The area of the region (x,y):x^2 y 8-x^2,y 7 is :

Question

The area of the region ${(x,y):{x}^{2}\le y\le 8-{x}^{2},y\le 7}$ is :

Accepted Answer

The given curves are

$x^2 \leq y, y \leq 8-x^2 ; y \leq 7$

On solving, we get $x^2=8-x^2$

$$
\begin{aligned}
& \Rightarrow x^2=4 \\
& \Rightarrow x= \pm 2
\end{aligned}
$$

$$
\text { So, area }=2\left[\int_0^4 \sqrt{y} d y+\int_4^7 \sqrt{8-y} d y\right]
$$

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

$$
\begin{aligned}
& =2 \times \frac{2}{3}\left\{\left[4^{3 / 2}-0\right]+\left(-(1)^{3 / 2}+(4)^{3 / 2}\right)\right\} \\
& =\frac{4}{3}\{8-1+8\}=\frac{4}{3} \times 15=20 \text { sq. units }
\end{aligned}
$$

Solutions · 1