Question

Find the area of the region bounded by the curve 4x² + y² = 36 using integration. 

Answer 1

Given curve is 4x² + y² = 36 

or, `x^2/9 + y^2/36 = 1`

or, `x^2/3^2 + y^2/6^2 = 1`

Since the ellipse is symmetrical along

x-axis and y-axis

Area of ellipse = Area of ABCD

= 4 × Area of OBC

= `4 xx int_0^3y  dx`

= `4 xx int_0^3 (2sqrt(9 - x^2)) dx`  

= `8int_0^3 sqrt(9 - x^2) dx`    [since OBC is above x-axis]

= `8 int_0^3 sqrt (3^2 - x^2) dx`

= `8[x/2 sqrt(3^2 - x^2) + 3^2/2 sin^(-1)  x/3]_0^3`

= `8[x/2 sqrt(9 - x^2) + 9/2 sin^(-1)  x/3]_0^3`

= `8[0 + 9/2 sin^(-1) (1) - (0 + 0)]`

= `8 xx 9/2 xx pi/2`

= 2 × 9π

= 18π sq. units