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Question Number 87298 by ajfour last updated on 03/Apr/20

I f y = \sin x, x = 0 t o x = 2 π i s

r e v o l v e d a b o u t t h e x - a x i s, f i n d

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

r e v o l u t i o n .

Answered by ajfour last updated on 04/Apr/20

A = 2 \int_{0}^{π} (2 π y) d s

d s = d x \times \sec θ

\tan θ = \frac{d y}{d x} = \cos x

A = 4 π \int_{0}^{π} (\sin x) (\sqrt{1 + \cos^{2} x} d x)

l e t \cos x = t \Rightarrow - \sin x d x = d t

A = 4 π \int_{- 1}^{1} \sqrt{1 + t^{2}} d t

= 8 π \int_{0}^{1} \sqrt{1 + t^{2}} d t

= 8 π {\frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \ln ∣ t + \sqrt{1 + t^{2}} ∣} ∣_{0}^{1}

= 8 π [\frac{1}{\sqrt{2}} + \frac{1}{2} \ln (1 + \sqrt{2})]

A = 4 π [\sqrt{2} + \ln (1 + \sqrt{2})] .

Commented by mr W last updated on 04/Apr/20

y o u a r e r i g h t s i r! t h a n k s!

Commented by ajfour last updated on 04/Apr/20

Y o u m i g h t l i k e Q . 87296 S i r .

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