Volume - Cross-sections
Volume; there?s lots of different ways to find volume in Calculus. We hinted at one of them back in the episode on the area between two lines. Let me show you what I mean. Let?s say you have two different curves and you want to go and find the area between them. Well, you get the area of a little section. If you make this three dimensional, that?ll be going straight downwards. What you?ll get is a shape that?s made up of stacks and stacks of shapes that all have the same area. You can use that to find the volume of that shape.
Another method involves taking discs and adding them. We covered that in episode on volume using discs. There?s another one with shells that we?re not going to cover, because typically not on the AP test; lots of methods. The one we?re going to cover now is volume using cross sections.
Volume with cross sections is incredibly useful. Engineers can use this stuff in all sorts of applications to find the volume of very strange shapes. Let?s have a look at one. Cross section problems are usually phrased like this. Cross sections are squares perpendicular to the x axis. The base is determined by the x axis and the function. So the information that we can extract from this is what the base of the squares are. We know that we?re going to use cross sections; so imagine having stacks of little squares that you?re adding up to make the volume. Let?s draw some of these in. Another I didn?t tell you to watch out for is what the bases are perpendicular to, very important. They can be perpendicular to the x axis, or to the y axis.
These are perpendicular to the x axis and the base is determined by the x axis and the function; so going from the function to the x axis that would be the base of our rectangle. Now let me draw a rectangle here. You?ve got to imagine this is 3-dimensions now that?s sticking up and outwards from the shape; so like that. If I draw another one in, this one doesn?t have as a big a base. It?s going to be a slightly smaller square that looks like so.
Let?s draw in another. Here is the square for that one. It?s a smaller square. It looks like that. Draw in another; we?re just making stacks of these. Stacks and stacks of them. Draw out here, draw another out here. Draw another, Now another. So you have to imagine that this looks like a deck of cards. They?re a sheath of paper; though the sizes of the sheaths of paper are changing as you go further and further out.
Now I?m going to connect up the corners; this corner to that corner. You get a curving shape in three dimensions. This would be like the bottom surface of our volume. We cover that in nice and dark, so you can?t see the cross sections any more. Over in this corner here, connect this corner to that corner, to that corner all the way up. You get another weird curve. So imagine a shape like I can say call those if you are a carpenter, or like the nose of an eagle. If you look in the bonus materials, you?ll see a link to some really neat websites that do a great animations of these shapes.
Time to go through an example involving the simplest type; the ones where with cross sections or squares. We?re going to begin by going through an explanation of the steps. Then we?re actually going to do it. This one asked you to find the volume of the shape. It gives you a domain. You usually need to be given a domain, although there are ones where they could be giving you intercepts and you have to find your on domain.
Using cross sections perpendicular to the x axis and the cross sections are squares; you?re given a function to help determine those bases. First you want to graph and drawing on it would be a good idea. I?m going to draw in just a couple of those bases. Here is one of them. The distance from here to here is the height of the function minus 0. So the nice thing here is that the base of the square is just that, square root minus 0. So you?ve got the square root of ?x plus 6. I won?t bother with the -0, because it doesn?t change anything. There is the base of one of them.
If I go and draw upwards, then this is a square. In a square, the base and the height are the same. We can go onto this next step; use the base formula to get the area. Well, like we just, the area of a square is base times height, but the base and height are identical. So the area on each individual square is going to be the square root of ?x plus 6 quantity squared. Base times height, and they?re the same, the square gets through to the square root. I?m lucky there. The area of any one cross section is just ?x plus 6. If we drew in another one of those cross section that are squares, yeah it will be a smaller square. It would have less area, but this formula will give us the area.
So area of any individual cross section is just ?x plus 6. Area is negative of x plus 6. We are just about ready to finish this problem. The really weird thing conceptually, is that if you have a square, and you find its area, it doesn?t have any thickness, so it doesn?t have any volume. It?s infinitely thin. But if you stack up a whole bunch of squares, all those areas of 0 actually add up to a volume, weird, though it works.
