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Question Number 87296 by ajfour last updated on 03/Apr/20

I f e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

i s r o t a t e d a b o u t 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 mr W last updated on 04/Apr/20

l e t μ = \frac{b}{a}

x = a \cos θ

y = b \sin θ

d x = - a \sin θ d θ

d y = b \cos θ d θ

d s = \sqrt{{(d x)}^{2} + {(d y)}^{2}} = \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ} d θ

S = 2 \times 2 π \int_{0}^{\frac{π}{2}} b \sin θ \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ} d θ

= 4 π a b \int_{0}^{\frac{π}{2}} \sin θ \sqrt{\sin^{2} θ + μ^{2} \cos^{2} θ} d θ

= 4 π a b \int_{\frac{π}{2}}^{0} \sqrt{1 - (1 - μ^{2}) \cos^{2} θ} d (\cos θ)

= 4 π a b \int_{0}^{1} \sqrt{1 - {k t}^{2}} d t w i t h k = 1 - μ^{2}, t = \cos θ

= \dots

= 4 π a b {[\frac{\sin^{- 1} \sqrt{k} t}{2 \sqrt{k}} + \frac{t \sqrt{1 - {k t}^{2}}}{2}]}_{0}^{1}

= 4 π a b [\frac{\sin^{- 1} \sqrt{k}}{2 \sqrt{k}} + \frac{\sqrt{1 - k}}{2}]

= 2 π a b [\frac{\sin^{- 1} \sqrt{1 - μ^{2}}}{\sqrt{1 - μ^{2}}} + μ] w i t h μ = \frac{b}{a}

w i t h a = b \Rightarrow s p h e r e a = b = R, μ = 1

S = 2 π R^{2} (1 + 1) = 4 π R^{2}

Commented by ajfour last updated on 04/Apr/20

E x a c t l y, t h a n k s S i r .

S = 2 π b^{2} + 2 π a b (\frac{\sin^{- 1} λ}{λ})

λ = \frac{\sqrt{a^{2} - b^{2}}}{a} .

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