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Question Number 7314 by Rasheed Soomro last updated on 23/Aug/16

Commented by Rasheed Soomro last updated on 23/Aug/16

$T o p a n d b o t t o m o f a s o l i d a r e c o n g r u e n t c i r c l e s .$ $A b o v e i s t h e c r o s s - s e c t i o n o f t h e s o l i d, w h i c h$ $i s o b t a i n e d b y c u t t i n g t h e s o l i d i n t w o e q u a l p a r t s .$ $F i n d o u t t h e v o l u m e o f t h e s o l i d .$
	Answered by Yozzia last updated on 24/Aug/16
	$Define the circle w with cartesian equation x^2 +(y−r_1 )^2 =r_2 ^2 where r_1 >r_2 >0. ⇒y=r_1 ±(√(r_2 ^2 −x^2 )) {∣x∣≤r_2 }. The problem given can be viewed as finding the volume V of the figure generated when a part of w, whose y coordinates satisfy 0<y≤r_1 , is rotated 2π radians about the x−axis. ∴ y=r_1 −(√(r_2 ^2 −x^2 )) {∣x∣≤r_2 }. Using V=π∫_(−r_2 ) ^( r_2 ) y^2 dx we get V=2π∫_0 ^r_2 (r_1 −(√(r_2 ^2 −x^2 )))^2 dx {y^2 is even}. V=2π∫_0 ^r_2 {r_1 ^2 +r_2 ^2 −x^2 −2r_1 (√(r_2 ^2 −x^2 ))}dx Let x=r_2 sinθ 0≤θ≤2π.⇒dx=r_2 cosθdθ. At x=0,θ=0 and at x=r_2 , θ=(π/2). ∴V=2πr_2 ∫_0 ^(π/2) {r_1 ^2 +r_2 ^2 −r_2 ^2 sin^2 θ−2r_1 (√(r_2 ^2 −r_2 ^2 sin^2 θ))}cosθdθ V=2πr_2 [∫_0 ^(π/2) (r_1 ^2 +r_2 ^2 )cosθ dθ−r_2 ^2 ∫_0 ^(π/2) sin^2 θcosθdθ−2r_1 r_2 ∫_0 ^(π/2) cos^2 θdθ] V=2πr_2 [r_1 ^2 +r_2 ^2 −{((r_2 ^2 u^3 )/3)}_0 ^1 −r_1 r_2 ∫_0 ^(π/2) (1+cos2θ)dθ] V=2πr_2 [r_1 ^2 +r_2 ^2 −(r_2 ^2 /3)−r_1 r_2 {θ+(1/2)sin2θ}_0 ^(π/2) ] V=2πr_2 [r_1 ^2 +(2/3)r_2 ^2 −((r_1 r_2 π)/2)] V=((πr_2 )/3)[6r_1 ^2 +4r_2 ^2 −3πr_1 r_2 ] V=((2πr_2 )/3)(3r_1 ^2 +2r_2 ^2 )−π^2 r_1 r_2 ^2 V=2πr_1 ^2 r_2 +((4πr_2 ^3 )/3)−π^2 r_2 ^2 r_1 −−−−−−−−−−−−−−−−−−−−−−−−−− By Pappus−Guldin theorem, V=2πAy^� where A=area bounded between y=r_1 −(√(r_2 ^2 −x^2 )), lines y=0, x=±r_2 and y^� =distance of centroid of A from axis of rotation (y=0). ∴ y^� =(V/(2πA)). It can be shown that A=((r_2 (4r_1 −πr_2 ))/2). ∴ y^� =((((πr_2 )/3)(6r_1 ^2 +4r_2 ^2 −3πr_1 r_2 ))/(2π((r_2 (4r_1 −πr_2 ))/2))) or y^� =((6r_1 ^2 +4r_2 ^2 −3πr_1 r_2 )/(3(4r_1 −πr_2 )))$
