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# Volume - Disc Method

###### John Postovit

###### John Postovit

**University of North Dakota**

M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

[0:00:00]

Today we are going to be talking about volume. Volume is a subject which shows up all over in AP Calculus. The method we are doing today, unlike the one we've done before with cross sections, is going to be one of the methods for finding volumes of objects that are rounded. We are going to be using the disc method. There are a couple of different methods for doing volumes of round objects. This method and another one called the shell method. The shell method is one that is often covered in calculus classes but we are not going to concentrate much on it because it's not required for the AP test. I know the fact that we are not going to be doing the shell method makes you feel all hollow inside like a shell but you'll get over it. That was a joke. Turn on the laugh light.

No matter how complex the problem looks, one thing is the same every time. What you're always doing in some form is taking a function, rotating it around a line and finding the volume of the shape that you get when you rotate it. Let me show you one. For example if you take a nice little triangle right here. I tipped a couple of pencils in this high tech demonstrator. The pencils are the axes. If I rotate it around this axis, you can see that it's describing a shape. If you spin it faster and faster, it might be possible for you to imagine the shape a little bit better. it's a couple of cones put together. Now, another. Let's say for example that you want to take the function 2Â² of x and rotate it around the x axis? There is our axis that we are going to rotate on and rotate this.

[0:02:00]

Notice the shape, rotate it around its axis. Can you see that shape now? It's kind of like a bowl. When you rotate something, what shape do you get? Let me see if I can do this nicely. You see the circle that it's describing? That's the heart of this, circles. And there is this line which is the radius of the circle, goes around, it basically crosses all the places that make up the area of the circle. Circles. This is the formula and it really comes from circles. There is pi f(x). f(x) is the function. Sometimes people will call this r(x) instead. The reason they call it r(x) is because it's really the radius. This is rÂ².

There is something funny though, pi rÂ² isn't volume. pi rÂ² is area and that's where the weirdness involved in this happens. You got your curve and you rotate it around its axis, rotate it like so, you get a circle. That circle has got an area. Go a little bit further over and rotate another spot around its center. Go a little further over, rotate around its radius. You can see the shape that's starting to emerge? This is in the front here.

[0:04:00]

There is the cone shape. But the weird thing is again, all you really have is stacks of discs. Each of those discs is infinitely thin, so it's got no volume. But if you add up an infinite number of discs across that region, that are infinitely thin, you'll still get volume. That's where the integration comes in.

The integration has to happen from one end of the shape that's being rotated, that's A to the other end of the shape, that's B. So the limits for this always have to be along the direction that you're rotating on. You can rotate on the x axis, you can rotate around other x lines, or you can rotate on the y axis. If you're rotating on the y axis or a y line, this has to be dy and your limits have to be y numbers.

So let's try it. We're going to rotate the shape that we were just looking at around the x axis. The fact that rotating around the x axis makes this a relatively simple problem. Find the volume of the shape obtained when y equals 2Â² of x is rotated around the x axis and use the limits between 1 and 5.

Here is one of the limits at 1, here is another limit at 5. If I have a disc at 1 and rotate it, it looks something like that. There is the radius. If I have a disc at 5 and rotate it, it's something like that. There is a couple of representative discs.

You notice that I changed the formula this time on the last limit said f(x) there instead of r(x). But remember what I said, radius is what counts. Radius is the distance from wherever you're rotating, to wherever the function is. Now, don't make the mistake though of thinking that it's always going to be found the same way.

[0:06:00]

There is a lot of variations possible in this, that's where the confusion comes in with finding the volume by discs. This one is relatively simple though, the distance from the x axis to the function just is the function. So, r(x), easy enough, it just is 2Â² of x. If I wanted to be fancy about this I'd say, it's rotating around the axis y equals 0, which is the x axis. So the distance from there is distance you know by subtracting, there is all subtracting whether it's shown. We're subtracting the function from where it's being measured from, or from the axis. So 2Â² of x minus 0 and it's just really 2Â² of x.

Later on we will have some problems where we'll be subtracting and you will get an axis that's different, a radius that is different from the function. We'll set this up now. Remember that you've got to set up the integral with the formula. But I feel like making some mistakes here today. So let's see if you can catch them. There's going to be a couple buried in here. Volume we have to do r{x) and r(x) is 2Â² of x. Dx, have a look through here, there's two mistakes buried in there. The pi is missing, that's one of them. It's easy to forget it. But you really are just doing areas and adding up areas to get volume. So you have to have pi rÂ². That's the other one, there is no square. It's not pi rÂ² yet. So I have to take this quantity and I have to square it.

