Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $x^2\le y, y\le 8-x^2 \text{and} y\le 7$

The point of intersection of the curves $x^2\le y \text{and} y\le 8-x^2$ is obtained by,

$\Rightarrow x^2=8-x^2$

$\Rightarrow x=\pm 2$ and $y=4$.

Hence the points are $\left(2,4\right),\left(-2,4\right)$.

The points of intersection of the curves $y\le 8-x^2 \text{and} y\le 7$ is obtained by,

$\Rightarrow 8-x^2=7$

$\Rightarrow x=\pm 1$ and $y=7$

Hence the points are $\left(1,7\right),\left(-1,7\right)$.

The required graph is 

The required area is symmetrical about $y-$axis.

Hence required area is $A=2\left[{\int }_0^4\sqrt{y}dy+{\int }_4^7\sqrt{8-y}dy\right]$

$A=2\left[{\left[\frac{y^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^4-{\left[\frac{{\left(8-y \right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_4^7\right]$

$=\frac{4}{3}\left[{\left(4\right)}^{\frac{3}{2}}-{\left(0\right)}^{\frac{3}{2}}-\left\{{\left(8-7\right)}^{\frac{3}{2}}-{\left(8-4\right)}^{\frac{3}{2}}\right\}\right]$

$=\frac{4}{3}\left(8-0-1+8\right)$

$=20$ sq units.

Hence, the required area is $20$ sq units.