	$D e f i n e t h e c i r c l e w w i t h c a r t e s i a n$ $e q u a t i o n x^{2} + {(y - r_{1})}^{2} = r_{2}^{2} w h e r e r_{1} > r_{2} > 0 .$ $\Rightarrow y = r_{1} \pm \sqrt{r_{2}^{2} - x^{2}} {∣ x ∣⩽ r_{2}} .$ $T h e p r o b l e m g i v e n c a n b e v i e w e d a s f i n d i n g$ $t h e v o l u m e V o f t h e f i g u r e g e n e r a t e d$ $w h e n a p a r t o f w, w h o s e y c o o r d i n a t e s$ $s a t i s f y 0 < y ⩽ r_{1}, i s r o t a t e d 2 π r a d i a n s$ $a b o u t t h e x - a x i s .$ $∴ y = r_{1} - \sqrt{r_{2}^{2} - x^{2}} {∣ x ∣⩽ r_{2}} .$ $U s i n g V = π \int_{- r_{2}}^{r_{2}} y^{2} d x w e g e t$ $V = 2 π \int_{0}^{r_{2}} {(r_{1} - \sqrt{r_{2}^{2} - x^{2}})}^{2} d x {y^{2} i s e v e n} .$ $V = 2 π \int_{0}^{r_{2}} {r_{1}^{2} + r_{2}^{2} - x^{2} - 2 r_{1} \sqrt{r_{2}^{2} - x^{2}}} d x$ $L e t x = r_{2} s i n θ 0 ⩽ θ ⩽ 2 π . \Rightarrow d x = r_{2} c o s θ d θ .$ $A t x = 0, θ = 0 a n d a t x = r_{2}, θ = \frac{π}{2} .$ $∴ V = 2 π r_{2} \int_{0}^{π / 2} {r_{1}^{2} + r_{2}^{2} - r_{2}^{2} {s i n}^{2} θ - 2 r_{1} \sqrt{r_{2}^{2} - r_{2}^{2} {s i n}^{2} θ}} c o s θ d θ$ $V = 2 π r_{2} [\int_{0}^{π / 2} (r_{1}^{2} + r_{2}^{2}) c o s θ d θ - r_{2}^{2} \int_{0}^{π / 2} {s i n}^{2} θ c o s θ d θ - 2 r_{1} r_{2} \int_{0}^{π / 2} {c o s}^{2} θ d θ]$ $V = 2 π r_{2} [r_{1}^{2} + r_{2}^{2} - {\frac{r_{2}^{2} u^{3}}{3}}_{0}^{1} - r_{1} r_{2} \int_{0}^{π / 2} (1 + c o s 2 θ) d θ]$ $V = 2 π r_{2} [r_{1}^{2} + r_{2}^{2} - \frac{r_{2}^{2}}{3} - r_{1} r_{2} {θ + \frac{1}{2} s i n 2 θ}_{0}^{π / 2}]$ $V = 2 π r_{2} [r_{1}^{2} + \frac{2}{3} r_{2}^{2} - \frac{r_{1} r_{2} π}{2}]$ $V = \frac{π r_{2}}{3} [6 r_{1}^{2} + 4 r_{2}^{2} - 3 π r_{1} r_{2}]$ $V = \frac{2 π r_{2}}{3} (3 r_{1}^{2} + 2 r_{2}^{2}) - π^{2} r_{1} r_{2}^{2}$ $V = 2 π r_{1}^{2} r_{2} + \frac{4 π r_{2}^{3}}{3} - π^{2} r_{2}^{2} r_{1}$ $- - - - - - - - - - - - - - - - - - - - - - - - - -$ $B y P a p p u s - G u l d i n t h e o r e m,$ $V = 2 π A \bar{y}$ $w h e r e A = a r e a b o u n d e d b e t w e e n$ $y = r_{1} - \sqrt{r_{2}^{2} - x^{2}}, l i n e s y = 0, x = \pm r_{2}$ $a n d \bar{y} = d i s t a n c e o f c e n t r o i d o f A f r o m$ $a x i s o f r o t a t i o n (y = 0) .$ $∴ \bar{y} = \frac{V}{2 π A} .$ $I t c a n b e s h o w n t h a t A = \frac{r_{2} (4 r_{1} - π r_{2})}{2} .$ $∴ \bar{y} = \frac{\frac{π r_{2}}{3} (6 r_{1}^{2} + 4 r_{2}^{2} - 3 π r_{1} r_{2})}{2 π \frac{r_{2} (4 r_{1} - π r_{2})}{2}}$ $o r \bar{y} = \frac{6 r_{1}^{2} + 4 r_{2}^{2} - 3 π r_{1} r_{2}}{3 (4 r_{1} - π r_{2})}$
	Commented by Rasheed Soomro last updated on 24/Aug/16