So, now if I simplify it a little bit, I have pi times integral from 1 to 5. Squaring 2Â² of x gives you 4x.

[0:08:00]

4x and there's still the dx. Integrating that gives you 2pi xÂ² integrated from 1 to 5. Remember the integral of x is xÂ², but you need the correction factors. So check it by doing the derivatives. Do it once in a while even if you're confident. Derivative of xÂ² is 2x, times that 2 gives you the 4 that you have in there. A couple of substitutions and we are done.

I can have that 2pi factor pulled out front, why not? It will save me some time. And then I'll substitute the 5 in to the xÂ² so that's 5xÂ². I'm using the first fundamental theorem of calculus which says substitute the first thing in and then substitute the second thing in and then substitute the results. So I've got 2 times pi times 5Â² minus 1Â². 5Â² is 25, 1Â² is 1. 25 minus 1, 24. 24 times 2 pi gives us our final answer a volume of 48 pi.

Onwards and upwards. The last example we did involved rotating around the x axis because the disc method is really oriented toward rotating around the x axis. It does that naturally easily. Rotating around around the other axes is a little harder but we need to try. It's not that much harder. So we are going to try rotating around the y axis. Why? Why not? The y axis. Yeah that was a math joke. I as rotating around a vertical axis. Find the volume of this shape obtained when y equals 1/3x is rotated around the y axis and use limits of 0 to 15.

[0:10:00]

Rotate around the y-axis. We're going to get a cone shape if we do that. Notice that I changed the volume formula bit. Not a lot but a bit, but I had to. Because the direction that you integrate on has to be the one which the discs are stacked up. These are piled up along they y axis, so you have to integrate along the stacks of discs going along the y axis. So that's why it's r(y) dy. And our next problem is that this isn't a function of y, it's a function of x. So I need to change it before I can find the radius and make it a function of y. That's really easy to do though especially for this one.

Y equals 1/3 of x so if that's the case then x has to be 3y, and you should multiply both sides by 3. x equals 3y and now that's a function of y, because you can find x by using a y number. We need to have the radius though. We have the radius in terms of y and the radius is the distance from the axis to what's being rotated. What's being rotated is this. There is the axis, it's going around that vertical axis. Well, if you're measuring from that axis out of the shape, you really are just using the function.

To be technically right though, I'd say in this case that I'm going to take the function and subtract out the axis from it, and then you get a radius of y equals 3y. I've got to tell you something that you have to remember, because it's going to come up near the end of the end of the episode. On the last two that we did, we had the function and we subtracted the axis. So you might be thinking by now, yeah I know what to do, I'm only going to take the function and I'm going to subtract the axis. No, it's not that simple.

[0:12:00]

Sometimes you have to reverse it in kind of strange ways. There's really no predictable way to say what you are going to do because there is a lot of variations possible. The only thing you can say for certain is radius. It's always radius. And you do whatever adding or subtracting is necessary, in whatever order is necessary to find the distance between your axis and the thing that you're rotating. Let's go back to the problem and I'll set it up now.

The volume is the integral of pi and I feel like making mistakes again, bad me. Of course my purpose in making these mistakes, is to see if you can catch them. Because if you can catch then, you're well in your way of understanding what's going on. There's only one mistake in here. Yeah, it's this. 0 and 15. Those are x numbers. We're integrating on y, we need y numbers. When x is 0, 1/3 of 0 is 0 so by coincidence the y number is the same but when x is 15, y is 1/3 of 15 which is just 5. Let's get rid of that. The real limits are 0 to 5. I want to square before I do the integral and while I'm at it to make things a little simpler I know that when I square 3y I'm going to get 9yÂ². I'll put the 9 out front as a factor. That's a nice trick for doing integrals because it makes it a little less confusing if you don't have these constants in there. So we have to integrate that yÂ², dy, and the integral of yÂ² is one power higher, that's y cubed. But of course there is the correction factor of 1/3.

[0:14:00]

So when it's all said and done, we're going to have 9 times the 1/3 which is 3, times pi times y to the 1/3 power. And this one has to be evaluated from 0 up to 5. And I'll leave that for you to try. It's really a pretty quick thing to do. The final result for this is 375 pi.

