Let us consider the general case of an ellipse of semi-major axis #a# and semi-minor axis #b#,
Equation of ellipse:#\qquad x^2/a^2+y^2/b^2=1#
Area of the ellipse: Because the ellipse is symmetrical about both the x and y axes, the total area can be written as #4# times the area of the first quadrant,
#A = 4\int_{x=0}^{x=a}\int_{y=0}^{y=b/a\sqrt{a^2-x^2}} dxdy#
#A = 4\int_{x=0}^{x=a}[\int_{y=0}^{y=b/a\sqrt{a^2-x^2}}dy]dx#
#A = 4\int_{x=0}^{x=a}b/a\sqrt{a^2-x^2}.dx#
#A = (4b)/a\int_{x=0}^{x=a}\sqrt{a^2-x^2}dx#
#x\equiv a\sin\theta; \qquad dx = a\cos\thetad\theta#
Limits:
#x=0 \qquad \rightarrow \qquad \theta=0; \qquad x=a \qquad \rightarrow \qquad \theta = \pi/2#
#A = (4b)/a\int_{0}^{\pi/2}[\sqrt{a^2-a^2\sin^2\theta}].a\cos\thetad\theta#
#A = 4ab\int_0^{\pi/2}\cos^2\theta.d\theta = 4ab\int_0^{\pi/2}1/2{1+\cos2\theta}d\theta#
#A = 2ab\int_0^{\pi/2}{1 + \cos2\theta}.d\theta#
#A = 2ab[\pi/2+\int_0^{\pi/2}\cos2\theta.d\theta] = 2ab[\pi/2 + {cancel{(\sin2\theta)/2}_0^{\pi/2}}]#
#A = \piab#
