# find area of a surface

label Calculus
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schedule 1 Day
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May 24th, 2015

x=(e^y+e^-y)/2

l∫ 6* π * x * ds

ds = √( 1 + (dx/dy)^2 )

dx/dy = (1/2) [ e^y - e^-y ] = (1/2) [ e^y - (1 / e^y) ] = (1/2) [ (e^2y -1) / e^y) ]

( dx/dy )^2 = [ (1/2) [ (e^2y -1) / e^y) ] ]^2 = [ (1/4) * [ (e^2y -1)^2 / (e^y)^2 ]

( dx/dy )^2 = [ (1/4) * [ (e^4y - 2e^2y + 1) / (e^2y) ]

1 + ( dx/dy )^2 = 1 + [ [ (e^4y - 2e^2y + 1) / (4e^2y) ] ====> LCD

1 + ( dx/dy )^2 = [ 1*(4e^2y) / (4e^2y) ] + [ [ (e^4y - 2e^2y + 1) / (4e^2y) ]

1 + ( dx/dy )^2 = [ 4e^2y / 4e^2y ] + [ [ (e^4y - 2e^2y + 1) / (4e^2y) ]

1 + ( dx/dy )^2 = [ 4e^2y + e^4y - 2e^2y + 1 / (4e^2y) ]

1 + ( dx/dy )^2 = [ 2e^2y + e^4y + 1 / (4e^2y) ]

1 + ( dx/dy )^2 = [ e^4y + 2e^2y + 1 / (4e^2y) ]

1 + ( dx/dy )^2 = [ (e^2y + 1)^2 / (4e^2y) ]

√( 1 + (dx/dy)^2 ) = √( [ (e^2y + 1)^2 / (4e^2y) ] )

√( 1 + (dx/dy)^2 ) = +/- (e^2y + 1) / (2e^y) ===> we are going to take the positive because our limit starts from 0

√( 1 + (dx/dy)^2 ) = (e^2y + 1) / (2e^y)

ln(6)
∫ 2 * π * x * ds

ln(2)
∫ 2 * π * [ (e^y+e^-y) / 2 ] * [ (e^2y + 1) / (2e^y) ] dy

ln(2)
∫ 2 * π * [ e^y(1+e^-2y) / 2 ] * [ (e^2y + 1) / (2e^y) ] dy

ln(2)
∫ 2 * π * [ (1+e^-2y) / 4 ] * [ (e^2y + 1) ] dy

ln(2)
∫ π * [ (1+e^-2y) * (e^2y + 1) / 2 ] dy

ln(2)
∫ π * [ (e^2y + 1 + 1 + e^-2y) / 2 ] dy

ln(2)
∫ π *(1/2) * (e^2y + 2 + e^-2y) dy

. . .. . . .. . . . . . .. ..... . .. . . .. . .. . . . . .. . . .ln(2)
π *(1/2) * [ (1/2)e^2y + 2y + (-1/2)e^-2y ]
. .. . . . .. . . .. . . . . .. . . . . .. . . .. . . .. . . .. 0

π *(1/2) * [ (1/2) * [e^2(ln(2)) - e^2*0] + 2[ln(2)-0] + (-1/2) [ e^-2(ln(2)) - e^-2(*0) ] ]

π *(1/2) * [ (1/2) * [e^(ln(2^2)) - 1] + 2ln(2) + (-1/2) [ e^(ln(2^-2)) - 1] ]

π *(1/2) * [ (1/2) * [ 2^2 - 1] + 2ln(2) + (-1/2) [ 2^-2 - 1] ]

π *(1/2) * [ (1/2) * [ 3] + 2ln(2) + (-1/2) [ (1/4) - 1] ]

π *(1/2) * [ (3/2) + 2ln(2) + (-1/2) [ (-3/4)] ]

π *(1/2) * [ (3/2) + 2ln(2) + (3/8) ]

π *(1/2) * [ (15/8) + 2ln(6) ]

π [ (15/16) + ln(6) ]

May 24th, 2015

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May 24th, 2015
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May 24th, 2015
Nov 18th, 2017
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