Integrals Archives | Jake's Math Lessons

How to Create an Euler’s Method Differential Equations Calculator

Euler’s method is a useful tool for estimating a solution to a differential equation initial value problem at a specific point. In this post, I’m going to show you how to apply Euler’s method both on a piece of paper doing calculations by hand, and in an Excel spreadsheet.

Before we jump into the first example I just want to mention that this uses one of the formulas on my Calculus 2 Study Guide (Integral Calculus Cheat Sheet). It’s available for instant download so you can start using it today. You can learn more about that cheat sheet and buy your copy today by clicking here.

How to Apply Euler’s Method With Differential Equations

We will go ahead and start with this first example here. Use Euler’s method with a step size of 0.2 to estimate $y(1)$ , where $y(x)$ is the solution of the initial-value problem $y'=xy-x^2, \ y(0)=1$ . If you’d prefer to see this example in video form you can watch it here.

I like to solve these problems using a table. I think that’s the easiest way to keep everything you’re doing organized. And then we’re going to use the formula on my study guide to fill in the table row by row until we get to our answer. So let’s just go ahead and start with the formula that’s on my study guide first, and then I’ll show you what I mean by setting up the table.

This is just the information that we would need to be given. We know that we have some initial value problem, where we have $y' = F(x,y)$ . And then we have the initial condition. So, we know that if we plug in $x=0$ , into the solution to the initial value problem, we would get out $y=1$ . Since we know that in this example, we have $y' = xy \ - x^2$ , that tells us right there that we’re going to have

$$F(x, y) = xy \ – x^2$$

To apply this Euler’s method formula, what we need to do is set this up in a table. And you can kind of think of this table like an Euler’s method differential equation calculator. I will show you how to use a computer to make this easier. But it is important to know how it works so that you can do it manually too.

How to Set Up the Euler’s Method Table

We’re going to need a few different columns in our table to keep track of all the calculations. We’re going to start with a column where we keep track of what n we’re on. We also need columns for our $x_{n-1}$ , and our $y_{n-1}$ .

Then we also want to calculate what we get when we plug into, our $F(x_{n-1}, y_{n-1})$ based on the $F(x,y)$ we figured out already. Finally we are going to use all these pieces to figure out our $y_n$ based on the formula discussed earlier from my study guide.

It really is just up to personal preference. If you don’t like keeping of all these columns, you don’t really have to. I like to break it down into the smallest possible pieces, and keep track of each individual piece so that you don’t get lost. I do this and recommend this for you because it’s really easy to get lost when you’re trying to keep track of all these different things.

I like to break it down into, at the smallest possible elements of this formula, and keep track of all those, so that when we put it all together, it’s a lot easier to figure out what’s going on. Doing this will give us the following columns to fill in.

First we want to figure out what you need in the n column. The point that we’re given that we start at is, $x=0$ . So we’re starting at $y(0)$ . And what we’re trying to estimate is $y(1)$ . That tells us using the given step size of 0.2, we’re going to start with $x=0$ and use Euler’s method to first estimate, what the y value is of this solution when $x=0.2$ . Then we’re going estimate what the y value is when $x=0.4$ , then 0.6, then 0.8, and finally when $x=1$ .

To put it in a more formulaic approach, we would take our ending x value minus the starting x value and divide by the step size.

$$n=\frac{x_n – x_0}{step \ size}=\frac{1-0}{0.2}= 5$$

Well, doing this, is going tell us that we need five steps to get from our starting point to our end. So our n column will have one, two, three, four, and five, because that would represent the five, individual steps that we have to take to get up to $x=1$ from $x=0$ .

How to Apply Euler’s Method

Now that we have set up our table, we can start applying Euler’s Method to fill the table out. First of all, we need to start with the x and y value that you’re given. We know when $x=0$ , $y=1$ . So you’re just going to start with those in the $n=1$ row and the $x_{n-1}$ and $y_{n-1}$ columns.

Then what you can do, is plug these two numbers, $x=0$ and $y=1$ into the function that we figured out earlier.

$$F(x, y) = xy \ – x^2$$

$$F(0, 1) = (0)(1) \ – (0)^2 = 0$$

And this will go in the $F(x_{n-1}, y_{n-1})$ column.

