= weight density
= cross-sectional area of a representative "slice"
= lift distance for a representative "slice"
= "slice" thickness
integrating sums all the "slices" residing from y = 0 to y = 25 ...
try it again.
Problem on review sheet for my son's calculus final:
A right circular canonical tank of height 25 and radius 5 has its vertex at ground level and a vertical axis has oil weighing 50 lbs/cubic foot. Set up an integral that represents the amount of work done in pumping the oil up to the top of the tank. Include an approximation for the work required to move one disk of water to the top, but do not evaluate the integral.
The way we tried to solve it (imitating an example in the textbook):
Force is weight times volume. Divide the canonical tank into circular slices of height delta y:
A circle with radius 5, centered at (0,0) has the form
which can be solved for
I don't completely understand how (or why) they picked the center for the example we're working from, so we just chose (0,0).
The distance to the top of the cone is height of the cone minus height of the oil:
Next, the work incremental is the force incremental times the distance
Integrate from height of 0 to 25, so the work must be
But that doesn't equal the answer given to us.
What's wrong please?
Is it a property of cones that the ratio of the height and radius is constant? So this cone's base's radius is 5 and it's height is 25, or 1:5.
A new question: What would this equation look like if the cone were oriented with the vertex at the top and the base at the bottom. That would require a different amount of work to empty (more oil further from the hole and less oil close to it). Where does that show up in this equation?
thus the radius of each differential circular slice would be:
and the work would then be:
Do you see how inverting the tank replaced y with 25 - y in the squared term? It takes 3 times as much work to empty the tank now.