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

Explanation

Given equation of circle is `4x^(2)+4y^(2)=9` and equation of parabola is `x^(2)=4y`.

Now, solving the equations of two curves,
`4(4y)+4y^(2)=9`
` implies4y^(2)+16y-9=0`
`impliesy=(-16+-sqrt(256+144))/(2xx4)`
`= (-16+-sqrt(400))/(8)`
`=(-16+-20)/(8)=(1)/(2),-(9)/(2)`
but y gt 0, so two curves will meet when `y=(1)/(2)`,
` :. x^(2) =4xx(1)/(2)=2`
`implies x=+-sqrt(2)`
` :." Required area"=2int_(0)^(sqrt(2)) (y_(2)-y_(1))dx`
`={int_(0)^(sqrt(2))sqrt((9-4x^(2))/(4))dx-int_(0)^(sqrt(2))(x^(2))/(4)dx}`
`=2int_(0)^(sqrt(2))sqrt(((3)/(2))^(2)-x^(2))dx-(2)/(4)[(x^(3))/(3)]_(0)^(sqrt(2))`
`=2[(x)/(2)sqrt(((3)/(2))^(2)-x^(2))+((3//2)^(2))/(2)"sin"^(-1)((x)/(3//2))]_(0)^(sqrt(2))-(1)/(6)[(sqrt(2))^(3)-0]`
`=[xsqrt((9)/(4)-x^(2))+(9)/(4)"sin"^(-1)((2x)/(3))]_(0)^(sqrt(2))-(2sqrt(2))/(6)`
`=[sqrt(2)sqrt((9)/(4)-2)+(9)/(4)"sin"^(-1)((2sqrt(2))/(3))]-{0-0}-(2sqrt(2))/(6)`
`=(sqrt(2))/(2)+(9)/(4)"sin"^(-1)((2sqrt(2))/(3))-(2sqrt(2))/(6)`
`=((sqrt(2))/(2)-(sqrt(2))/(3))+(9)/(4)"sin"^(-1)((2sqrt(2))/(3))`
`=(sqrt(2))/(6)+(9)/(4)"sin"^(-1)((2sqrt(2))/(3))` sq. units.