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Find the area of the circle 4x^2+4y^2=9 ...

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

Text Solution

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Let the given equations Equation of the circle is `4x^2+4y^2=9`...(i)
And equation of parabola is `x^2=4y`
`y = x^2/(4)`....(ii)
from equation (i) and (ii)
`4x^2 +4(x^2/(4) )^2 =9`
`x^4 +16x^2 -36 =0`
`(x^2+18)(x^2-2)=0`
`x= sqrt(2) =sqrt(-18)`
Required area `= 2int_0^sqrt(2) (y_1 -y_2)dx`
`2[ int_0^sqrt(2) { sqrt(9/(4)-x^2} - (x^2)/(4)} dx]`
`= 2[ (x/2) sqrt((9/(4)-x^2 ))+9/(8) sin^-1((2x)/(3)) - x^3/(12)]_0^(sqrt2)`
solve the limits we get A`=[ sqrt(2)/(6) +9/(4) sin^-1(( 2sqrt2)/3)]` sq. units
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