Question

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


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Answer 1

Answer:

Let the given equations

Equation of the circle is

4x 2 +4y 2 =9 ------ (i)

And equation of parabola is

x 2 =4y

y= 4x 2 ---- (ii)

From (i) and (ii)

4x 2 +4( 4x2 ) 2 =9x 4 +16x 2 −36=0

(x 2 +18)(x 2−2)=0

x=± 2

= −18

Required area

=2∫ 0 2 (y 1 −y 2)dx

=2[∫ 0 2 { 49 −x 2 − 4x 2 }dx](y

1 :x 2 +y 2 ,y 2 :x 2 =4y)=2

2x49 −x 2 + 89 sin −123x − 12x 3

02=2[ 42 + 89 sin −1322 − 62 ]

=( 62 + 49 sin −132 2)sq.units