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Question Number 120864 by mnjuly1970 last updated on 03/Nov/20

Commented by math35 last updated on 03/Nov/20

$p l e a s e s i r h o w w a s t h e$ $d i a g r a m m a d e$
Commented by mnjuly1970 last updated on 03/Nov/20

$h e l l o s i r T a n m a y :$ $I t o o k a p h o t o o f t h e c h a r t$ $a n d s e n t i t . u s i n g t h e a t t a c h i m a g e$ $o p t i o n . h a v e a n i c e t i m e . . .$
Commented by TANMAY PANACEA last updated on 03/Nov/20

$p l s c h k m y a n s w e r . . . i d o n o t r e m e m b e r f o r m u l a$
	Answered by TANMAY PANACEA last updated on 03/Nov/20
	$D is origin(0,0) C(−4,0) eqn of big semi circle→x^2 +y^2 =4^2 mid point C and D (−2,0) eqn of small semi circle centre(−2,0) and radius 2 (x+2)^2 +y^2 =2^2 eqn another small semi circle x^2 +(y−2)^2 =2^2 solve x^2 +4x+4+y^2 =4 x^2 +y^2 −4y+4=4 −4x−4y=0→x=−y so x^2 +4x+4+x^2 =4→2x^2 +4x=0 2x(x+2)=0 so point B(−2,2) eqn AB is y=2 point A ...solve y=2 and x^2 +y^2 =4^2 x^2 =16−4=12→x=±2(√3) point A(−2(√3) ,2) required area {∫_(−4) ^(−2(√3)) (√(16−x^2 )) dx+∫_(−2(√3) ) ^(−2) 2dx}−{∫_(−4) ^(−2) (√(4−(x+2)^2 )) dx} pls chk ∫(√(a^2 −x^2 )) dx=(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a)) I_1 =∣(x/2)(√(16−x^2 )) +((16)/2)sin^(−1) ((x/4))∣_(−4) ^(−2(√3) ) =(((−2(√3))/2)(√(16−12)) +8sin^(−1) (((−2(√3))/4)) =−(√3) ×2+8(−(π/3))=−2(√3) −((8π)/3) I_2 =2∣x∣_(−2(√3)) ^(−2) =2(−2+2(√3) )=4(√3) −4 ★∫_(−4) ^(−2) (√(4−(x+2)^2 )) dx (x+2)=p ∫_(−2) ^0 (√(2^2 −p^2 )) dp=∣(p/2)(√(2^2 −p^2 )) +(4/2)sin^(−1) ((p/2))∣_(−2) ^0 =(0)−{((−2)/2)(√(4−4)) +2sin^(−1) (((−2)/2))} =0−0−2sin^(−1) (−1)=−2×((−π)/2)=π rewuired answer I_1 +I_2 −I_3 =(−2(√3) −((8π)/3))+(4(√3) −4)−(π) =2(√3) −4−((11π)/3)$
	$D i s o r i g i n (0, 0) C (- 4, 0)$ $e q n o f b i g s e m i c i r c l e \to x^{2} + y^{2} = 4^{2}$ $m i d p o i n t C a n d D (- 2, 0)$ $e q n o f s m a l l s e m i c i r c l e c e n t r e (- 2, 0) a n d r a d i u s 2$ ${(x + 2)}^{2} + y^{2} = 2^{2}$ $e q n a n o t h e r s m a l l s e m i c i r c l e x^{2} + {(y - 2)}^{2} = 2^{2}$ $s o l v e x^{2} + 4 x + 4 + y^{2} = 4$ $x^{2} + y^{2} - 4 y + 4 = 4$ $- 4 x - 4 y = 0 \to x = - y$ $s o x^{2} + 4 x + 4 + x^{2} = 4 \to 2 x^{2} + 4 x = 0 2 x (x + 2) = 0$ $s o p o i n t B (- 2, 2)$ $e q n A B i s y = 2$ $p o i n t A . . . s o l v e y = 2 a n d x^{2} + y^{2} = 4^{2}$ $x^{2} = 16 - 4 = 12 \to x = \pm 2 \sqrt{3}$ $p o i n t A (- 2 \sqrt{3}, 2)$ $r e q u i r e d a r e a$ ${\int_{- 4}^{- 2 \sqrt{3}} \sqrt{16 - x^{2}} d x + \int_{- 2 \sqrt{3}}^{- 2} 2 d x} - {\int_{- 4}^{- 2} \sqrt{4 - {(x + 2)}^{2}} d x}$ $p l s c h k$ $\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{x}{a})$ $I_{1} =∣ \frac{x}{2} \sqrt{16 - x^{2}} + \frac{16}{2} {s i n}^{- 1} (\frac{x}{4}) ∣_{- 4}^{- 2 \sqrt{3}}$ $= (\frac{- 2 \sqrt{3}}{2} \sqrt{16 - 12} + 8 {s i n}^{- 1} (\frac{- 2 \sqrt{3}}{4})$ $= - \sqrt{3} \times 2 + 8 (- \frac{π}{3}) = - 2 \sqrt{3} - \frac{8 π}{3}$ $I_{2} = 2 ∣ x ∣_{- 2 \sqrt{3}}^{- 2} = 2 (- 2 + 2 \sqrt{3}) = 4 \sqrt{3} - 4$ $★ \int_{- 4}^{- 2} \sqrt{4 - {(x + 2)}^{2}} d x$ $(x + 2) = p$ $\int_{- 2}^{0} \sqrt{2^{2} - p^{2}} d p =∣ \frac{p}{2} \sqrt{2^{2} - p^{2}} + \frac{4}{2} {s i n}^{- 1} (\frac{p}{2}) ∣_{- 2}^{0}$ $= (0) - {\frac{- 2}{2} \sqrt{4 - 4} + 2 {s i n}^{- 1} (\frac{- 2}{2})}$ $= 0 - 0 - 2 {s i n}^{- 1} (- 1) = - 2 \times \frac{- π}{2} = π$ $r e w u i r e d a n s w e r$ $I_{1} + I_{2} - I_{3}$ $= (- 2 \sqrt{3} - \frac{8 π}{3}) + (4 \sqrt{3} - 4) - (π)$ $= 2 \sqrt{3} - 4 - \frac{11 π}{3}$
	Commented by TANMAY PANACEA last updated on 03/Nov/20

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