If y=sin x , x=0 to x=2π is revolved about the x-axis, find the surface of the solid of revolution.


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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
	$A=2∫_0 ^( π) (2πy)ds ds=dx×sec θ tan θ=(dy/dx) = cos x A=4π∫_0 ^( π) (sin x)((√(1+cos^2 x)) dx) let cos x=t ⇒ −sin xdx=dt A=4π∫_(−1) ^( 1) (√(1+t^2 )) dt =8π∫_0 ^( 1) (√(1+t^2 )) dt = 8π{(t/2)(√(1+t^2 ))+(1/2)ln ∣t+(√(1+t^2 ))∣ }∣_0 ^1 = 8π[(1/(√2))+(1/2)ln (1+(√2))] A=4𝛑[(√2)+ln (1+(√2))].$
	$A = 2 \int_{0}^{π} (2 π y) d s$ $d s = d x \times \sec θ$ $\tan θ = \frac{d y}{d x} = \cos x$ $A = 4 π \int_{0}^{π} (\sin x) (\sqrt{1 + \cos^{2} x} d x)$ $l e t \cos x = t \Rightarrow - \sin x d x = d t$ $A = 4 π \int_{- 1}^{1} \sqrt{1 + t^{2}} d t$ $= 8 π \int_{0}^{1} \sqrt{1 + t^{2}} d t$ $= 8 π {\frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \ln ∣ t + \sqrt{1 + t^{2}} ∣} ∣_{0}^{1}$ $= 8 π [\frac{1}{\sqrt{2}} + \frac{1}{2} \ln (1 + \sqrt{2})]$ $A = 4 π [\sqrt{2} + \ln (1 + \sqrt{2})] .$
	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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