I am stuck on a calculus question regarding "work"

A trough is 9 feet long and 7 foot high. The vertical cross sections of the trough parallel to its ends are right triangles with height of 7 feet and horizontal width of two feet. The trough is completely full of water. Find out how much work is required to empty the trough two feet above the top of the trough. Note the weight of the water is 62.5 pounds per cubic foot.
So after countless tries of this problems, I'm hoping to get some help. I took a vertical cross section and use l*w*h where l=9 and h=is equal to delta (x) and I found w through through similar triangles. But that got me a negative number and so far I don't how to do it. A nice detailed explanation would really help

So, you are sucking all the water up to a height two feet above the top of the trench. That will be a constant amount of work to go from zero to two feet. Find this first.
Now you have to integrate the product of mass and distance down from the top. Compute the mass of the water for a thin slice (decreasingly small dy) at distance Y from the top, and integrate this function from 0 to 7. Add this to first result.
You should be taking a HORIZONTAL cross section, and finding the work needed to lift it vertically.

Work is a force through a distance...distance is [ y + 2 ] , y in [0,7] ;

force is the water weight...[ 62.5] δy [ 9 ] [ width] , where y : 9 as width : 7

or width = 7y / 9...so integrate { 62.5 x 9 x 7y / 9 x ( y +2) δy } over y in [0,7 ];

I assumed the right angle was at the top of the trough