Question

Find the area of the smaller region bounded by the ellipse and the line

Answer 1

We have to find the area of the smaller region enclosed by the ellipse whose equation is x 2 a 2 + y 2 b 2 =1 , and the line x a + y b =1 . Draw the graphs of the equations and shade the smaller common region.

Figure (1)

The area of the region ACBA is,

Area of the region ACBA=Area of the region OACBO−Area of the region OABO

To find the area bound by the straight line x a + y b =1 with the x-axis, assume a vertical strip of infinitesimally small width and integrate the area.

Area of the region OABO= ∫ 0 a b( 1− x a ) dx =b [ ( x− x 2 2a ) ] 0 a =b[ ( a )− ( a ) 2 2a −( ( 0 )− ( 0 ) 2 2a ) ] = ab 2

Similarly find the area bound by the ellipse with x-axis,

Area of the region OBACO= ∫ 0 a y dx

From the equation of ellipse, find the value of y in terms of x and substitute in above integral.

Area of the region OBACO= ∫ 0 a b ( 1− x 2 a 2 ) dx = b a ∫ 0 ( ( a ) 2 − x 2 ) dx

Use the identity ∫ a 2 − x 2 dx= x 2 a 2 − x 2 + a 2 2 sin −1 x a +C to solve the above equation.

Area of the region OBACO= b a [ x 2 a 2 − x 2 + a 2 2 sin −1 x a ] 0 a = b a [ a 2 a 2 − a 2 + a 2 2 sin −1 a a −( [ 0 2 a 2 − 0 2 + a 2 2 sin −1 0 a ] ) ] = b a [ a 2 2 ( π 2 ) ] = abπ 4

Area of the region ACBA=Area of the region OACBO−Area of the region OABO = abπ 4 − ab 2 = ab 4 ( π−2 )

Thus, the area enclosed by the ellipse whose equation is x 2 9 + y 2 4 =1 , and the line x 3 + y 2 =1 is ab 4 ( π−2 )sq units .