Cal 2

Find the volume of a frustum of a right circular cone with height 25, lower base radius 30 and top radius 13.

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  1. Volume = from 0 to 4 of ∫ pi [f(y)]²dy where f(y) = (-3/4)y + 5

    ∫pi ((-3/4)y + 5)² dy = pi ∫ ((9/16)y² - (15/2)y + 25) dy

    = pi[(3/16)y^3 - (15/4)y² + 25y + C]

    evaluated from 0 to 4 gives:
    pi[(3/16)(4^3) - (15/4)(4²) + 25(4) - (0-0+0)]
    =pi[12 - 60 + 100]
    =52pi

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  2. no Calculus needed.

    extend the sides to form the original cone.
    let the height of the added cone be h
    by ratios
    h/13 = (h+25)/30
    30h = 13h + 325
    17h = 325
    h = 325/17

    volume of fulcrum
    = (1/3)π(30^2)(25 + 325/17) - (1/3)π(13^2)(325/17)
    = (1/3)π[900(25+325/17) - 169(325/17) ]
    = (1/3)π (36475)
    = 36475π /3 cubic units

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