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 =$\displaystyle \rho= 9810 N/m^3$)

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

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

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):

$\displaystyle h=\frac{3}{2}b-3$

but, I want to go down a positive depth, so I negated the equation:

$\displaystyle h=-\frac{3}{2}b+3$

$\displaystyle b= (h-3)\frac{-2}{3}$

$\displaystyle b=\frac{2}{3}(3-h)$---------------------------------(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= $\displaystyle V=\frac{1}{2}bhL$ (b=base, h=height, L=length)

But, because my equation for b is only taking one half of the triangle, I double it:

$\displaystyle V=2(\frac{1}{2}bhL)=bhL$-------------------------------(2)

substituting (1) into (2) I get

$\displaystyle V=\frac{2}{3}(3-h) hL=$ L=6m so

$\displaystyle V=4(3h-h^2)$

therefore

$\displaystyle W=\int^3_1 F(x)dx=\int^3_1 V\rho dx=\int^3_1 4(3h-h^2)(9810) dx$

$\displaystyle W=4(9810)(\frac{3h^2}{2}-\frac{h^3}{3})\mid^3_1$

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