -> x^2;

> parab := plot(f(x),x=a..b,thickness=2,color=blue):

> seg := plot([[a,f(a)],[b,f(b)]],thickness=2,color=magenta):

> triang := plots[polygonplot]([[a,f(a)],[(a+b)/2,f((a+b)/2)],[b,f(b)]],
color=clr):
plots[display]([parab,seg,triang],scaling=constrained,axes=none);
end;

> exhaustit := proc(a,b,n,clrs)
movie := drawit(a,b,clrs[1]):
frame := movie:
h := b-a:
for i from 2 to n do
frame := plots[display]([frame,seq(drawit(a+j*h/2^(i-1),
a+(j+1)*h/2^(i-1),clrs[i]),j=0..2^(i-1)-1)]):
movie := movie,frame od;
plots[display]([movie],scaling=constrained,insequence=true) end;

Warning, `movie` is implicitly declared local

Warning, `frame` is implicitly declared local

Warning, `h` is implicitly declared local

Warning, `i` is implicitly declared local

> exhaustit(-1,1.4,4,[yellow,red,magenta,blue]);

>

>

So, qV = VQ. Next, draw a line that is tangent to Q and extent VP until both lines intersect.

Now, PV = PT. Next, draw a line parallel to PV that includes q. Extend QT until the two lines intersect.



Since qV = VQ and VT is parallel to qE, VT = ˝ Eq. Finally, draw a triangle, QqP inside the parabola.

Proposition 16 states that the area of parabola is 1/3 the area of the triangle EqQ (area of parabola = 1/3 EqQ). Archimedes proves this by contradiction. First, he proves that the area under the segment could not be greater than 1/3 EqQ. Then, he proves that the area could not be less than 1/3 EqQ. Therefore, it must be 1/3 EqQ. His method of disproving each instance comes from propositions 6 - 15, which deal with the principles of levers. These propositions are very difficult to understand and also very, very lengthy so the proof will not be restated here. Once the area of the parabola is established to be 1/3 EqQ, the area of the parabola can be found to be 4/3 PqQ as follows:

> PqQ = 1/2*Qq*PV;

> PV = PT ;

> PV = 1/2*VT;

> PqQ = 1/2*Qq* (1/2*VT);

> VT = 1/2* qE;

> PqQ= 1/2* Qq* 1/2* ( 1/2 * qE);

> PqQ= 1/4*( 1/2* Qq * qE);

In the above equations, you notice that 1/2*Qq*qE equals the area of the triangle.

> PqQ= 1/4 * QqE;

> QqE = 4*PqQ;

From Proposition 16,
(area of parabola) = 1/3 QqE

> A:=1/3*QqE;

Substituting QqE = 4* PqQ,
*(area of parabola) = 4/3 * PqQ

> A:=4/3*PqQ;

Archimedes second way of proving that the (area of parabola) = 4/3 * PqQ, is as follows:

First, draw the same parabola as before with a point P at the maximum height and V bisecting qQ. Then, bisect Vq and VQ with lines that extend parallel to PV.

In proposition 21, Archimedes proves that the area of the triangle PQq = 8 times the area of the triangle PRQ (PQq = 8PRQ). In the following proposition, Archimedes showed that PQq = 8(Pqr) = 4(PQr + Pqr) since PQr = Pqr. In other words, the area of the triangle inscribed in the parabolic segment is eight times the area of the triangle inscribed by the parabola with one of the sides of the triangle as the base of the new parabola. Each time a new triangle is drawn, the side of the triangle can be the base of a new parabola and the previous relationship is attained. This is done an infinite amount of times thus exhausting the parabola with triangles.

It is interesting that Archimedes next step was to show that if ’Given a series of areas of A, B, C, D, … Z, of which A is the greatest, and each is equal to four times the next in order, then A+B+C+…+Z+1/3Z = 4/3A.´ Graphically, this looks as follows:

So by propositions 21 and 22, the area of the parabola is 4/3 PQq which is actually the infinite sum of the areas of triangles, each of which inscribe the parabola. However, Archimedes represented this infinite series as 4/3 PQq, which was easier to compute, and a more acceptable equation form for the time.

Today, we have found an much easier way to determine the area of a parabola. With closers examination, we see that the parabola is nothing more than a curve which can be integrated in merely a few steps.

> plot([-x^2+4,0], x=-2..2, color=[red,blue], axes=none);

Instead of going through all the steps that Archimedes performed with the triangles, we simply construct what seems to be an infinate series of rectangles that help us produce an approximation. This approximation is based on the idea of eliminating space under the curve and derviving the area. The formula today goes as follows.

> eqn:=int(F(x),x=-2..2);

> F(x)=-x^2+4;

The above equations show that the curve goes from -2 to 2 and that the equation of the curve is -x^2+4. From here we then perform the calculus on the equation.

> F(x)=-1/3*x^3+4*x;

Then you simply take the new equation with repsect to the bounds, or in Archimedes case, the length of the base of his triangle. When we perform our computations, we find that the area of the parabola turns out to be 32/3. With Archimedes ideas, you would take 4/3 of the area of the triangle. In this case, the base of the triangle would be 4 and the height would be 4.

> plot([-x^2+4,0,-2*x+4,2*x+4], x=-2..2,y=0..4,color=[red, green,blue,blue]);

> base=4;

> height=4;

> area:=1/2*base*height;

> area:=1/2*4*4;

Now we simply take 4/3 of that area.

> 4/3*area;

As you can see, the quantity comes out to be the same.

You can use Maple to do the more general calculation, for any a < b.

> restart;

> area_of_parabolic_seg := factor((b-a)*(b^2+a^2)/2- int(x^2,x=a..b) );

> base := sqrt((b^2-a^2)^2 + (b-a)^2);

> v := [(b-a)/2,((b+a)/2)^2-a^2];

> u := [b-a,b^2-a^2];

Use some vector algebra to get the height of the triangle

> height := sqrt((v[1]^2+v[2]^2)-(v[1]*u[1]+v[2]*u[2])^2/(u[1]^2+u[2]^2))\;

> sqr_area_of_triangle := (1/2*height*base)^2;

> simplify(%);

> factor(%);

> area_of_triangle := 1/8*(b-a)^3;

> area_of_parabolic_seg/area_of_triangle;

Voila!

>

The methods of today and the way that Archimedes calculated the area of a parabola are not that far apart. Much of what we use to derive our answer comes from what he did. Archimedes used triangles and bisectors to prove his theories. We today use rectangles and approximations to derive ours. Through his work, our ability to calculate the area of a parabola is rather simple compared to what he performed. His 24 propositions proved what he stated as being the area, the 4/3 of the inscribed triangle. Each of the propostition were detailed and oftened called upon one another to support themselves. When we take our calculations and plug them into what he determined, we get the same results.

Need references. Obviously, you went back to the original paper of Archimedes here.