Find the area of the surface obtained by rotating the curve
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Find the area of the surface obtained by rotating the curve

[From: ] [author: ] [Date: 11-05-07] [Hit: ]
. . . . . .......
y=1+2x^2, from x=0 to x=1 around the y-axis

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around y-axis

y = 1 + 2x^2 =====> x^2 = (1 - y)/2 ====> x = √( (1 - y)/2 ) ====> x = ( (1 - y)/2 )^(1/2)

dx/dy = (1/2) * ( (1 - y)/2 )^(-1/2) * -1/2

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

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

√ [ 1 + (dy/dx)^2 ]

√[ 1 + ((-1/4) * [ 1 / √( (1 - y)/2 ) ])^2 ]

√ [ 1 + (1/16) * [ 1 / ( (1 - y)/2 ) ] ]

√ [ 1 + (1/16) * [ 2 / (1 - y) ]

√ [ 1 + (1/8) * [ 1 / (1 - y) ]

√ [ 1 + [ 1 / 8(1 - y) ]

√[ 1*8(1 - y)/8(1 - y) + [ 1 / 8(1 - y) ]

√[ 8(1 - y)/8(1 - y) + [ 1 / 8(1 - y) ]

√[ (8(1 - y) + 1 ) / 8(1 - y) ]

√[ (8 - 8y + 1) / 8(1 - y) ]

√[ (9 - 8y) / 8(1 - y) ]

1
∫ 2π x √ [ 1 + (dy/dx)^2 ] dy
0

1
∫ 2π * √( (1 - y)/2 ) * √[ (9 - 8y) / 8(1 - y) ] dy
0

1
∫ 2π * √((1 - y)/(√2 ) * √[ (9 - 8y) / 8(1 - y) ] dy
0

1
∫ √(2)π * √(1 - y) * √(9 - 8y) * (1 / ( √8√(1 - y) ) ] dy
0

1
∫ √(2)π * √(9 - 8y) * (1 / (√8) dy
0

1
∫ [ √(2) / √8 ] π * √(9 - 8y) dy
0

1
∫ [ √( 2 / 8) ] π * √(9 - 8y) dy
0

1
∫ [ √( 1 / 4) ] π * √(9 - 8y) dy
0

1
∫ (1/2) * π * √(9 - 8y) dy ===> u-sub
0

u = 9 - 8y
du = -8 dy ===> dy = du/-8

1
∫ (1/2) * π * √(u) ( du/-8 )
0

1
∫ (-1/16) * π * √(u) du
0

. . . . .. . . . . . . . . . . . . . .1
(-1/16) * π * (u)^(3/2)/(3/2) ]
. . . . . . . . . . . . . . . . . . . .0

. . . . .. . . . . . . . . . . . . . .1
(-1/16) * π * (2/3) * (u)^(3/2)]
. . . . . . . . . . . . . . . . . . . .0

. . . . .. . . . . . . . . . . . .1
(-1/24) * π * (9 - 8y)^(3/2)]
. . . . . . . . . . . . . . . . . .0

(-1/24) * π * [ (9 - 8*1)^(3/2) - (9 - 8*0)^(3/2) ]

(-1/24) * π * [ (9 - 8)^(3/2) - (9 - 0)^(3/2) ]

(-1/24) * π * [ (1)^(3/2) - (9)^(3/2) ]

(-1/24) * π * [ 1 - (729)^(1/2) ]

(-1/24) * π * [ 1 - 27 ]

(-1/24) * π * ( - 26 )

(26/24) * π

(13/12) * π

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Hello,
The formual for surface area of revolution is: integrate from 0 to 1 : 2pi * y * sqrt( 1 + (dy/dx)^2) or
2pi(1+2x^2)*(sqrt(1 + (dy/dx)^2)
The integral is ∫ f dx = (31*asinh(4*x)+sqrt(16*x^2+1)*(128*x^3+1…
I'll let you put interval from 0 to 1.

Hope This Helps!
1
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