When a piece has been scooped out: apparent capacity against actual

0:00A juice seller has a glass. It is a cylinder, 5 centimetres across on the inside and 10 centimetres tall. You can see straight through it, and you can see how much it holds. Except that you cannot. The bottom of this glass is not flat. It rises into the glass as a smooth dome of solid glass. The dome is glass, and glass is not juice. So there are two different numbers here, and only one of them is what you are paying for.

0:33The first is what the outline of the glass promises. The second is what is actually inside it. This video is about the gap between those two numbers, and about the fact that finding it needs no new rule at all. Start with the outline, and forget for a moment that anything is in the way. A cylinder 5 across has a radius of 2.5. Notice that the 5 was a diameter and the radius is half of it, because that step is where most of the mistakes in this topic happen.

1:10The outline holds 62.5 times pi. With pi taken as 3.14, that is 196.25 cubic centimetres. Call that the apparent capacity: what the shape appears to hold. It is a perfectly good number. It is just not the answer to the question anybody asked. Now the dome. It is half a sphere, and it fills the glass right across, so its radius is the glass's radius: 2.5 again. A hemisphere of radius 2.5 takes up 125 twelfths of pi.

1:53At 3.14 that is 32.71 cubic centimetres, near enough. That is 32.71 cubic centimetres of glass sitting where juice would otherwise be. You are not being cheated exactly. The glass is honest about its shape. It is just that the shape is not what you assumed. So what does the glass really hold? 196.25, less 32.71. 163.54 cubic centimetres. That is the actual capacity, and the arithmetic took one line.

2:35But look at what that line is. When two solids are joined, their volumes add. When one is taken out of another, the volume is subtracted. Those are not two rules. They are one rule, and the second is the first with a minus sign in front of a term. Which is worth more than the answer, because it turns three different-looking situations into the same one. Here is why the minus is legitimate, and it is the same reason the plus was.

3:08A solid is the set of points inside its outline and outside its hollows. That is the whole definition. So take the outline, cut it into little cells you can count, and sort every cell into one of two piles: in the hollow, or not. Every cell goes into exactly one pile. None goes into both, and none is left over. The room the remaining solid occupies is therefore the outline's room, less the cells the hollow took. Counted, not argued.

3:44But now watch, because there is a condition hiding in that sentence, and it is the same condition the adding rule had. The hollow must lie inside the outline. Bore a hole right through a block and out the far side, and part of your hollow was never in the block to be removed. Count it: the hollow holds 768 cells, and only 384 of them were ever in the block. Subtract the whole 768 and you have taken away material that was not there.

4:20So the honest statement is: subtract what the hollow and the outline share. When the hollow is inside, that is the hollow, and that is why the simple version works. Back to the glass, and to a much better question than what it holds. What fraction of the glass is the customer losing? The dome is 125 twelfths of pi. The outline is 125 halves of pi. Divide one by the other and the pi cancels. Both numbers carried it, so neither answer needs it.

4:58The customer loses one sixth of the glass. Exactly one sixth. And that is not special to these measurements. A hemisphere rising from the base of a cylinder always takes two thirds of r cubed, against the cylinder's r squared h. The fraction is 2r over 3h, every time. Here that is 5 over 30, which is a sixth. Notice what that means. Whether you take pi as 3.14, or as 22 sevenths, or exactly, the customer loses a sixth.

5:35The value of pi decides how many cubic centimetres. It has no opinion at all about the fraction. That is the version of this result worth carrying out of the room. Same rule, different material. Here is a wooden block, 15 centimetres by 10 by 3.5, with four conical hollows drilled into the top to hold pens. Each hollow has a radius of half a centimetre and a depth of 1.4.

6:09The block, before drilling, is 15 times 10 times 3.5, which is 525 cubic centimetres. One cone, with pi as 22 sevenths, comes to 1.1 divided by 3. Four of them come to about 1.4667. So the wood left is 523.53 cubic centimetres. One outline, four hollows, one subtraction. The number of hollows never mattered to the rule. Now compare those two subtractions. The dome took a sixth of the glass. That is 16.67 per cent.

