# How do I find the area or the volume (or a formula) for a region that is rectangular but has a cone slope to a drain?

This is a quick (not to scale) drawing of what I mean: the floor gradually slopes from the wall to the drain at about 7/16( 0.4375) inch per foot from the short sides of the wall (the 7 ft from wall to the drain) and exactly 3/16(.1875") of a inch per foot on the long sides ( the 16ft from the wall to the drain).

So the base is a rectangle 14'x32' but the shape is most related to a cone ,.... but all cone formulas assume the base is circular. I'm at a loss of how to pull out only the rectangular portion out from the middle of a cone shape. I hope this made it more visual it has 3" of Drop in the Center, length is 32' and width is 14'. just looking for a formula to calculate it. Any help in the right direction would be appreciated.

-

It looks like this is a truncated rectangular pyramid or pyramidal frustum. This is probably what frustrated your searches.

The volume for a square base is given in mathworld.

For the rectangular case, you can compute the area of the full pyramid and then of the top that you cut off; the formulas are also available at mathworld. A quick Google search also provided this page.

It can be done with Calculus fairly straightforward (a standard Calculus II exercise/problem); but you don't need it. The egyptians already knew the formula.

-
"It can be done with Calculus ... but you don't need it. The Egyptians already knew the formula." - I had to smile at this. :D – J. M. Dec 18 '10 at 11:03

The volume of a pyramid is one-third the height times the area of the base. In your case you have sliced off a tiny (approximate) pyramid from the top, to create the drain hole, so subtract that volume from the total.

-

I am not completely sure about the actual geometry of the problem, so I'll make a few assumptions. Correct me if the assumptions are wrong.

• You have a rectangle of 168" by 384".
• The center of the rectangle is depressed by 3"
• (most importantly) The shape is such that if you draw a line segment from the "drain" to the "wall", it will be straight (not curved in the vertical direction).

The third assumption may be different from what you want, since this assumption will require there be something like a "fold/corner" along the diagonals. To get more precise answers, you need to specify exactly what the shape of the floor is (how it curves towards the drain).

Then in this case you don't need calculus at all.

Each "trianglular piece" in your drawing can be computed directly. For the triangles with base on the short sides, the base is 168". The "height" can be found using Pythagorean theorem applied to the vertical dip:

$$h^2 = 192^2 + 3^2 \implies h \sim 192.0234$$

so the area of the triangle piece is

$$A = h \times 168 \sim 16130.0 \textrm{sq in}$$

Similarly the area of the triangle piece based on the long side of the wall is roughly 16130.6 sq in.

So the total area is roughly 64521 sq in, which is 9 square inches more than if you don't factor in the vertical drop to the drain hole.

-
so what would be the Cubic inches ? or is that what you meant? i assume that square inches are surface area and cubic inches would be volume. and on your assumption of it being a strait line from the edge of the rectangle to the drain is ok,... – Daniel P Dec 17 '10 at 18:34
For the volume, you can just use the volume of a pyramid: one-third base-area times height. Base area is 64512 sq in. Height here is 3 in. So the total volume is 64512 cu in. This assumes your drain hole is a single point. If your drain hole takes up some area, you will have to correct for that. – Willie Wong Dec 17 '10 at 19:23