# Thread: pumping out a liquid - work

1. ## pumping out a liquid - work

A tank in the shape of an inverted right circular cone has height meters and radius meters. It is filled with 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

$
A = \pi r^2
$

$
V = \pi r^2 \triangle{x}
$

$
\frac{r}{h} = \frac{13}{8}
\frac {r}{8-x} = \frac{13}{8} r=\frac{13}{8}(8-x)
$

$
V = \pi (\frac{13}{8})^2 (8-x)^2 \triangle x
$

$
F = ma = V*p*g
$

$
F = \pi (\frac{13}{8})^2 (8-x)^2 (1550)(9.8) \triangle x
$

$
W = \int^8_1 [\pi (\frac{169}{64}) (8-x)^2 (1550)(9.8)] dx
$

$
W = \frac{2567110\pi}{64} \int^8_1 (8-x)^2 dx
$

$
W = 14407454.03 J
$

Can anyone see where i went wrong?

Thanks for any help

2. Originally Posted by mybrohshi5
A tank in the shape of an inverted right circular cone has height meters and radius meters. It is filled with 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
let the bottom of the tank be the origin.

one sloping side of the tank runs along the line $y = \frac{8}{13}x$

volume of a representative horizontal cross-section is ...

$\pi x^2 \cdot dy = \frac{169\pi}{64}y^2 \, dy$

weight-density = $\left(1550 \, \frac{kg}{m^3}\right) \cdot \left(9.8 \, \frac{N}{kg}\right) = 15190 \, \frac{N}{m^3}
$

lift distance for a representative horizontal cross-section is $(8-y)$

$W = \int_0^7 15190 \cdot (8-y) \cdot \frac{169\pi}{64}y^2 \, dy$

$W = \frac{1283555 \pi}{32}\int_0^7 8y^2-y^3 \, dy$

$W = 3.96 \times 10^7 \, J$