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Question Number 215789 by universe last updated on 18/Jan/25

Commented by mr W last updated on 18/Jan/25

	Answered by mr W last updated on 18/Jan/25

	Commented by mr W last updated on 18/Jan/25
	$radius of cylinder r=1 radius of sphere R=2r=2 ρ=2r cos (θ/2) ϕ=(θ/2) R cos φ=ρ=2r cos (θ/2) cos φ=((2r)/(R )) cos (θ/2)=cos (θ/2) ⇒φ=(θ/2) dA=2ρϕRdφ=2r^2 θ cos (θ/2) dθ S_1 =∫_0 ^π 2r^2 θ cos (θ/2) dθ =4r^2 ∫_0 ^π θ dsin ((θ/2)) =4r^2 {[θ sin ((θ/2))]_0 ^π −∫_0 ^π sin (θ/2) dθ} =4r^2 {π+2 [cos (θ/2)]_0 ^π } =4(π−2)r^2 h=ρ tan φ=2r sin (θ/2) dA=hrdθ=2r^2 sin (θ/2) dθ S_2 =2∫_0 ^π 2r^2 sin (θ/2) dθ =8r^2 [cos (θ/2)]_π ^0 =8r^2 S=2πR^2 −S_1 +S_2 +πR^2 −πr^2 =8πr^2 −4(π−2)r^2 +8r^2 +4πr^2 −πr^2 =(7π+16)r^2 =7π+16 ≈37.99 ✓$
	$r a d i u s o f c y l i n d e r r = 1$ $r a d i u s o f s p h e r e R = 2 r = 2$ $ρ = 2 r \cos \frac{θ}{2}$ $φ = \frac{θ}{2}$ $R \cos ϕ = ρ = 2 r \cos \frac{θ}{2}$ $\cos ϕ = \frac{2 r}{R} \cos \frac{θ}{2} = \cos \frac{θ}{2}$ $\Rightarrow ϕ = \frac{θ}{2}$ $d A = 2 ρ φ R d ϕ = 2 r^{2} θ \cos \frac{θ}{2} d θ$ $S_{1} = \int_{0}^{π} 2 r^{2} θ \cos \frac{θ}{2} d θ$ $= 4 r^{2} \int_{0}^{π} θ d \sin (\frac{θ}{2})$ $= 4 r^{2} {{[θ \sin (\frac{θ}{2})]}_{0}^{π} - \int_{0}^{π} \sin \frac{θ}{2} d θ}$ $= 4 r^{2} {π + 2 {[\cos \frac{θ}{2}]}_{0}^{π}}$ $= 4 (π - 2) r^{2}$ $h = ρ \tan ϕ = 2 r \sin \frac{θ}{2}$ $d A = h r d θ = 2 r^{2} \sin \frac{θ}{2} d θ$ $S_{2} = 2 \int_{0}^{π} 2 r^{2} \sin \frac{θ}{2} d θ$ $= 8 r^{2} {[\cos \frac{θ}{2}]}_{π}^{0}$ $= 8 r^{2}$ $S = 2 π R^{2} - S_{1} + S_{2} + π R^{2} - π r^{2}$ $= 8 π r^{2} - 4 (π - 2) r^{2} + 8 r^{2} + 4 π r^{2} - π r^{2}$ $= (7 π + 16) r^{2}$ $= 7 π + 16$ $\approx 37.99 ✓$
	Commented by universe last updated on 18/Jan/25

	$g i v e n a n s i s 8 (\frac{π}{2} - 1)$
	Commented by mr W last updated on 18/Jan/25

	$t h e n w h a t d o e s t h e q u e s t i o n w a n t t o$ $a s k ?$ $8 (\frac{π}{2} - 1) = 4 (π - 2) = S_{1}$ $t h i s i s o n l y t h e a r e a o f t h e s p h e r e$ $p a r t w h i c h i s c u t o f f b y t h e c y l i n d e r .$ $m y a n s w e r i s t h e t o t a l a r e a o f t h e$ $i n t e r s e c t i o n o b j e c t .$ $p l e a s e m a k e c l e a r w h a t t h e q u e s t i o n$ $w a n t s t o a s k!$
	Commented by universe last updated on 18/Jan/25
	Set s only contains those points which satisfy sphere EQ therefore we can't find surface area regarding cylinder we have to find the area regarding sphere Points lying in cylinder will not satisfy sphere eq
	Commented by universe last updated on 18/Jan/25

	$f o r e x a m p l e 0, 0, 1 p o i n t s a t i s f y c y l e n d e r$ $e q u a t i o n b u t 0^{2} + 0^{2} + 1^{2} \neq 4 s o (0, 0, 1) d o e s$ $n o t b e l o n g s t o s e t S$
	Commented by mr W last updated on 18/Jan/25

	$i h a v e c a l c u l a t e d S_{1} a n d S_{2} . w i t h$ $t h e m o n e c a n f i n d t h a t w h a t h e$ $w a n t s t o k n o w . y o u c a n t a k e S_{1} a s$ $a n s w e r .$
	Commented by mr W last updated on 18/Jan/25

	Commented by universe last updated on 18/Jan/25

	$t h a n k u s i r$
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