6:54The four hollows took 0.28 per cent of the block. The same reasoning, in one case removing a sixth of the object and in the other removing a fortieth of a per cent more than nothing. About sixty times the difference. There is a temptation, seeing 0.28 per cent, to say the hollows do not matter and skip them. Resist it. Whether a correction is negligible is something you compute, not something you assume.

7:26You cannot tell which case you are in by looking. The glass and the pen stand look equally like objects with small bits missing. Only the arithmetic separates them. There is one thing about hollowing that catches almost everybody, and it is worth meeting head on. Take a block. Scoop a hemisphere out of its top face. The volume goes down. Obviously: material left. The surface area goes up. Count it on cells and watch both numbers at once. The block holds 3888 cells and shows 1512 faces to the air.

8:10Scoop the hollow: 456 cells gone, and 224 more faces showing. Room down, skin up, off one and the same object. The reason is not mysterious. You removed a flat disc of surface and put a curved bowl in its place, and the bowl is bigger. So a hollow subtracts from one measurement and adds to the other. If you have just learned that scooping raises the area, do not let that carry over. It lowers the volume. Different measurement, different sign.

8:48Third situation, and it does not look like the other two at all. A container is full of water to the brim. Lower a solid into it, and water spills over the side. How much? The water that stays is the water that can still find room, and the solid is now occupying part of the container. So the water left is the container's room, less the room the two of them share.

9:16That is the same sentence as before, with the solid playing the part of the hollow. If the solid goes entirely under, what leaves is exactly the solid's own volume. If it sticks out over the rim, less leaves. Count it: a ball of 280 cells, half of it above the surface, pushes out 140. So an immersed solid displaces its own volume, and it does so because of the same set-counting, not because of a separate law about water.

9:52That gives you a way to count things you never counted. A vessel shaped as a cone, point down, 8 centimetres deep, with a circular top of radius 5. It is filled to the brim. Drop in small lead spheres of radius half a centimetre, until a quarter of the water has spilled out. How many spheres went in? The vessel holds 200 thirds of pi. A quarter of that is 50 thirds of pi, and that is the volume that left.

10:25The water that left equals the room the spheres took. That is the physical step, and no amount of arithmetic will hand it to you. One sphere is one sixth of pi. Divide: 50 thirds of pi, by one sixth of pi. The pi cancels again. The answer is exactly 100. Not about 100. Exactly, because the only thing left after the cancelling was a ratio of whole numbers. One more, and it is the one where reading the question is most of the work.

11:02A solid is a cone 120 centimetres tall of radius 60, standing on a hemisphere of radius 60. It is stood upright in a cylinder of the same radius, 180 centimetres tall, full of water. First, does it fit? 120 plus 60 is 180, which is exactly the cylinder's height. It fits with nothing to spare, and noticing that is part of reading the question. Now measure. The cone is 144000 pi. The hemisphere is 144000 pi as well.

11:45Those two being equal is worth a second look. A cone whose height is twice its radius holds two thirds of r cubed, and so does a hemisphere of that radius. The solid together is 288000 pi. The cylinder is 648000 pi. So the water still standing is 360000 pi. With pi as 22 sevenths, that is about 1.13 cubic metres. Finish with a question that hands you an answer and asks whether to believe it.

12:23A glass vessel: a sphere 8.5 centimetres across, with a cylindrical neck 8 centimetres long and 2 across. A child measures how much water it holds and reports 345 cubic centimetres. Work it out. The neck is 8 pi. The round part is 4913 forty-eighths of pi. Together, with pi as 3.14, that is 346.51. So the report is low, by about one and a half cubic centimetres. Close. And not correct.

13:03That is the whole topic in one line: apparent and actual, hollow and outline, water in and water out, are one rule with a sign in it. A solid is the points inside its outline and outside its hollows. Everything else was arithmetic.