Time for one more elaboration. Recent AP tests have included problems as complicated as the one we are abut to do so it's best to be prepared. We're going to add two different things in this time. One is that we're going to add in something called washers. Washers are what they sound like. It's a shape that looks like, this, it's a circle with a circle carved in the center, just like the kind of washers that they use in assembling things. So when you're doing this, you're basically doing one shape and subtracting another one. The other elaboration we are going to add is that the axis we rotate on isn't one of the coordinate axis any more.

Let's look at it. Find the volume of the shape obtained when the region defined by all of these is rotated round the line y equals -1. So it's rotated around a y line, that tells me right there that I have to use the version of the formula that integrates on x. Look at this other stuff here. It's not the same formula as it was. It's now capital RÂ² minus little rÂ². In other words, we're doing the volume of the big one and then doing the volume of the little one, and subtracting them out.

Notice there are squares on both of the rs.

[0:16:00]

If you write this, (R-r)Â² and integrate that, you won't get the same answer. You really could do the two volumes separately and subtract them. But it's often faster to square them, subtract them and then integrate, so let's get rid of this. Drop in all of the axes.

This problem didn't give you the x or y limits, but what it did was give you enough formulas, that you can lock down what the shape is. We have x equals 0, that's this vertical line. We have x equals 3. That's this vertical line. We have f(x) equals xÂ² + 5, it's this parabola. And g(x) equals x + 2, it's that tilted line. So the region that's trapped in there is shaped like that and it's going to be rotated around the axis which is down here. I'm going to draw in some of the shapes now.

If you rotate the outer function, it gives you something like this. If you rotate the inner function, you get something like this. And then you have a nice little washer. You can have lots of these washers, another one like there and there. And if you integrate all those washers up, you're going to get the volume. So, final result is going to be a shape that looks something like that. We've got to get to work now. Remember the last two problems, for the radius, you just use the functions. That's it. It's more complicated this time. Radius is radius. So radius is the distance from the function to the axis.

[0:18:00]

So it's the function minus the axis. Big one first, capital R. R(x), that's the distance from here the axis, to here, the outer function. It's the bigger radius, that's R. Now, if the axes are just the x axis, the distance would be the function, would be xÂ² + 5. But xÂ² + 5, goes something like this. The axis goes something like this. The function result is that length and then you have to add in a bit more because it's farther from the axis from this axis y equals -1 than it is from the function.

So, function minus axis, the function is xÂ² + 5 and I have to subtract the axis so that's -1. It's further away. It's a -1 further way. So our final result for that radius is xÂ² + 6. We'll save that for the set up. Now, we need to find the smaller radius. Small radius is the same thing, it's just a different function minus the same axis. For this one, the smaller function is x + 2. By smaller I mean closer to the axis. And I have to subtract the distance to the axis, so that's minus -1. There's our small one. It is x + 2 minus -1 which is x + 3.

[0:20:00]

So there we got the two parts that we need to go on to the next part. Volume. Both our radii were the large one. The large one was xÂ² + 6, the small one was x + 3 and now we can set it up. Our limits of integration were defined by that left or right vertical lines, so we're ready to go.

Volume is the integral. One of the vertical lines is at 0, the other one was at 3 and I just feel like making mistakes today, see if you can catch it. See if you can catch them all. Look back on your notes for the formula. What did I forget? Something involving areas, it's the squares. And remember, they have to be squared individually like that, and I should really put a big bracket around the whole thing because the whole thing has to be integrated with dx.

Well now, let's go to town on this. We have to square that, square that, subtract the results. So we have pi, integral from 0 to 3. Nice thing though is that, this is just a binomial so when you square xÂ² + 6, you get the first term squared, that's x to the 4th. First term, second terms doubled, that's 12xÂ². You should really memorize the binomial shortcut if you haven't yet. It's super handy. Plus the last thing, squared which is 36 minus.

[0:22:00]

And I really need to have the next parenthesis because if you just square this and leave the minus in front you'll likely to lose some of the minus signs that you need. Squaring x + 3 gives you the first term squared plus the first term last term double that is 6x plus the last term squared, which is 9 and now we're almost up.

So we've got pi times the integral from 0 to 3, put our like terms together. Before I do that I'm going to take that baby step that I think you really should take, even though it seems silly. Of taking that negative in front of the parenthesis and distributing it to everything inside just so you don't mess up a sign. Plus, minus, minus, minus. So x to the 4th, there is no x to the 4th there. 12xÂ² plus -xÂ² is 11xÂ². We have the the plus -6x, we don't want to forget that. And now the 36. So you get 36 plus -9, it is going to be 25. Look at all that stuff. Again, you need to integrate.

