calculus

A reservoir is in the form og the frustum of a cone with upper base of radius 8ft and lower base radius of 4 ft and altitude of 10ft. The water in the reservoir is xft deep. If the level of the water is increasing at 4ft/min., how fast is the volume of water in the reservoir is increasing when its depth is 2ft.?

Answer: 100pi ft^3/min..

Help please, i need to see the solution...

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  1. when the water is x feet deep, the surface has radius 4+(8-4)x/10 = 4+x/5

    The volume of the cut-off part of the cone is 1/3 pi *16*10 = 160/3 pi

    So, the volume of water is
    v = 1/3 pi (4+x/5)^2 (x+10) - 160pi/3
    = pi/75 x^3 + 2pi/3 x^2 + 32pi/3 x

    dv/dt = (pi/25 x^2 + 4pi/3 x + 32pi/3) dx/dt

    when x=2,
    dv/dt = (4pi/25 + 8pi/3 + 32pi/3)(4)
    = 4048/75 pi

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  2. 100pi ft^3/min

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  3. Solution pls

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