Find the area of the region bounded by the ellipse `x^2 / a^2 + y^2 / b^2 = 1`

Explanation

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO. Clearly for this region, the limits of integration are 0 and a.
From the equation of the ellipse,
` ( y ^2) /(b^2) = 1 - ( x^2)/ ( a ^ 2 ) = ( a^ 2 - x^2) /(a^(2)) " " therefore y ^(2) = ( b ^(2))/(a ^(2)) ( a ^(2) - x ^(2)) `
In the first quadrant, ` y gt 0 " " therefore y = ( b ) /(a) sqrt ( a ^(2) - x ^(2)) `
` therefore 4 int _ 0 ^ a y dx = 4 int _ 0 ^a ( b )/(a) sqrt ( a ^(2) - x ^(2)) dx `
` = ( 4b) /(a) int _ 0 ^( a ) sqrt ( a ^2 - x ^2) dx `
` = ( 4b ) /(a) [ ( x ) /(2) sqrt ( a ^(2) - x ^(2)) + ( a ^2)/(2) sin ^(-1) (( x ) /(a)) ] _ 0 ^ a `
` = ( 4b) /(a) [ { (a)/(2) sqrt ( a ^2 - a ^2) + ( a^ 2 ) /(2) sin ^ ( -1 ) ( 1) } - { ( 0 ) /(2) sqrt ( a^2 - 0 ) + ( a^2) /(2) sin ^(-1) (0)}] `
` = ( 4b ) /(a) ( (a ^(2))/(2) xx ( pi ) /(2)) = pi ab ` sq units.