Remember this will integrate to be 1/5 of x to the 5th, this will integrate to be 11/3 of x cubed. This will integrate to be -3xÂ², that will integrate to 25x. And after you've substituted all those numbers, all those powers, all those fractions, it does take a while. And if this is on the free response calculator section, you wouldn't bother with this, you'd just type it into the calculator.

[0:24:00]

But if you have to do this manually, this is the result you should get, 1008/5 pi, which is a really big piece of pi, 201.6. We've got enough time to look at one more thing. Take a look.

This one is asking you to find the volume of the shape that's obtained when the region is defined by all that stuff, is rotated around y equals 4. The reason I threw this one in is because, finding the radius on it can be kind of confusing. Let's make a little sketch for this first.

So x, y axis, I need to have a definition of x equals 0, that's one of our boundaries. We have y equals 16, that's another of our boundaries. That's going to be way out here. And the function is the square root of x, and we need one more definition of this boundary. I only have several bounds on that. I don't have quite enough. I don't have a top or bottom bound on there. So I'm going to add that to the problem. I'm going to add that to the problem.

I'm going to add the bound y equals 0. That's just a fancy way of saying the x axis. So here is the shape. One of these bounds is really unnecessary if you think about it. That x equals 0. Because that's the furthest this function can go anyhow. Now let's draw in the axis, the line y equals 4. Here is where the interesting part comes. See when x is 16, you put that into the function, the square of 16 is 4. So height on this spot right here is 4.

[0:26:00]

That means that the axis just touches it and even worse, the axis is above the shape instead of below it. That can make it trickier. You might think, wait a minute. This little problem here, this is just going to be a disc but actually this one is a washer. Believe it or not. Because there's 2 functions that have to be rotated around the axis. There's this one which gives you a nice big circle like so and then there is the smaller one. Imagine if you will the shape that we are getting here is basically a big cylinder and then it's been hollowed out on the inside. It's been hollowed out by this square wood shape.

In a way, imagine that how that square shape is kind of like the bell of a trumpet and then take the bell of the trumpet and then drop it into a cylinder, the volume we're finding is between the bell of the trumpet and the outside of the cylinder. Let's do the set up. The outer unction is the distance between the axis which is at 4. Distance between the axes and this function. This function is just x equals 0 but the radius is still going to be the distance between the axis and the function so in this case, you don;t have to think a lot about that. The distance between y equals 4 and y equals 4. That's the whole outer function. That's okay. Sometimes, they can be that simple.

The inner one. This one can cause you another bit of confusion. On the inner function, it's the distance between the axis and the function. Let me draw another little sketch here so you can see this a little more clearly.

[0:28:00]

Here is our axis at 4, there is our function and little r has to be between the axis and the function. If you use the function, it's not going to give you that, it's going to give you this. That's the function. This is the little radius. But together the function in the little radius make up a total distance of 4, because that's how the axis was. You can get the radius by taking 4 and subtracting the function. Weird, it can happen. It's not always going to be as simple as just taking the function and using the axis. Sometimes you have to do this. 4 minus the function and the function was the square root of x.

Now, I'm going to have this one written out. I'm going to put it into the supplementary materials with all the steps along with the other problems that I've shown you, so you can see everything that's involved in working out the answers.

Today we covered volumes or rotation, which are the volumes that you get when you rotate a shape defined by a certain region around an axis. The first one we did rotated around the x axis, a relatively simple problem. We moved on to one that was a little bit harder, rotating around the y axis. Rotating around the y axis required us to find out the function in terms of a function of y, instead of a function of x.

Next we went to a further elaboration, where we used the washer method. And our axis was not even along the shape. The axis was removed from the shape giving you something like that. Basically the volume minus the volume.

[0:30:00]

The last one we looked at, was one where the axis was attached to the shape and it wasn't on the x axis. So that one again was just one step harder. Well, you are guaranteed to see volumes of rotation on the AP test. Most likely in the free response section. They may ask you one as difficult as where you need to find the intersection between the two functions in order to find the limits of integration. Look in the bonus materials for a practice problems along those lines, including all the steps.

You might also believe it or not, find one of these in the multiple choice. If you do, it might be as simple as a problem that asks you to choose between 5 different possible set ups for finding the volume, in which case, it's pretty quick, you just need to write it out. I suggest writing it out on your own before you even look at the multiple choice answers, because sometimes seeing a multiple choice answer can influence you in the wrong direction. Again, thanks for tuning in and remember keep turning.

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###### John Postovit

University of North Dakota

M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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