Let?s set this up. When you do this, you have to integrate the cross sections. They?re also going to be set up the same way. It?s always going to be the integral of the area, and maybe dx or maybe dy. This one is going to be dx. So we have the integral. You integrate across the direction that the squares are perpendicular to, in this case x. So 0 to 6 is my domain. That?s the limit of my integral. I?m going to integrate ?x plus 6dx. So there is the set-up for that one, nice and simple.
Let?s go over here. The area then is the integral of ?x is -1/2x². Integral of 6 is 6x. Integrate that from 0 to 6, that?s a quick going easy. I?ll leave the substitutions for you, nice and simple. I wrote area there. I was mistaken. That?s the volume. This volume was the result of doing this integral. The volume of this one is going to be just 18.
Our next problem is a couple of steps harder than the first. One thing that?s going to make it harder is that the areas that we?re integrating up aren?t going to be squares any more. They?re going to be rectangles. That?s much harder, but it is something you have to think about. The second thing that?s going to make this one harder is that instead of defining the bases as the distance between the function and the x axis, it?s going to be defined as the distance between two different functions.
Find the volume of the shape. Our domain is going to be between 0 and 3 using cross sections perpendicular to the x axis. This time the cross sections are rectangles though. Their base is determined by the functions y equals x² minus 9 down here. Y equals 2x minus 6 up here. The rectangles heights are going to be twice their bases. That?ll be the base of one of those rectangles.
Now if I want to find out how far that is, the graph helps me see that it?s that function minus that function; the linear function minus the quadratic. So let?s do that. One base equals look at the linear that?s -2x plus 6, minus, because distance is always done by subtracting, x² minus 9; minus x² minus 9. Let me simplify a little bit. The base then equals negative of x² minus 2x. 6 minus -9 is 15. It might seem like all those negatives are going to give you bases that don?t have positive results, but they will.
If you put 0 for example onto this, -0² minus 2 times 0 plus 15gives you a height of 15 for the base between here and here. Good deal, no problem. Think back to the last problem. After we found the base, we had to do something before we integrated. What do you integrate? Areas. So you need areas. Now we?ve got rectangles. Whenever you do one of these problems, you have to be aware of what kind of a shape we?re integrating up. That last one was squares, this one was rectangles. Any Geometric shape is possible.
The areas on this since it?s a rectangle is base times height. The heights are twice the bases. The bases, base times height. If the heights are twice the bases, then you have base times, I can replace the height with 2B. You don?t necessarily need to show much work. You can probably figure this out in your head. We?re going to do that times itself. Base times base and, then double it.
We would have 2 times the parenthesis. Base is -x² minus 2x, plus 15. Now you?re saying oh!
My aching head. You?re going to have to square that trinomial, because remember it was base times base. Each base is -x² minus 2x plus 15. So to save a little bit of time, I already did this. I?m going to be lazy. I?m going to write this down. So the area after that whole mess has been distributed is 2x to the fourth, plus 12x³, minus 52x², minus 120x, plus 450.
Well, before I go think about what you?re going to do. These problems have a lot of steps. It?s easy to get caught up in a step and then panic. Once again we found the size of our base, then we calculated the area of individual rectangles. Now we?re going to integrate up all those areas. So this time our volume is the integral we have to integrate from 0 to 3. 0 to 3 of that whole mess; 2x to the fourth, plus 12x cubed, minus 52x², minus 120x, plus 450. Long as that looks, it?s a relatively straight forward integral. You?re going to have to do a lot of substitution, a lot of calculation. To save a little bit of time, I did that again off-screen. If do this integral properly, you might want to try it for practice. You?re going to get a volume of 3,006 over 5 which is 601.2.
Remember what I said, cross sections can be any Geometric shape. This time we?re going to try semi-circles. I?m going to throw in another twist. Find the volume with domains between 0 and 3 using cross sections perpendicular to the y axis. The function is determined on x. We?re going to have to deal with that.
Let me draw it out first though. We want to use a domain from 0 to 3.
X equals 0 to x equals 3. There is x equals 3. The cross sections are going to have to integrate. Go to the y axis, perpendicular to the y axis. In other words the base is the distance between here and here. We?re going to have to change this to be a function of y, so that we can find the distance.
