# Work to empty an inverted pyramid

I have the question below that I am attempting to solve. I have correctly performed work related questions for other shapes (cylinder, cone, trapezoid) but have yet to deal with a pyramid. I have searched the Internet and have not turned up any relevant information.

Question: A water tank has the shape of an inverted square pyramid and is filled with water. Find the work performed in pumping all water out of the pyramid. The pyramid has a height of 8m and sides of 3m.

** I have changed the scenario up from the original question. It has different parameters and is pumped through a raised pipe instead of straight out top. **

I have the knowledge to compute everything once I can figure out the proper method of computing the area of the slice. I have yet to be able to figure out the initial formula for the change.

• If you have done the cone, the pyramid should be relatively easy. The only difference is that the horizontal slices are all squares instead of circles. How did you solve the "area of the slice" problem for the cone, and what difficulty did you encounter when trying to apply the same principles to the pyramid? (It is probably better to add this information to the question rather than try to explain such details in a comment.) Dec 4, 2016 at 14:51
• The horizontal slice is the issue I am having going from a circle to a square. In the cone I computed the generic radius by the slope-intercept form y=mx+b and solved for x. This was added into the integral along with the constants for water and gravity and the computed distance to travel out of the tank. For the pyramid not sure how to start on computing this for a square. Dec 4, 2016 at 14:58

It seems to me that part of the difficulty is that your cone method was (slightly) more complicated than necessary.

Your use of the equation $y=mx+b$ was perfectly correct for the cone, but let's step back a moment and see where it comes from.

If you take a vertical cross-section of the cone from the apex to a diameter of the base, you get a triangle. The base diameter is the base of the triangle, and the other two sides are lines along the surface of the cone. Take any line segment parallel to the base, connecting the other two sides of the triangle: that segment is a diameter of the horizontal slice at a certain distance from the apex of the cone. And the triangle formed by this slice's diameter and the apex of the cone is similar to the triangle formed by the base diameter and the apex.

So we get this essential fact about the cone: the linear size of the circular slice parallel to the base is exactly proportional to the distance from the apex. "Linear size" can refer either to the radius or the diameter--they're proportional to each other, so if one is proportional to the distance from the apex then they both are.

Expressing this as an equation, with $y$ as the distance from the apex and $x$ as the radius, we have $x=ky.$ (This is what I meant by your method being slightly more complicated than it needed to be: if you actually get the formula from first principles, there is no constant term $b$ and the equation already has $x$ in terms of $y$ without needing to be solved for $x$.)

Now consider this: a square pyramid is generated from an apex and a square base exactly the same way a circular cone is generated from an apex and a circular base. You can take a vertical cross-section of the pyramid and you will have a triangle, just like in the cone. The only thing really different is that the shape of the triangle depends on which cross-section you take: the base of the triangle could be a diagonal of the square, it could be parallel to a side, or it could transect the square at some other angle.

So choose your preferred cross-section, and observe how the linear measure of each horizontal cross-section relates to the distance from the apex. Then you can develop a formula for the linear measure as a function of $y$ and get the area of the square horizontal cross-section from that.

• I am still lost. I know that it is a simple concept that I am missing. I have taken Calculus before, and have consulted with a friend in a calculus class currently, and we still can not figure it out. What formula is needed for the pyramid. In the cone I used (π)(r^2). Dec 4, 2016 at 17:18
• The formula $\pi r^2$ is the area of a circle. What is a formula for the area of a square? Dec 4, 2016 at 17:30
• I figured it out. Thanks I will update shortly with the answer so that others needing the information can find it Dec 4, 2016 at 17:58