# Integration; work; missing 2

1. Apr 24, 2010

### Tclack

A vat (shown in attachment) contains water 2 m deep. Find work required to pump all water out of top of vat. (weight density of water density= 9810 N/m^3)

$W=\int^b_a F(x)dx$

weight of water=$F(x)=V\rho$

put my figure on the coordinate axis so I came up with general equation for the base, with respect to height (using a y=mx+b format):

$h=\frac{3}{2}b-3$
but, I want to go down a positive depth, so I negated the equation:
$h=-\frac{3}{2}b+3$
$b= (h-3)\frac{-2}{3}$
$b=\frac{2}{3}(3-h)$---------------------------------(1)

As you can see, the above equation follows correctly, at a depth(h) of 0, we have a base of 2, at a depth(h) of 3 we have a base of 0, so the equation is valid, so far no mess-ups... I hope

Now, the general equation for a triangular volume is: Volume= $V=\frac{1}{2}bhL$ (b=base, h=height, L=length)
But, because my equation for b is only taking one half of the triangle, I double it:
$V=2(\frac{1}{2}bhL)=bhL$-------------------------------(2)
substituting (1) into (2) I get
$V=\frac{2}{3}(3-h) hL= L=6m so$
$V=4(3h-h^2)$

therefore
$W=\int^3_1 F(x)dx=\int^3_1 V\rho dx=\int^3_1 4(3h-h^2)(9810) dx$
$W=4(9810)(\frac{3h^2}{2}-\frac{h^3}{3})\mid^3_1$
W=130800
...which is exactly half of the answer, I just can't seem to find that missing 2 despite being so concise. I remembered to double the volume of my second figure to take into account the full triangle base. Please help me find this... it's so close

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2. Apr 24, 2010

### Tclack

Last edited: Apr 24, 2010