	$THANKS! I a m t r y i n g t o u n d e r s t a n d . A t t h e m o m e n t$ $I a m n o t m u c h i n v o l v e d i n a n a l y t i c a p p r o a c h .$
	Commented by Yozzia last updated on 24/Aug/16

	Commented by Yozzia last updated on 24/Aug/16

	$T h e v o l u m e y o u s e e k c a n b e f o u n d$ $b y r o t a t i n g t h e l o w e r s e c t i o n o f t h e$ $c i r c l e x^{2} + {(y - r_{1})}^{2} = r_{2}^{2} 360 ° a b o u t$ $t h e x - a x i s . I h a v e u s e d f u l l i n k t o$ $s h o w t h e l o w e r s e c t i o n a n d t h e u p p e r$ $s e c t i o n i s i n b r o k e n i n k .$
	Commented by Rasheed Soomro last updated on 24/Aug/16

	$TH α n k S A g a i n! T h e g r a p h a n d t h e c o m m e n t$ $i s g r e a t h e l p i n u n d e r s t a n d i n g y o u r a n s w e r!$ $I h a d a n o t h e r i d e a :$ $V o l u m e o f c y l i n d e r - s e m i - c i r c l e g a p e a r o u n d$ $S e m i - c i r c l e g a p e = \frac{π r_{2}^{2}}{2} \times c i r c u m f e r e n c e .$ $I w a s i n c o n f u s i o n w h i c h c i r c l e - c i r c u m f e r e n c e$ $s h o u l d b e u s e d . C i r c l e w i t h r a d i u s r_{1} o r c i r c l e$ $w i t h r a d i u s r_{1} - r_{2} o r a v e r a g e o f t h e b o t h .$ $P e r h a p s I a m n o t s u c c e s s f u l t o c l e a r m y i d e a .$
	Commented by Yozzia last updated on 24/Aug/16

	$- - - - - - - - - - - - - - - - - - - - - - - - -$ $C e n t r o i d o f s e m i c i r c u l a r a r e a i s \frac{4 r_{2}}{3 π} f r o m t h e c e n t e r o f s t r a i g h t e d g e .$ $∴ R e q u i r e d d i s t a n c e = (r_{1} - r_{2}) + (r_{2} - \frac{4 r_{2}}{3 π}) = r_{1} - \frac{4 r_{2}}{3 π}$ $C i r c u m f e r e n c e = 2 π (r_{1} - \frac{4 r_{2}}{3 π}) = 2 π r_{1} - \frac{8 r_{2}}{3} = c$ $v o l u m e = \frac{π r_{2}^{2}}{2} (2 π r_{1} - \frac{8 r_{2}}{3}) = π^{2} r_{1} r_{2}^{2} - \frac{4 π r_{2}^{3}}{3}$ $\Rightarrow R e q u i r e d v o l u m e = 2 π r_{1}^{2} r_{2} - c = 2 π r_{1}^{2} r_{2} + \frac{4 π r_{2}^{3}}{3} - π^{2} r_{1} r_{2}^{2} a s w a s f o u n d .$
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