The following tank is completely filled with water. Find the total amount of work done in pumping water out of the outlet. Note that the density of water is 1000 kg/m$^3$
I feel like I am headed in the right direction to solving this problem, however, I am unsure of where to go from here (if I'm doing it right!). This is where I'm at:
By similar triangles within the cross-section:
$$\frac{r}{10-x}=\frac{3}{10}$$
$$r=\frac{30-3x}{10}$$
$$A=LW\Rightarrow A=20\big[2\big(\frac{30-3x}{10}\big)\big]$$ $$V=AH\Rightarrow V=(120-12x)\Delta x$$ $$w=mg \Rightarrow w=DVG$$
$$\int(1000)(9.8)(120-12x)xdx$$
The spout thing is throwing me off. If I am doing this problem correctly, would the bounds for the integral be $a=3,b=13$?
EDIT: Would the answer be $3.1556\times 10^8$J?
You are definitely on the right track, but I think you want to change a couple of things:
The distance a slice of water is being pumped is given by $x+3$ instead of $x$, so you want to replace the $x$ by $x+3$ before the dx.
I'm not sure where you are getting the factor of (1000)(9.8); in place of this you want the weight-density of water (which should be given in the problem, or in your book at least). $\;\textbf{Edit:}$ This is right (I looked it up).
The limits of integration should correspond to where the water actually is; in your coordinate system, that is from 0 to 10.
(A couple of other remarks: you might find this slightly easier to set up if you let $x=0$ correspond to the bottom of the tank and $x=10$ correspond to the top.
I think you want to have $V$ instead of $A$ where you have $A=(120-2x)\Delta x$.)