Question-81485

Question Number 81485 by ajfour last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

I f a r e a o f r e g i o n A i s e q u a l t o

t h a t o f B, f i n d e q . o f p a r a b o l a .

Commented by jagoll last updated on 13/Feb/20

area B = \frac{1}{4} π

Commented by ajfour last updated on 13/Feb/20

h o w c a n y o u b e s o q u i c k, M j S

S i r ?, t h a n k s f o r a n s w e r .

C a n w e h a v e a n e q u a t i o n i n

a y i e l d i n g t h i s a n s w e r, S i r ?

Commented by MJS last updated on 13/Feb/20

approximation leads to

y = {a x}^{2} with a \approx . 427395

Commented by ajfour last updated on 13/Feb/20

a n d {(b - 1)}^{2} + {a b}^{2} = 1

Commented by jagoll last updated on 13/Feb/20

area B = {\int_{0}}^{b} (\sqrt{1 - (1 - x^{2})} - {ax}^{2}) dx = \frac{π}{4}

Commented by jagoll last updated on 13/Feb/20

typo

Answered by mr W last updated on 13/Feb/20

c i r c l e : y = \sqrt{1 - {(x - 1)}^{2}}

p a r a b o l a : y = {A x}^{2}

i n t e r s e c t i o n a t (h, {A h}^{2})

{A h}^{2} = \sqrt{1 - {(h - 1)}^{2}} = \sqrt{h (2 - h)}

\Rightarrow A = \sqrt{\frac{2 - h}{h^{3}}}

\int_{0}^{h} (\sqrt{1 - {(x - 1)}^{2}} - {A x}^{2}) d x = \frac{π}{4}

\int_{- 1}^{h - 1} \sqrt{1 - u^{2}} d u - \frac{{A h}^{3}}{3} = \frac{π}{4}

\frac{1}{2} {[\sin^{- 1} u + u \sqrt{1 - u^{2}}]}_{- 1}^{h - 1} - \frac{{A h}^{3}}{3} = \frac{π}{4}

\frac{1}{2} [\sin^{- 1} (h - 1) + (h - 1) \sqrt{h (2 - h)}] - \frac{h \sqrt{h (2 - h)}}{3} = 0

\Rightarrow \sin^{- 1} (h - 1) = \frac{(3 - h) \sqrt{h (2 - h)}}{3}

\Rightarrow h \approx 1.446798

\Rightarrow A \approx 0.427396

Commented by mr W last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

T h a n k s S i r, p e r f e c t!

Answered by ajfour last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

y e l l o w l i n e m a k e s a n ∠ \approx 63.46 °

w i t h x - a x i s .

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