Question

Find the area of the circle 4 x 2 + 4 y 2 = 9 which is interior to the parabola x 2 = 4 y

Answer 1

We have to find the area bounded by the circle interior to the parabola. The equation of circle is 4 x 2 +4 y 2 =9 and equation of parabola is x 2 =4y. Draw the graph of these equations and shade the region bounded by the circle and parabola.

Figure (1)

We solve the equation of circle and parabola to find the point of intersections, B and D.

The given equations are 4 x 2 +4 y 2 =9and x 2 =4y.

Substitute 4y for x 2 in equation of circle.

4( 4y )+4 y 2 =9 4 y 2 +16y−9=0

Use quadratic formula to find the solutions to the above quadratic equation.

y 1,2 = −b± b 2 −4ac 2a = −16± 16 2 −4⋅4( −9 ) 2⋅4 = −16±20 8 y= 1 2 ,y=− 9 2

The x coordinate of point B is,

x=2 y =2 1 2 = 2

Draw a perpendicular from point B to x-axis.

From Figure (1), it can be observed that the shaded area is symmetric about y-axis, so the area of BCDO is twice the area of OBCO.

The area of OBCO is the difference of the area bounded by the circle with the x-axis and area bounded by the parabola with the x-axis up to the point M.

The area of OMBCO is calculated as,

Area of the region OMBCO= ∫ 0 2 ydx = ∫ 0 2 ( 9−4 x 2 ) 4 dx

The area of OMBO is calculated as,

Area of the region OMBO= ∫ 0 2 ydx = ∫ 0 2 x 2 4 dx

The equation for the area of OBCO is,

Area of the region OBCO=Area of the region OMBCO−Area of the region OMBO = ∫ 0 2 ( 9−4 x 2 ) 4 dx − ∫ 0 2 x 2 4 dx = 1 2 ∫ 0 2 ( ( 3 ) 2 − ( 2x ) 2 ) dx − 1 4 ∫ 0 2 x 2 dx = 1 2 ∫ 0 2 ( ( 3 ) 2 − ( 2x ) 2 ) dx − 1 12 [ x 3 ] 0 2

Simplify further,

Area of the region OBCO= 1 4 [ x 9−4 x 2 + 9 2 sin −1 2x 3 ] 0 2 − 1 12 [ x 3 ] 0 2 ={ 1 4 [ 2 9−4 ( 2 ) 2 + 9 2 sin −1 2 2 3 −( 0⋅ 9−4 ( 0 ) 2 + 9 2 sin −1 2( 0 ) 3 ) ] − 1 12 [ ( 2 ) 3 ] } = 1 4 [ 2 9−8 + 9 2 sin −1 2 2 3 ]− 2 6 = 2 12 + 9 8 sin −1 2 2 3

The area of OBCDO is twice the area of OBCO,

Area of the region OBCDO=2×Area of the region OBCO =2×( 2 12 + 9 8 sin −1 2 2 3 ) = 2 6 + 9 4 sin −1 2 2 3

Thus, the required area of region OBCDO is ( 2 6 + 9 4 sin −1 2 2 3 ) sq units.