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Question Number 20230 by ajfour last updated on 24/Aug/17

Commented by ajfour last updated on 27/Aug/17

$F i n d t h e a r e a o f t h a t p a r t o f t h e$ $s u r f a c e o f t h e s p h e r e :$ $x^{2} + y^{2} + z^{2} = a^{2} w h i c h i s c u t o u t b y$ $t h e s u r f a c e o f t h e c y l i n d e r :$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b) .$
Commented by ajfour last updated on 26/Aug/17

$I {c o u l d n}^{'} t s o l v e t h i s, m e t h o d i s$ $c l e a r t o m e, b u t t h e i n t e g r a l i$ $c a n n o t s i m p l i f y . m r W 1 S i r,$ $p l e a s e h e l p a f t e r i a t t a c h a$ $d i a g r a m f o r i t s s o l u t i o n .$
Commented by ajfour last updated on 26/Aug/17

Commented by ajfour last updated on 26/Aug/17

$L e t A r e a o f s p h e r e o u t s i d e$ $c y l i n d e r b e A .$ $A = 2 \int_{- θ_{0}}^{θ_{0}} r (2 ϕ) a d θ$ $w h e r e a \cos θ_{0} = b$ $r^{2} = x^{2} + y^{2} s u c h t h a t$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 a n d a l s o r = a \cos θ$ $a n d \tan ϕ = \frac{x}{y} .$ $P l e a s e h e l p e v a l u a t i n g t h e i n t e g r a l . .$
Commented by ajfour last updated on 26/Aug/17

$x = y \tan ϕ$ $x^{2} + y^{2} = y^{2} (1 + \tan^{2} ϕ) = r^{2} . . . (i ($ $\frac{y^{2} \tan^{2} ϕ}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $y^{2} (\frac{\tan^{2} ϕ}{a^{2}} + \frac{1}{b^{2}}) = 1 . . . . (i i)$ $d i v i d i n g (i) b y (i i) :$ $\frac{1 + \tan^{2} ϕ}{\frac{\tan^{2} ϕ}{a^{2}} + \frac{1}{b^{2}}} = r^{2} \Rightarrow \tan^{2} ϕ = \frac{(\frac{r^{2}}{b^{2}} - 1)}{(1 - \frac{r^{2}}{a^{2}})}$ $1 + \tan^{2} ϕ = \frac{r^{2} (\frac{1}{b^{2}} - \frac{1}{a^{2}})}{(1 - \frac{r^{2}}{a^{2}})}$ $a s r = a \cos θ$ $\sec^{2} ϕ = \frac{\cos^{2} θ (a^{2} - b^{2})}{b^{2} \sin^{2} θ}$ $\Rightarrow \cos ϕ = \frac{b \tan θ}{\sqrt{a^{2} - b^{2}}}$ $\Rightarrow ϕ = \cos^{- 1} (\frac{b \tan θ}{\sqrt{a^{2} - b^{2}}})$ $A = 8 a \int_{0}^{θ_{0}} r ϕ d θ = 8 a (I)$ $I = \int_{0}^{θ_{0}} a \cos θ \cos^{- 1} (\frac{b \tan θ}{\sqrt{a^{2} - b^{2}}}) d θ$ $= a \sin θ \cos^{- 1} (\frac{b \tan θ}{\sqrt{a^{2} - b^{2}}}) ∣_{0}^{θ_{0}}$ $- \frac{- b}{\sqrt{a^{2} - b^{2}}} \int_{0}^{θ_{0}} \frac{(a \sin θ \sec^{2} θ) d θ}{\sqrt{1 - \frac{b^{2} \tan^{2} θ}{(a^{2} - b^{2})}}}$ $a n d a s a \cos θ_{0} = b$ $\tan θ_{0} = \frac{\sqrt{a^{2} - b^{2}}}{b}, s o$ $I = 0 + a b \int_{0}^{θ_{0}} \frac{\sec θ \tan θ d θ}{\sqrt{a^{2} - b^{2} - b^{2} \tan^{2} θ}}$ $= a \int_{0}^{θ_{0}} \frac{d (b \sec θ)}{\sqrt{a^{2} - b^{2} \sec^{2} θ}}$ $= a \sin^{- 1} (\frac{b \sec θ}{a}) ∣_{0}^{θ_{0}}$ $= a [\sin^{- 1} (\frac{b}{a \cos θ_{0}}) - \sin^{- 1} (\frac{b}{a})]$ $b u t a \cos θ_{0} = b, s o$ $I = a [\frac{π}{2} - \sin^{- 1} \frac{b}{a}]$ $a n d$ $A = 8 a^{2} [\frac{π}{2} - \sin^{- 1} \frac{b}{a}]$ $A = 4 π a^{2} - 8 a^{2} \sin^{- 1} (\frac{b}{a}) .$ $(s t i l l a n s w e r d o n t m a t c h)$ $H e l p!$
	Answered by ajfour last updated on 25/Aug/17

	$A r e a = 4 π a^{2} - 8 a^{2} \sin^{- 1} (\frac{\sqrt{a^{2} - b^{2}}}{a}) .$
	Commented by ajfour last updated on 26/Aug/17

	$T h i s i s t b e a n s w e r i n b o o k .$ $i f b \to a t h e n A r e a s h o u l d t e n d$ $t o z e r o . T h i s i s f u l f i l l e d b y m y$ $a n s w e r .$ $i f b \to 0 A r e a s h o u l d t e n d t o 4 π r^{2},$ $a g a i n t h i s i s f u l f i l l e d b y m y$ $a n s w e r . .$
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