The distance is going to be the function of y minus 0, because this spot is at an x coordinate of 0. So our cross sections are stacked up like this. We?ll worry about that 3 in a little bit. This one is not going to be a hard one to solve for y. Remember how you do this. It?s not too tough. You got f of x equals x². I?m going to replace f of x with y. That?ll make thing a little bit easier. Y equals x². Basically I?m making an inverse.
Now let?s see. We will do the square root of both sides. You get x is plus or minus the square root of y. Looking at this domain that we have though, we?re always going to have positive values for x, so we don?t want the minus there. We just want the positive version of it. One last thing; x equals the square root of y. If you write this in function notation, we no longer have a function of x; we have a function of y. The function of y is the square root of y. So if you pick any y value you want like say 4, the square root of 4 is 2. This location would be at 2; this location would be at 1. That distance is 2. 2 minus 0 is 2. That?s our base.
Now that we have a base, we?ll go on and just a second to make the cross sections. Any individual cross sections look something like this. It?s a semi-circle. Before we go on to do that, why don?t we figure out the limits? Remember what I said before. When you do your integration, integration has to be on the axis that you?re perpendicular to. This time, the shapes you?re integrating up, those are perpendicular to the y axis. So before I do my integral, I need to know the y limits, not the x limits. Fortunately, they?re not hard to find out.
Domain x equals 0, when x equals 0, y equals 0. 3 is the other number in the domain, but if three was, 3² would be 9. We?re going to be using 0 and 9 for the limits of our integration. We?re going to cut to the next screen now, because we need to work out the area formula. Remember f of y equals square root of y.
Semi-circular cross-sections. We know the distance from here to here is the square root of y. That?s a half of a circle. If you?re going to find the area of a circle, you need the radius. This is where you can really get caught up. That?s not the radius, that?s the diameter. So the radius of this is ½ the square root of y. The distance from there to there, ½ of the square root of y.
Now we can go on and use the area formula for our circle. Area equals pi r², but wait that?s not a circle; it?s a semi-circle so I need another half. This half has nothing to do with that one. This half was from turning diameter into radius. This half is from turning a full circle into half a circle. Now we?re ready to go. I have to work out what this area our formula is so I can put it into the integral, almost done with part.
We have ½ pi and the radius is ½ square root of y, so replace the r with ½ square root of y. That will be squared. Not as hard as it looks. Square of y, square root is y. ½ square root id ¼ times the half is 1/8. There we go. Area is 1/8 pi y. Now we?ve got to integrate them up.
Volume is the integral. This is a function of y, so remember our limits had to be y numbers. Not the 0 and 3 given in the problem, but the 0 and the 9 that you would get if you put these domain limits into your function. We?re stacking up areas. Each area is 1/8 pi y, easy enough. Integrate that and you would have let?s see one more power of y is y². I?ve done a correction factor. That correction factor is ½ end of the fraction. This 1/8 doesn?t change the fact that you need a correction factor.
So we have ¼ times ½ times y² from 0 to 9. With a little bit of substitution, you can get final answer easy enough. This one works out to be 81 pi over 16; a nice big piece of pi. There you have it, cross section.
We covered 3 different practice problems, 3 different set ups. The first one was a fairly basic one where the cross sections were squares that were perpendicular to the x axis. The second one was rectangles perpendicular to the x axis with a little extra complication, and making the bases a little harder to find. The third one was semi-circles perpendicular to the y axis.
Now a little bit of follow up. I?d like to recommend you do two different things. One is to go after that second problem again; the one that involves cross section of rectangles. Set it up in your calculator. You might see one of these in the calculator section of the AP test in which case you wouldn?t want to do all those powers and all those distributing for that, way too many chances of making a mistake.
You?d want to go back as far as you can before you put it in the calculator. So again I?d recommend that you try that one out again on your calculator. I have the full set up for it in the bonus problems. The other thing I?d recommend is that in the bonus problems there is one involving equilateral triangles that I think you should give a try. Well, in the meantime remember time flies like an arrow, flies like a banana.