Now with this final column here, $y_n$ , what we can do is use this formula that we have here, which is on my calculus two study guide. This will use the previous columns along with our given step size of 0.2, which is denoted by h.

$$y_n = y_{n-1}+hF(x_{n-1},y_{n-1})$$

$$y_1 = 1+(0.2)(0) = 1$$

And then we can put this in our final column of this first row.

Now that we figured out this, we can just carry this piece down into the next column. Whatever your previous $y_n$ was, is just gonna be your $y_{n-1}$ in the next column.

To figure out your next $x_{n-1}$ , all you have to do is take your previous $x_{n-1}$ and just add whatever your step size is. In this case, our step size is 0.2. We’re just going get 0.2 for $x_{n-1}$ in the second row of our table. Doing both of these will give us:

Now Repeat This Process a Few Times

Once you fill out your entire first row, then get the $x_{n-1}$ and $y_{n-1}$ in your second row, the only thing to do is repeat this process. You will now plug in these two x and y values into the $F(x_{n-1}, y_{n-1})$ to get the value in the third column. Then plug those into the formula for $y_n$ to get the forth column. Then figure out the $x_{n-1}$ and $y_{n-1}$ in your third row and continue repeating until your table is full and all 5 rows are filled out. Doing this should leave you with this:

So, after we iterate through this process, all the way up to $n=5$ , we end up getting in 1.194942 in our $y_n$ column. And that should be the answer that they wanted us to find because that should estimate $y(1)$ .

How to Apply Euler’s Method in Excel

Turns out, you can also do these Euler’s method problems using Excel or any other spreadsheet software. And it’s a lot easier and faster than doing it by hand. Watch the videos below to see how you can do this yourself to create an Euler’s method calculator to apply Euler’s method in Excel. And don’t forget to get yourself a copy of my integral calculus cheat sheet to help make the rest of your homework and exams easier and smoother!

Center of Mass Example Problems Using the Center of Mass Equation Integral

Today, we’re going to be going through some center of mass example problems. So, we’re going to sketch the region bounded by the curves and then find the exact coordinates of the centroid of that region. And the curves that we have in this first example are:

$$y=4-x^2, \ and$$ $$y=0$$

If you would prefer to watch a video of this problem, you can do so here:

Centroid vs Center of Mass

Before we jump into this first example, I do want to just point out one thing. And that is the difference between a centroid vs center of mass. The reason I wanna bring that up is because, you can see this problem is asking us to find the coordinates of the centroid. But I said I was going to show you center of mass example problems. Well, they’re pretty much the same thing in most contexts. In this context specifically, they’re going to be the same.

Really, the only difference when you’re looking at a centroid versus a center of mass is that a center of mass is referring to the center point of some actual physical object that actually has a mass. The context that you’d usually see for that is when you have a thin plate, which is described by some functions or some region, and you want to find the center of mass of that plate. Whereas, a centroid usually comes into play where you are just described some region that exists bound between two curves on an x-y-plane.

Like this case here for example. We were just given these functions and we are looking at the region that is trapped between those functions. Since we’re looking at some region, “centroid” would be the proper terminology there. But if you’re ever given some uniform density object like a thin plate, for example, the centroid and the center of mass are actually going to be the same thing. So you would get the same point whether you were thinking of it in either context. They are going to use the same formulas to figure those out.

So that brings me to the center of mass equation integrals, which are two equations that are on my calculus 2 study guide. If you haven’t checked that out you can click here to learn more about that. You can go download that right away. It’s only a few bucks, it’s pretty affordable. And I highly recommend you grab yourself a copy of that. It’s available right away, so you can go start using that today. Before I show you how to apply the formulas on my study guide though, let’s go ahead and start with graphing the given curves.

Sketch the Given Curves and the Bounded Region

First we will start with $y=4-x^2$ . This is just going to be a downward facing parabola with the vertex at the point (0, 4). Then, the line $y=0$ is going to be a horizontal line on the x-axis. Therefore, we would get a region like the one below.

