Write and evaluate an integral involving a function to prove

Write and evaluate an integral involving a function to prove that the formula for the area of an ellipse of semi-major-radius a and semi-minor-radius b is (pi ab) . Show all steps (including the proper technique of integration) clearly and neatly.

Solution

LET the ellipse ((x^2)/(a^2))+((y^2)/(b^2))=1 be divided into 4 equal halves in each quadrant. one such quadrant is OABO where O is origin, OB=b and OA= a Area of the ellipse =4(area of OABO) =4 * f (y) dx where f represents the integral sign ((y/b)^2) = 1-((x/a)^2) y^2 = ((b/a)^2)*(a^2-x^2) y=+(b/a)*((a^2-x^2)^1/2) since y is in the first quadrant area = 4 * f (b/a)*((a^2-x^2)^1/2) dx lower limit=0 upper limit=a => 4*(b/a) f ((a^2-x^2)^1/2) dx => 4*(b/a)*( ((x/2)*((a^2-x^2)^1/2 + ((a^2)/2) * (sin -1 (x/a)) ) (where sin^-1 represents sin inverse) putting the limits :- =>4*(b/a)* ( 0+(a^2)/2 * sin ^-1(a/a) ) - (0+0) = 2(ba) * sin^-1 (1) = (pi)ab sq. units since sin^-1(1) = (pi)/2