Question
Area bounded by the curve $$x{y^2} = {a^2}\left( {a - x} \right)$$ and $$y$$-axis is :
A.
$$\frac{{\pi {a^2}}}{2}{\text{ sq}}{\text{. units}}$$
B.
$$\pi {a^2}{\text{ sq}}{\text{. units}}$$
C.
$$3\pi {a^2}{\text{ sq}}{\text{. units}}$$
D.
None of these
Answer :
$$\pi {a^2}{\text{ sq}}{\text{. units}}$$
Solution :
$$x{y^2} = {a^2}\left( {a - x} \right) \Rightarrow x = \frac{{{a^3}}}{{{y^2} + {a^2}}}$$
The given curve is symmetrical about $$x$$-axis, and meets it at $$\left( {a,\,0} \right).$$
The line $$x = 0,$$ i.e., $$y$$-axis is an asymptote
$$\therefore $$ Area is
$$\eqalign{ & = \int\limits_0^\infty {x\,dy} \cr & = 2\int\limits_0^\infty {\frac{{{a^3}}}{{{y^2} + {a^2}}}dx} \cr & = 2{a^2}\frac{1}{a}\left[ {{{\tan }^{ - 1}}\frac{y}{a}} \right]_0^\infty \cr & = 2{a^2}\frac{\pi }{2} \cr & = \pi {a^2}{\text{ sq}}{\text{. units}} \cr} $$
$$x{y^2} = {a^2}\left( {a - x} \right) \Rightarrow x = \frac{{{a^3}}}{{{y^2} + {a^2}}}$$
The given curve is symmetrical about $$x$$-axis, and meets it at $$\left( {a,\,0} \right).$$
The line $$x = 0,$$ i.e., $$y$$-axis is an asymptote
$$\therefore $$ Area is
$$\eqalign{ & = \int\limits_0^\infty {x\,dy} \cr & = 2\int\limits_0^\infty {\frac{{{a^3}}}{{{y^2} + {a^2}}}dx} \cr & = 2{a^2}\frac{1}{a}\left[ {{{\tan }^{ - 1}}\frac{y}{a}} \right]_0^\infty \cr & = 2{a^2}\frac{\pi }{2} \cr & = \pi {a^2}{\text{ sq}}{\text{. units}} \cr} $$