Hello victorxz90044 Originally Posted by

**victorxz90044** (b1+b2)/2 y = A

pump distance= 2-y

y =0 , b=2

y=2, b=3

-b2 = (b1+(1/2)y)/2

6*b2*dy

6(b1+(1/2)y)dy

dw = int 62.5 [6(2+2y)dy] (2-y)

now what

Each cross-section is a rectangle whose length is 6 feet. If the breadth is $\displaystyle b$ at height $\displaystyle y$ above the base, then, as you say:When $\displaystyle y = 0, b = 2$

When $\displaystyle y = 2, b = 3$

and the relation is clearly linear, so:$\displaystyle b = 2+\tfrac12y$

So the area of cross-section is $\displaystyle 6b = 12 + 3y$ square feet, and the volume of this thin slice is $\displaystyle (12+3y)\delta y$ cubic feet. Its mass is therefore $\displaystyle 62.5(12+3y)\delta y$ pounds. It is raised through a distance $\displaystyle (2-y)$ feet. So the work done is:$\displaystyle \delta W = 62.5(12+3y)\delta y(2-y)$ foot pounds. (I don't know whether that's the correct unit; I usually work in SI.)

$\displaystyle \Rightarrow \frac{dW}{dy}=62.5(12+3y)(2-y)$

$\displaystyle \Rightarrow W = \int_0^262.5(12+3y)(2-y)\;dy$

Grandad