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February 1, 2015

February 1, 2015

Posted by **Casablanca** on Tuesday, March 13, 2012 at 11:40am.

- Calculus (Solid of Revolution) -
**Reiny**, Tuesday, March 13, 2012 at 12:10pmy=x^2 and x=2 intersect at (2,4)

the radius of the rotated region = 2 - x

= 2 - √y

V = π∫ (2-√y)^2 dy from 0 to 4

= π∫ (4 - 4√y + y) dy from 0 to 4

= π[ 4y - (8/3)y^(3/2) + (1/2)y^2 ] from 0 to 4

= π (16 - (8/3)(8) + 8 - 0)

= 8π/3

check my arithmetic and thinking.

- Calculus (Solid of Revolution) -
**Steve**, Tuesday, March 13, 2012 at 12:13pmy=x^2 intersects x=2 at (2,4)

using shells,

v = ∫2πrh dx [0,2]

where

r = 2-x

h = y

v = 2π∫(2-x)(x^2)dx [0,2]

= 2π∫2x^2 - x^3 dx [0,2]

= 2π(2/3 x^3 - 1/4 x^4)[0,2]

= 2π(16/3 - 4)

= 2π(4/3)

= 8π/3

using discs,

v = ∫πr^2 dy [0,4]

= π∫(2-√y)^2 dy [0,4]

= π∫(4 - 4√y + y)dy [0,4]

= π(4x - 8/3y√y + 1/2 y^2)[0,4]

= π(16 - 64/3 + 8)

= 8π/3

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