Question

Find the area of the region {(x, y) : y² ≤ 8x, x² + y² ≤ 9}.

Answer 1

$$\text{ Let }R=\{(x,y):y^2\le 8x,x^2+y^2\le 9\}$$
$$R_1=\{(x,y):y^2\le 8x\}$$
$$R_2=\{(x,y):x^2+y^2\le 9\}$$
$$\text{ Thus, }R=R_1\cap R_2$$
$$\text{ Now, }{y}^2=8x\text{ represents a parabola with vertex O(0, 0) and symmetrical about }x-\text{ axis }$$
$$\text{ Thus, }{R}_1\text{ such that }{y}^2\le 8x\text{ is the area inside the parabola }$$
$$\text{ Also, }{x}^2+y^2=9\text{ represents with circle with centre O(0, 0) and radius 3 units .}$$
$$\text{ The circle cuts the x axis at C(3, 0) and C'( - 3, 0 ) and }Y-\text{ axis at B(0, 3) and B'(0, - 3 ) }$$
$$\text{ Thus, }{R}_2\text{ such that }{x}^2+y^2\le 9\text{ is the area inside the circle }$$
$$\Rightarrow R=R_1\cap R_2=\text{ Area OACA'O }=2\left(\text{ shaded area OACO }\right)...\left(1\right)$$
The point of intersection between the two curves is obtained by solving the two equations

$$y^2=8x\text{ and }{x}^2+y^2=9$$
$$\Rightarrow x^2+8x=9$$
$$\Rightarrow x^2+8x-9=0$$
$$\Rightarrow (x+9)(x-1)=0$$
$$\Rightarrow x=-9\text{ or }x=1$$
$$\text{ Since, parabola is symmetric about + ve }x-\text{ axis, }x=1\text{ is the correct solution }$$
$$\Rightarrow y^2=8$$
$$\Rightarrow y=\pm 2\sqrt{2}$$
$$\text{ Thus, A}(1,2\sqrt{2})\text{ and A' }(1,-2\sqrt{2})\text{ are the two points of intersection }$$
$$\text{ Area OACO = area OADO + area DACD }...\left(2\right)$$
$$\text{ Area OADO }={\int }_0^1\sqrt{8x}dx.............[\text{ Area bound by curve }{y}^2=8x\text{ between }x=0\text{ and }x=1]$$
$$=2\sqrt{2}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$$
$$\Rightarrow \text{ Area OADO }=\frac{4\sqrt{2}}{3}...\left(3\right)$$
$$\therefore \text{ Area DACD = area bound by }{x}^2+y^2=9\text{ between }x=1\text{ to }x=3$$
$$\Rightarrow A={\int }_1^3\sqrt{9-x^2}dx$$
$$={[\frac{1}{2}x\sqrt{9-x^2}+\frac{1}{2}9{\sin }^{-1}\left(\frac{x}{3}\right)]}_1^3$$
$$=0+\frac{9}{2}si{n}^{-1}\left(\frac{3}{3}\right)-\frac{1}{2}\sqrt{9-1^2}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)$$
$$=\frac{9}{2}si{n}^{-1}1-\frac{1}{2}\sqrt{8}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)$$
$$=\frac{9}{2}\frac{\pi }{2}-\frac{1}{2}2\sqrt{2}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)$$
$$\Rightarrow \text{ Area DACD }=9\frac{\pi }{4}-\sqrt{2}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)...\left(4\right)$$
$$\text{ From }\left(1\right),\left(2\right),\left(3\right)\text{ and }\left(4\right)$$
$$R=\text{ Area OACA'O }$$
$$=2(\frac{4\sqrt{2}}{3}+9\frac{\pi }{4}-\sqrt{2}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right))$$
$$=2(\frac{4\sqrt{2}}{3}-\sqrt{2}+9\frac{\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right))$$
$$\therefore \text{ Area OACA'O }=2(\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right))\text{ sq . units }$$