Question

the area bounded by the curve x2/ a2 + y2 /b2 =1

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

represents the equation of ellipse intersects the x- axis at (a, 0) and (- a, 0) and intersects the y - axis at (0, b) and (0, - b)

Again,

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

$\rm :\longmapsto\:\dfrac{ {y}^{2} }{ {b}^{2} } = 1 - \dfrac{ {x}^{2} }{ {a}^{2} }$

$\rm :\longmapsto\:\dfrac{ {y}^{2} }{ {b}^{2} } = \dfrac{ {a}^{2} - {x}^{2} }{ {a}^{2} }$

$\rm :\longmapsto\: {y}^{2} = {b}^{2} \bigg[\dfrac{ {a}^{2} - {x}^{2} }{ {a}^{2} }\bigg]$

$\bf\implies \:\boxed{ \tt{ \: y = \frac{b}{a} \sqrt{ {a}^{2} - {x}^{2}} \: \: }}$

So, required area bounded by the given curve is

$\rm \: = \:4\displaystyle\int_0^a \: ydx$

$\rm \: = \:4\displaystyle\int_0^a \: \frac{b}{a} \sqrt{ {a}^{2} - {x}^{2} } dx$

$\rm \: = \:\dfrac{4b}{a}\displaystyle\int_0^a \sqrt{ {a}^{2} - {x}^{2} } \: dx$

$\rm \: =\dfrac{4b}{a}\bigg[\dfrac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{x}{a} \bigg]_0^a$

$\rm \: =\dfrac{4b}{a}\bigg[\dfrac{a}{2} \sqrt{ {a}^{2} - {a}^{2} } + \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{a}{a} - 0 - {sin}^{ - 1}0 \bigg]$

$\rm \: =\dfrac{4b}{a}\bigg[\dfrac{ {a}^{2} }{2} {sin}^{ - 1}1\bigg]$

$\rm \: =\dfrac{4b}{a}\bigg[\dfrac{ {a}^{2} }{2} \times \dfrac{\pi}{2} \bigg]$

$\rm \: = \:\pi \: ab \: square \: units$