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Question Number 46103 by rahul 19 last updated on 21/Oct/18

F i n d t h e a r e a e n c l o s e d b e t w e e n c u r v e s

y^{2} (2 a - x) = x^{3} a n d l i n e x = 2 a b o v e t h e

x - a x i s ?

G r a p h i n g c a l c u l a t o r s a r e n o t a l l o w e d . .

Commented by rahul 19 last updated on 21/Oct/18

I^{'} m s t u c k i n d r a w i n g g r a p h o f a b o v e

c u r v e \dots . p l s t e l l s t e p s t o d r a w i t .

Commented by ajfour last updated on 21/Oct/18

a n y i n f o a b o u t a ?

Commented by rahul 19 last updated on 21/Oct/18

S i r, h e r e {i t}^{'} s a + v e c o n s t a n t .

Answered by ajfour last updated on 21/Oct/18

Commented by ajfour last updated on 21/Oct/18

y^{2} (2 a - x) = x^{3}

y^{2} = \frac{x^{3}}{2 a - x} > 0

i f x > 0 \Rightarrow 2 a - x > 0

\Rightarrow x < 2 a

i f x < 0 \Rightarrow 2 a - x < 0

\Rightarrow x > 2 a (n o t p o s s i b l e)

2 y \frac{d y}{d x} = \frac{3 x^{2} (2 a - x) + x^{3}}{{(2 a - x)}^{2}}

2 y \frac{d y}{d x} = \frac{2 x^{2} (3 a - x)}{{(2 a - x)}^{2}}

r e q u i r e d a r e a = \int_{0}^{2} \frac{x \sqrt{x}}{\sqrt{2 a - x}} d x

l e t x = 2 a \sin^{2} θ

f o r x = 2, θ_{0} = \sin^{- 1} \frac{1}{\sqrt{a}}

A r e a = \int_{0}^{θ_{0}} \frac{2 a \sqrt{2 a} \sin^{3} θ (4 a \sin θ \cos θ) d θ}{\sqrt{2 a} \cos θ}

= 2 a^{2} \int_{0}^{θ_{0}} {(1 - \cos 2 θ)}^{2} d θ .

Commented by rahul 19 last updated on 21/Oct/18

Thank you sir ! ��