JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 9)
The area of the region $$\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}$$ is :
18
24
20
21
Explanation
The given curves are
$x^2 \leq y, y \leq 8-x^2 ; y \leq 7$
On solving, we get $x^2=8-x^2$
$$ \begin{aligned} & \Rightarrow x^2=4 \\\\ & \Rightarrow x= \pm 2 \end{aligned} $$
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$$ \text { So, area }=2\left[\int_0^4 \sqrt{y} d y+\int_4^7 \sqrt{8-y} d y\right] $$
$$ =2\left\{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4+\left[\frac{-(8-y)^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^7\right\} $$
$$ \begin{aligned} & =2 \times \frac{2}{3}\left\{\left[4^{3 / 2}-0\right]+\left(-(1)^{3 / 2}+(4)^{3 / 2}\right)\right\} \\\\ & =\frac{4}{3}\{8-1+8\}=\frac{4}{3} \times 15=20 \text { sq. units } \end{aligned} $$
$x^2 \leq y, y \leq 8-x^2 ; y \leq 7$
On solving, we get $x^2=8-x^2$
$$ \begin{aligned} & \Rightarrow x^2=4 \\\\ & \Rightarrow x= \pm 2 \end{aligned} $$
$$ \text { So, area }=2\left[\int_0^4 \sqrt{y} d y+\int_4^7 \sqrt{8-y} d y\right] $$
$$ =2\left\{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4+\left[\frac{-(8-y)^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^7\right\} $$
$$ \begin{aligned} & =2 \times \frac{2}{3}\left\{\left[4^{3 / 2}-0\right]+\left(-(1)^{3 / 2}+(4)^{3 / 2}\right)\right\} \\\\ & =\frac{4}{3}\{8-1+8\}=\frac{4}{3} \times 15=20 \text { sq. units } \end{aligned} $$
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