The oval described seems to be an ellipse, the equation of which is x^2/a^2+y^2/b^2=1 where a and b are the "radii", the semi-major and semi-minor axes.
a=100/2=50 and b=72/2=36 so the equation is x^2/2500+y^2/1296=1.
An approximate answer for the circumference is given by 2π√((a^2+b^2)/2)=273.3 in.
METHOD USING CALCULUS
To find the circumference we need calculus, or something that gives us a good approximation.
First, find the eccentricity, e: b^2=a^2(1-e^2) so 1296=2500(1-e^2), 1-e^2=0.5184, e=√0.4816=0.6940 approx.
An ellipse can be cut into 4 pieces of equal size so we only need the length of the arc of one to find the whole circumference. Consider a point P(x,y) on the ellipse. A small arc (segment of the circumference) is created as x and y change. If x changes by an amount dx and y by dy, then the arc ds=√(dx^2+dy^2) by Pythagoras.
The circumference is the sum of all such ds or ∫ds. ds/dx=√(1+(dy/dx)^2), so ∫ds=∫(√(1+(dy/dx)^2)dx).
x^2/a^2+y^2/b^2=1, x^2/a^2+y^2/(a^2(1-e^2))=1 (1)
y^2/(a^2(1-e^2))=1-x^2/a^2=(a^2-x^2)/a^2 for all ellipses with centre at the origin (0,0).
y^2=a^2(1-e^2)(a^2-x^2)/a^2=(1-e^2)(a^2-x^2), y=√((1-e^2)(a^2-x^2)).
Differentiate (1): 2x/a^2+(2y/(a^2(1-e^2)))dy/dx=0; dy/dx=-x/a^2 * (a^2(1-e^2))/√((1-e^2)(a^2-x^2));
dy/dx=-x√((1-e^2)/(a^2-x^2)) and (dy/dx)^2=x^2(1-e^2)/(a^2-x^2).
1+(dy/dx)^2=1+x^2(1-e^2)/(a^2-x^2)=(a^2-x^2+x^2-e^2x^2)/(a^2-x^2)=(a^2-e^2x^2)/(a^2-x^2).
s=∫ds=∫(√((a^2-e^2x^2)/(a^2-x^2))dx)=∫(√((1-e^2x^2/a^2)/(1-x^2/a^2))dx).
[Let x=asin(t) then dx/dt=acos(t). s=∫((√(1-e^2sin^2(t))/cos(t))dx)=a∫(√(1-e^2sin^2(t))dt. The whole circumference is 4s=4a∫(√(1-e^2sin^2(t))dt. The lower limit for t is x/a when x=0, so t=0; the upper limit is x/a when x=a so t=90° or π/2 radians.]
[a^2=2500, e^2=0.4816, s=∫ds=∫(√((1-0.4816x^2/2500)/(1-x^2/2500))dx)=∫√((1-0.00019264x^2)/(1-0.0004x^2))dx).]
The limits of this integral for a quadrant of the ellipse are x=0 to x=a=50.
Rather than attempting to integrate this directly, we may be able to reduce the expression under the square root to a series:
By algebraic division, (1-e^2(x/a)^2)/(1-(x/a)^2)=1+(x/a)^2(1-e^2)+(x/a)^4(1-e^2)+(x/a)^6(1-e^2)+...
Let z=(x/a)^2(1-e^2)+...+(x/a)^2n(1-e^2)+... =(1-e^2)((x/a)^2+(x/a)^4+(x/a)^6+...).
This is a geometric progression with common ratio (x/a)^2. So z/(1-e^2)=∑(x/a)^2i for 1≤i≤n.
The sum, Sn, to n terms of A+Ar+Ar^2+...+Ar^(n-1), where A is the first term and r is the common ratio is found as follows:
rSn=Ar+Ar^2+Ar^3+...+Ar^n, so rSn-Sn=Ar^n-A and Sn=A(r^n-1)/(r-1).
Put A=R=(x/a)^2: z/(1-e^2)=(x/a)^2((x/a)^2n-1)/((x/a)^2-1).
Unfortunately this series does not converge quickly because x/a starts off at 0 but finishes up as 1.
Nevertheless we can attempt to use it as follows:
s=∫ds=∫((1+z)^½dx)=∫((1+z/2-z^2/8+z^3/16-5z^4/128+...)dx)=
∫((1+(x/a)^2(1-e^2)/2+(x/a)^4(1-e^2)/2+...)-((x/a)^2(1-e^2)+(x/a)^4(1-e^2)+...)^2/8+...)dx).
If we ignore all terms beyond (x/a)^4 we end up with ∫((1+(x/a)^2(1-e^2)/2+(x/a)^4(1-e^2)/2-(x/a)^4(1-e^2)^2/8)dx)=
(x+x^3(1-e^2)/6a^2+x^5(1-e^2)/10a^4-x^5(1-e^2)^2/40a^4) for 0≤x≤50. When x=0, the lower limit, this expression is zero.
This simplifies to a(1+(1-e^2)/6+(1-e^2)/10-(1-e^2)^2/40) when we put x=a, the upper limit.
So putting a=50 and 1-e^2=0.5184 we have s=50(1+0.5184/6+0.5184/10-0.5184^2/40)=
50(1+0.0864+0.05184-0.006718)=56.576 inches.
This is the arc of a quadrant so the circumference of the ellipse is approximately 4*56.576=226.3 inches.
This is shorter than the approximate circumference (273.3") given at the beginning, so there appears to be an error. The error is that we haven't taken sufficient terms, so we need to look for a better approximation method.