Class 12

Math

Calculus

Application of Integrals

Find the area of the circle $4x_{2}+4y_{2}=9$ which is interior to the parabola $x_{2}=4y$.

Solving the given equation of circle, $4x_{2}+4y_{2}=9$, and parabola, $x_{2}=4y$, we obtain the point of intersection as $B(2 ,21 )$ and $D(−2 ,21 )$.

It can be observed that the required area is symmetrical about $y$-axis.

$∴$ Area $OBCDO$ $=2×AreaOBCO$

We draw $BM$ perpendicular to $OA$.

Therefore, the coordinates of M are $(2 ,0)$.

Therefore, Area $OBCO$ $=$ Area $OMBCO$ - Area $OMBO$

$=∫_{0}4(9−4x_{2}) dx−∫_{0}4x_{2} dx$

$=21 ∫_{0}9−4x_{2} dx−41 ∫_{0}x_{2}dx$

$=41 [x9−4x_{2} +29 sin_{−1}32x ]_{0}−41 [3x_{3} ]_{0}$

$=41 [2 9−8 +29 sin_{−1}322 ]−121 (2 )_{3}$

$=42 +89 sin_{−1}322 −62 $

$=122 +89 sin_{−1}322 $

$=21 (62 +49 sin_{−1}322 )$

Therefore, the required area $OBCDO$ is

$(2×21 [62 +49 sin_{−1}322 ])=[62 +49 sin_{−1}322 ]$ sq. units