JEE Main 2023 — Mathematics Calculus

The given curves are ${x}^{2}\leq y,y\leq 8-{x}^{2}\text{and}y\leq 7$x2≤y,y≤8−x2andy≤7

The point of intersection of the curves ${x}^{2}\leq y\text{and}y\leq 8-{x}^{2}$x2≤yandy≤8−x2 is obtained by,

$$\Rightarrow {x}^{2}=8-{x}^{2}$$⇒x2=8−x2

$\Rightarrow x=\pm 2$⇒x=±2 and $y=4$y=4.

Hence the points are $(2,4),(-2,4)$(2,4),(−2,4).

The points of intersection of the curves $y\leq 8-{x}^{2}\text{and}y\leq 7$y≤8−x2andy≤7 is obtained by,

$$\Rightarrow 8-{x}^{2}=7$$⇒8−x2=7

$\Rightarrow x=\pm 1$⇒x=±1 and $y=7$y=7

Hence the points are $(1,7),(-1,7)$(1,7),(−1,7).

The required graph is

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

Hence required area is $\displaystyle A=2[{\int }_{0}^{4}\sqrt{y}dy+{\int }_{4}^{7}\sqrt{8-y}dy]$A=2[∫04​y​dy+∫47​8−y​dy]

$$A=2[{[\frac{{y}^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{4}-{[\frac{{(8-y)}^{\frac{3}{2}}}{\frac{3}{2}}]}_{4}^{7}]$$A=2[[23​y23​​]04​−[23​(8−y)23​​]47​]

$$=\frac{4}{3}[{(4)}^{\frac{3}{2}}-{(0)}^{\frac{3}{2}}-{{(8-7)}^{\frac{3}{2}}-{(8-4)}^{\frac{3}{2}}}]$$=34​[(4)23​−(0)23​−(8−7)23​−(8−4)23​]

$$=\frac{4}{3}(8-0-1+8)$$=34​(8−0−1+8)

$=20$=20 sq units.

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