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Question Number 25744 by rita1608 last updated on 13/Dec/17

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

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

t h e c y c l o i d x = a (t + s i n t),

y = a (1 + c o s t) a b o u t x a x i s

Answered by ajfour last updated on 14/Dec/17

\frac{d x}{d t} = y \frac{d y}{d x} = \frac{- \sin t}{(1 + \cos t)}

\sqrt{1 + {(\frac{d y}{d x})}^{2}} = \sqrt{1 + \frac{\sin^{2} t}{{(1 + \cos t)}^{2}}}

= \frac{\sqrt{2}}{1 + \cos t}

S = 2 \int_{0}^{π} (2 π y) \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x

= 4 π \int_{0}^{π} {[a (1 + \cos t)]}^{2} [\frac{\sqrt{2}}{1 + \cos t}] d t

= 4 \sqrt{2} π a^{2} \int_{0}^{π} (1 + \cos t) d t

= 4 \sqrt{2} π^{2} a^{2} .

Commented by rita1608 last updated on 14/Dec/17

w h y 2 i s m u l t i p l i e d, i n s o m e

q u e s t i o n s 2 i s n o t m u l t i p l i e d p l s s t e l l

m e w h e n 2 i s m u l t i p l i e d

Commented by ajfour last updated on 15/Dec/17

i f t h e r e e x i s t s a l i n e / p l a n e o f

s y m m e t r t r y w e c a n f i n d o n l y

h a l f t h e l e n g t h / s u r f a c e a r e a /

v o l u m e a n d d o u b l e i t .

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