The given equation of ellipse is x 2 4 + y 2 9 =1. Draw the graph of the ellipse and mark the region in the first quadrant as AOBA.
Figure (1)
To calculate the area of the region AOBA, we take a vertical strip in the region with infinitely small width, as shown in the figure above.
To find the area of the region AOBA, integrate the area of the strip.
Area of the region AOBA= ∫ 0 4 ydx (1)
The equation of the ellipse is x 2 4 + y 2 9 =1. From this equation find the value of y in terms of x and substitute in equation (1).
y 2 9 =1− x 2 4 y 2 =9( 1− x 2 4 ) y=3 1− x 2 4
Substitute 3 1− x 2 4 for y in equation (1).
Area of the region AOBA= ∫ 0 2 3 1− x 2 4 dx
From Figure (1), it can be observed that the ellipse is symmetric about x and y-axis. Thus, the area bounded by the ellipse is four times the area AOBA.
Area of the ellipse=4×Area bound by the region AOBA =4 ∫ 0 2 3 1− x 2 4 dx = 4×3 2 ∫ 0 2 4− x 2 dx =6 ∫ 0 2 ( 2 ) 2 − x 2 dx
Further, solve the above integral.
Area of the ellipse=6 [ x 2 ( ( 2 ) 2 − x 2 )+ ( 2 ) 2 2 sin −1 x 2 ] 0 2 =6[ 2 2 ( ( 2 ) 2 − 2 2 )+ ( 2 ) 2 2 sin −1 2 2 −( 0 2 ( ( 2 ) 2 − 0 2 )+ ( 2 ) 2 2 sin −1 0 2 ) ] =6[ 2( 0 )+2 sin −1 ( 1 )−0−2 sin −1 ( 0 ) ] =6[ 2( π 2 ) ]
Further simplify,
Area of the ellipse=6π sq units
Thus, the area bounded by the ellipse is 6π sq units.