A-circle-has-been-drawn-with-one-unit-opened-compass-on-surface-of-a-sphere-of-one-unit-radius-What-will-be-the-area-of-the-circle-

Question Number 5291 by Rasheed Soomro last updated on 05/May/16

A circle has been drawn with one

unit opened compass on surface

of a sphere of one unit radius .

What will be the area of the circle ?

Commented by Yozzii last updated on 05/May/16

A = 0.75 π

Commented by Yozzii last updated on 05/May/16

P l a c e t h e c o m p a s s o n t h e s p h e r e s u c h

t h a t i t s a m p l i t u d e o f 1 u n i t i s u n a l t e r e d .

C a l l t h e l i n e j o i n i n g t h e t w o p o i n t s

w h e r e t h e c o m p a s s m e e t s t h e s p h e r e

a s A B . S o ∣ A B ∣= 1 . L e t O b e t h e c e n t r e

o f t h e s p h e r e . I t f o l l o w s t h a t ∣ O A ∣=∣ O B ∣= 1

s i n c e p o i n t s A a n d B l i e o n t h e s p h e r e .

B u t, ∣ A B ∣= 1 a l s o . H e n c e, ∣ A B ∣=∣ O A ∣=∣ O B ∣

f o r △ A B O . A n d s o, △ A B O i s e q u i l a t e r a l .

\Rightarrow ∠ B O A = ∠ O A B = ∠ A B O = \frac{π}{3} .

L e t J b e a p o i n t o n t h e l i n e B O s u c h

t h a t A J ⊥ B O . S i n c e △ A B O i s e q u i l a t e r a l,

A J i s a p e r p e n d i c u l a r b i s e c t o r o f B O .

S o, w e h a v e ∠ B A J = \frac{1}{2} ∠ B A 0 = \frac{π}{6}

a n d ∣ A B ∣= 1 . \Rightarrow ∣ B J ∣=∣ A B ∣ c o s ∠ B A J

∣ B J ∣= 1 \times c o s \frac{π}{6} = \frac{\sqrt{3}}{2} . H e n c e, t h e r a d i u s

o f t h e c i r c l e i n q u e s t i o n i s r = \frac{\sqrt{3}}{2} .

U s i n g a r e a, A = π r^{2} \Rightarrow A = \frac{3 π}{4} s q u a r e u n i t s .

Commented by Rasheed Soomro last updated on 05/May/16

T h^{α} n k S! I am trying to understand!

I think that circle on a sphere is

concave surface . . . . Anyway I like your

approach very much . Not only in this

answer but I always like your approach!

Commented by Yozzii last updated on 05/May/16

A h y e s . T h a t s a s e c t i o n o f t h e s u r f a c e

o f t h e s p h e r e . I n t e g r a l C a l c u l u s c a n e a s i l y

d e r i v e t h e s u r f a c e a r e a g i v e n t h e

c e n t r a l a n g l e . {Y o u}^{'} d t h u s h a v e t o

i n r e g r a t e t h e f u n c t i o n y = \sqrt{1 - x^{2}}

b e t w e e n 0.5 ⩽ x ⩽ 1 u s i n g

S = \int_{0.5}^{1} 2 π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x .

T h a t s e c t i o n, t h e c r o s s - s e c t i o n

o f t h e s p h e r e, d e f i n e d b y x^{2} + y^{2} = 1

a n d 0.5 ⩽ x ⩽ 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 . T h e g e o m e t r y t h a t I h a v e

o u t l i n e d i s w h a t h e l p s u s f i n d t h e

l o w e r l i m i t o f x = 0.5 .

Commented by Yozzii last updated on 05/May/16

W h a t m y i n i t i a l a n s w e r g a v e w a s

t h e a r e a o f t h e r e g i o n o f a p l a n a r

c u t o f t h e s p h e r e i n t h e c i r c l e g e n e r a t e d

b y t h e c o m p a s s .

Commented by Rasheed Soomro last updated on 05/May/16

ThankS again!

Commented by Yozzii last updated on 05/May/16

y^{2} + x^{2} = 1 \Rightarrow 2 {y y}^{^{'}} + 2 x = 0 (i m p l i c i t d i f f e r e n t i a t i o n)

\Rightarrow y^{^{'}} = - \frac{x}{y} \Rightarrow {(y^{^{'}})}^{2} + 1 = \frac{x^{2} + y^{2}}{y^{2}} = \frac{1}{y^{2}} (y > 0 f o r 0.5 ⩽ x ⩽ 1)

∴ \sqrt{1 + {(y^{'})}^{2}} = \frac{1}{∣ y ∣} = \frac{1}{y} \Rightarrow y \sqrt{1 + {(y^{^{'}})}^{2}} = 1

∴ S = 2 π \int_{0.5}^{1} d x = π s q u a r e u n i t s .

Commented by Yozzii last updated on 05/May/16

{A r c h i m e d e s}^{'} t o m b s t o n e t h e o r e m i s

a p p l i c a b l e h e r e . S o t h e r e q u i r e d s u r f a c e

a r e a, a f t e r f i n d i n g t h a t ∣ J B ∣= 0.5,

A r e a o f s u r f a c e o f s p h e r e b e t w e e n J a n d B

e q u a l s t h e s u r f a c e a r e a o f t h e s u r f a c e, b e t w e e n

t w o p a r a l l e l p l a n e s c o n t a i n i n g t h e p o r t i o n o f s p h e r e, o f

a c y l i n d e r c i r c u m s c r i b i n g t h e e n t i r e

s p h e r e i n q u e s t i o n

∴ S = 2 π r h

w h e r e r = r a d i u s o f e n t i r e s p h e r e

h =∣ B J ∣= 0.5

\Rightarrow S = 2 π \times 1 \times 0.5 = π s q u a r e u n i t s a s w a s f o u n d

b y i n t e g r a l c a l c u l u s .

T h e P a p p u s - G u l d i n t h e o r e m g i v e s

a n a l t e r n a t i v e b u t p e r h a p s l o n g e r

m e t h o d s i n c e t h e y c o o r d i n a t e o f t h e c e t r o i d o f t h e r e g i o n

b o u n d e d b y x = 0.5, x = 1, y = 0 a n d

x^{2} + y^{2} = 1 n e e d s t o b e f o u n d .

\Rightarrow S = 2 π {\bar{y s}}_{0}

w h e r e s_{0} = a r c l e n g t h o f y = r θ = 1 \times \frac{π}{3} = \frac{π}{3} u n i t s,

f o r 0.5 ⩽ x ⩽ 1 .

Commented by Rasheed Soomro last updated on 06/May/16

LOT OF Knowledge!

LOT OF T han k S!