# 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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