What is the Work to Pump Water Out of a Horizontal Cylinder Tank with a Spout?

[nameVoid]

a tank the shape of a cylinder has base radius 3/2 m and height 6m lies horizontal and has a spout of length 1m. Find the work to pump all of the water out of the tank.

placing one end of the tank at the origin and using rectangular cross sections. I have attempted to first calculate the bottom half for negative y values and then add it to the top

$$C=9.8*10^3$$

$$C\int_{-\frac{3}{2}}^{0}12\sqrt{\frac{9}{4}-y^2}(\frac{5}{2}+y)dy +C\int_{0}^{\frac{3}{2}}12\sqrt{\frac{9}{4}-y^2}(\frac{5}{2}-y)dy$$
also similar case here, A tank with semicircle base radius 2ft and height 8ft lies horizontl and has spout 1ft
placing the center of the circle at the orgin
$$C=62.5$$
$$C\int_{-2}^{0}16\sqrt{4-y^2}(1+y)dy$$
it looks likes the text is taking the limits of integration here from 0 to 2 I don't understand why

 

[HallsofIvy]

If a cylindrical tank has length 6 and is lying horizontal, then one side of a rectangular cross-section will have length 6- a constant. With the central axis on the x-axis, radius3/2, a circular cross section has equation \(\displaystyle y^2+ z^2= 9/4\) or \(\displaystyle z= \sqrt{9/4- y^2}\).

The area of such a rectangular cross section will be \(\displaystyle 6\sqrt{9/4- y^2)\).

I don't know where you got the "5/2+ y" term from.

 

[nameVoid]

a cylinder with radius 3/2m and height 6m lies on its side (the side that's 6m) and has a spout 1m high above one of the circular ends. the equation of the base is x^2+y^2=9/4 yes but your forgetting about the spout which rises 1 meter so the distance the cross section on needs to travel is 5/2-y* for the positve portion and 5/2+y* for the negative at least that's what I am seeing