# Work - math

1. Mar 16, 2005

A tank in the shape of an inverted right circular cone has height 8 meters and radius 6 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is 1510kg/m^3

m = pi*r^2
m = 36*pi*1510
m = 54360*pi
distance = (8-x)

$$\int_{0}^{7} 54360*pi*(8-x) dx$$

this is clearly wrong, because i already tried to submit this, but it wouldnt accept it. can someone tell me what im doing wrong?

2. Mar 16, 2005

### xanthym

From problem statement:
{Height of Liquid Top Surface} = h
{Radius of Liquid Top Surface} = r = (6/8)*h = (3/4)*h
{Volume of Liquid in Tank} = (1/3)*π*r2*h = (1/3)*π{(3/4)*h}2*h = (0.589)*h3
{Mass of Liquid in Tank} = M = {Density}*{Volume of Liquid in Tank} =
= (1510)*(0.589)*h3 = (889.4)*h3

{dM/dh} = (3)*(889.4)*h2 = (2668)*h2
{Work Lifting Liquid Over Tank Top} = W = ∫ g*(8 - h)*(dM/dh)*dh

$$\ \ \ \ W = \int_{0}^{7} g(8 - h)(2668)h^{2} \ dh$$

$$\ \ \ \ W = (2668)g\int_{0}^{7} (8 - h)h^{2} \ dh$$

$$\ \ \ \ W = (26173)\int_{0}^{7} (8h^{2} - h^{3}) \ dh$$

$$\ \ \ \ \color{red} W = (26173)(314.4) = (8229000 \ J) = (8.229*10^{6} \ J)$$

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