APCALC-063Applications of Integration

These resources will help you better understand integrations and their role in calculus.

Overview of the Disk and Washer Method

This video tutorial explains how to use the disk method and the washer method to calculate the volume of a solid when the region enclosed by the curves are rotated around the x axis, y axis, and non axis or another line parallel to the x or y axis.

Video Transcript

In this video, we’re going to talk about how to calculate the volume by rotating a region around the X axis or the Y axis using the disk method and the washers method.

So let’s begin. Let’s say if we have a curve, and the curve looks like this, and we wish to calculate the volume of the curve from, let’s say, a to b.

One method that we can use is the disk method. And the disk method works by taking a cross-sectional area- this is going to be the radius. And if we take the cross-sectional area, we can turn it into a disk.

As we rotate the region about the x-axis, we could form basically something that looks like a disk. Now, I’m going to draw this disk the other way. Notice that this forms the shape of a cylinder.

The cylinder has a radius r . And it has a height h. The volume of a cylinder is basically the volume of the cross-sectional area times the height, the cross-sectional area being the circle on top, which is pi r^2.

Thus the volume of a cylinder is pi r^2 times height.

And so the volume of this blue disk, it has a height h, which you can also view it as dy/dx or delta x, and it has a cross-sectional area, which is pi r^2.

Now, to find the volume of the solid we need to add up all the little disks from a to b. If you make the other disk, there’s another disk that you can make.

My drawings are kind of terrible, but basically the idea is that we add up all these smaller disks to find the total volume of the object, or the solid, that is created by revolving about the x-axis.

But to find the volume, we’ve got to find the volume of, we have to integrate the cross-sectional area from a to b.

And the area is basically pi r^2, where r is going to be a function of x.

And so this is the equation to find the volume using the disk method when revolving about the x-axis.

The main idea you need to understand is that the cross-sections that are used for this technique are basically a circle, and the area of a circle is pi r^2.

Now sometimes you may need to find the area when it’s revolved or rotated about the y-axis.

Let’s just go over some basic things.

So if you want to find the volume rotated about the x-axis, this is going to be the shaded region, and it has to be in terms of x.

So the radius is between the x-axis, that’s where the solid is being rotated about, and it’s between the curve, so that’s the radius in terms of x.

So the volume is going to be from, you have to integrate the function from a to b, and this is going to be r^2 of x dx. Let’s say you find the volume of the solid rotated about the x-axis.

Now, let’s say if you want to find it about the y-axis. Instead of using a and b, we’re going to use c and d, which represent y-values.

The radius is going to be between the y-axis and the curve, since we’re rotating about the y-axis.

So the volume’s going to be pi integration from c to d, r of y^2 times dy.

So that’s how you can find the volume rotated about the y-axis, or a line parallel to the y-axis.

And this is for the x-axis or a line parallel to the x-axis.

Alright, let’s start with an example. Let’s say we have the function y is equal to the square root of x, and we want to find the volume when rotating this curve about the x-axis, and we want the portion of the curve where it’s bounded between x=0 and x=4.

So feel free to try this problem.

The first thing we want to do is you want to plot the function.

The square root of x looks like this. The radius is the distance between the curve and the axis of rotation. In this case, we’re rotating about the x-axis.

And we want the portion of the curve between x=0 and x=4.

So this is the region that we want.

The first thing we’re going to do is find out what r(x) is equal to. R(x) is equal to y, which, y is equal to square root x, so r(x) is equal to both of these things, but we want it to be in terms of x.

So we’re going to use square root x.

Now, let’s use the equation. The volume is equal to pi times the integration from a to b r(x) dx, and let’s not forget to square it.

So this is going to be pi, a is 0, b is 4, r(x) is the square root of x, and we’re going to square that, and then we have dx.

Root x^2 is basically x.

So let’s integrate this function. The anti-derivative of x is going to be x^2/2. You have to add one to the exponent and then divide by that result.

So now let’s find the value of this definite integral. So first, let’s plug in 4. So it’s going to be 4^2 over 2, and then we plug in 0.

4^2 is 16, 0^2 is just 0, and 16 divided by 2 is 8. So the final answer is 8 pi.

Now it’s your turn.

Find the volume of the solid generated by rotating the function y=1/x about the x-axis, bounded by the region x=1 and the line x=3.

So feel free to pause the video and work on this example. Use the disk method to calculate the volume.

So the first thing we need to do is plot the function. So the right side of 1/x, it looks like this. But we only want the portion from 1 to 3. So we only want this region.

And we’re gonna rotate it about the x-axis.

So the radius, r(x), is the distance between the x-axis, or the axis of rotation, and the curve. So r(x) is simply equal to the function 1/x.

The way you find it is you take the top part of the function, or the top part of this line which is 1/x, minus the bottom part, which is y=0, that’s the x-axis. So it’s 1/x minus 0, which is just 1/x.

Now, even though that process seems useless at this point, it’s useful when you’re rotating about an axis that is neither the X or Y axis, which we’ll cover later in this video.

But for now, r(x) is simply 1/x.

So the volume is gonna be pi integration from a to b, or 1 to 3, and then it’s gonna be r(x)^2 dx.

So this is pi, 1 to 3, (1/x)^2, dx.

(1/x)^2, we’ll need to rewrite it. We can write it as x^-2, and now let’s use the power rule, let’s add 1 to the exponent. -2 plus 1 is -1, and then we need to divide by -1.

So rewriting the function, we can bring the x back to the bottom to get rid of the negative exponent, so this is gonna be pi times -1/x, evaluated from 1 to 3.

So this is gonna be pi times (-⅓ minus -1/1). Always start with the top number 3, and then subtracted by the 1 on the bottom.

So this basically -⅓ plus 1/1. Now, we need to get common denominators, so let’s multiply this by 3/3.

So what we have is -⅓ plus 3/3. -1 plus 3 is 2, so the final answer is just 2 pi over 3.

Now, let’s try a few examples of finding the volume of a solid when it’s rotating about the y-axis.

So let’s say that y=x^2, and the curve is bounded by the lines x=0 and y=4, and we’re rotating about the y-axis.

So let’s begin by drawing a graph.

So this is the y-axis which we’re rotating about, and the graph y=x^2, the right side of it, looks like this.

Now, this is the line x=0, it’s basically the y-axis. And we have the line y=4, which is basically a vertical line.

So this is the shaded region. That’s the region that we’re interested in.

The radius is between the axis of rotation, which is the y-axis, and the curve. So that is basically, that’s r(y).

Now notice that r(y) is the same as x. X is the distance between the y-axis and the curve. So we could say that r(y) is equal to x.

Now, we need to get r(y) in terms of y, if we want to find the volume rotated about the y-axis. So we need to find out what x is equal to in terms of y.

So we have this equation: if y is equal to x^2, then if we take the square root of both sides, we can see that x is equal to the square root of y.

So let’s replace x with root y.

So the radius in terms of y is equal to the square root of y.

Now to find the volume of this region when it’s rotated about the y-axis, we can use this equation: v=pi integration from c to d, those are values on the y-axis, r(y) dy squared.

At the origin, the y value is 0, and at the line y=4, y is 4. So therefore c is 0, d is 4.

r(y), we know it’s the square root of y, and we have to square it.

The square root of y^2 is just gonna be y. So now we can integrate the function.

The integration of y to the first power (y^1) is y^2/2, evaluated from 0 to 4, and let’s not forget the constant in front.

So now let’s plug in those numbers. Let’s start with the top number 4.

So it’s 4^2/2 minus 0^2/2. 4^2 is 16, 0^2 is just 0, and 16 divided by 2 is 8. So the volume is gonna be 8 pi.

Now it’s your turn. Try this example: Let’s say y is equal to x raised to ⅔ (x^⅔) , and the curve is bounded by the lines x=0 and y=1. And we’re gonna rotate it about the y-axis. Find the volume of this solid that forms once you rotate this curve bounded by those lines about the y-axis.

Feel free to pause the video as you work on this example.

So let’s begin with a graph.

So the graph y=x^⅔ is an increasing function, and it increases at a decreasing rate and looks like that. The line x=0 is basically the x-axis, and y=1, that’s a horizontal line.

So we can see that the y values are 0 and 1.

And we’re rotating about the y-axis. So this is the shaded region, and the radius, that’s r(y).

Now once again r(y) is equal to x, and it turns out that r(x) is equal to y. Now let me explain.

So let’s say if we have this curve and we know this is x, x is the distance between the y-axis and the curve, and that’s also the radius relative to the y-axis.

So r(y) is always going to be equal to x.

Now in the first example when we had y equal x^2 and we were revolving it about the x-axis, r(x) was the distance between the x-axis and the curve. But notice that r(x) is parallel to the y-axis. So r(x) is equivalent to y.

But since y is equal to x^2, and if r(x) is equal to y, r(x) is equal to x^2. And that’s what we did in the first example.

It’s important to understand that r(x) is equal to y, and r(y) is equal to x. Now let’s go back to this problem.

So if r(y) is equal to x, what’s r(y) in terms of y?

So now let’s use the equation that we had in the beginning. So we know y is equal to x^⅔ so we need to solve for x. And the only way we can do that is by raising both sides to 3/2, which is the reciprocal of ⅔.

When you raise one exponent to another, you need to multiply the two exponents. ⅔ times 3/2 is 1. The 3s cancel and the 2s will cancel.

So you get x^1 or simply x. Therefore x is y^(3/2). And since r(y) is equal to x, then r(y) is also equal to y^(3/2). All we need to do is replace x with y^(3/2).

Now once we have the radius in terms of y, we can find the volume of the solid that is produced when rotating about the y-axis.

So now let’s use this formula.

So let’s replace c with 0 and d with 1, and let’s replace r(y) with y^(3/2), and let’s not forget to square it.

y^(3/2)^2, and we need to multiply the two exponents. 3/2 times 2, which is 2/1, the 2s cancel so you just get 3.

So what we now have is v=pi integration 0 to 1 y^3 dy. The anti-derivative of y^3 is y^4/4. Evaluated from 0 to 1, multiplied by pi. So let’s begin by plugging in 1.

So it’s 1 to the fourth over 4 (1^4/4) minus 0 to the fourth over 4 (0^4/4).

1^4 is basically 1, and 0^4 is just 0, so the final answer is pi over 4 (pi/4). And that’s the finished problem.

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