Source: https://www.wolframalpha.com/input/?i=graph+of+region+between+y%3D4-x%5E2+and+y%3D0

So once you’ve sketched your region and kind of given yourself a visualization of what we’re trying to do here, the best place to go from there is to just go straight into the center of mass equation integrals.

Center of Mass Equation Calculus

There’s a separate equation for the x-coordinate of the centroid and for the y-coordinate of the center of mass. Those equations, from my calculus 2 study guide are:

$$\bar{x} = \frac{1}{A} \int_a^b x f(x) \ dx$$ $$\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \Big[ f(x) \Big]^2 \ dx$$

Where the centroid of the region will be at the point $( \bar{x}, \ \bar{y} )$ .

Before I show you how to use these, I do just wanna point out the different pieces of these equations. First of all, we have A in both of these equations. A is just the area of the region whose centroid, or center of mass, we’re trying to find. We would first need to figure out the area between these two curves. And that’s what A would be. I’m not going to show you how to do that here, but you can see more about that by clicking here.

Then we have the bounds of our integrals, the a and b. Those bounds of the integrals are just going to be the x values that are the left and right edge of our region. So in this case, a will be -2. Meanwhile, b will be 2. And that’ll be true for both of those integrals.

Finally, we have f(x). That is just going to be the function that creates this region. These functions assume that the lower bound of the region will be formed by the line $y=0$ . Based on that, f(x) is just going to be the top function, $y=4-x^2$ .

You can simply use those pieces in the formulas above, and I’ll show you how to do that shortly. However, before we do that I do want to point out one thing.

Finding the Center of Mass of a Symmetrical Region

This is an interesting example, because you can see this region whose centroid we are looking for, is actually symmetrical in the x direction. This region is symmetrical to the right and to the left of the y-axis. Whenever you have symmetry in your region that actually saves you some work. Since we have symmetry in the x direction, that tells us that the x coordinate of our centroid must be on that line of symmetry, which is the line $x=0$ . As a result, we already know that we’ll have $\bar{x} = 0$ .

Since we already know $\bar{x}=0$ , all we need to do is use the above formula to calculate $\bar{y}$ .

How to Apply the Center of Mass Formula

If you use that method shown here, you would figure out that the area of this region is $\frac{32}{3}$ . And like I said earlier, the bounds of the integral are the left edge and the right edge of this area. So that gives us $a=-2$ and $b=2$ . I also mentioned above that we will know that $f(x)=4-x^2$ . Plugging this into the $\bar{y}$ equation tells us:

$$\bar{y} = \frac{1}{32/3} \int_{-2}^{2} \frac{1}{2} \Big[ 4-x^2 \Big]^2 \ dx$$

I do want to point out something important about the above equation. When you are applying these equations you put the f(x) all in brackets or parentheses, because we need make sure to square this whole function. From here we can simplify things a bit. When you have a fraction in the denominator of another fraction like we do here, you want to keep in mind that dividing is the same as multiplying by its reciprocal. Therefore, $\frac{1}{32/3}$ can be rewritten as $\frac{3}{32}$ .

Then we can expand and simplify the rest of the integral.

$$\bar{y} = \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^{2} (4-x^2)(4-x^2) \ dx$$

$$\bar{y} = \frac{3}{64} \int_{-2}^{2} 16-8x^2+x^4 \ dx$$

$$\bar{y} = \frac{3}{64} \Bigg[ 16x \ – \frac{8}{3}x^3+ \frac{x^5}{5} \Bigg]_{-2}^2$$

$$\bar{y} = \frac{3}{64} \Bigg[ \Bigg( 16(2) \ – \frac{8}{3}(2)^3+ \frac{(2)^5}{5} \Bigg) – \Bigg( 16(-2) \ – \frac{8}{3}(-2)^3+ \frac{(-2)^5}{5} \Bigg) \Bigg]$$

$$\bar{y} = \frac{3}{64} \cdot \frac{512}{15}$$

$$\bar{y} = \frac{8}{5}$$

That tells us that the y-coordinate of our centroid of this region is $\bar{y}=\frac{8}{5}$ . And we already figured out that the x-coordinate of our centroid was $\bar{x}=0$ . Therefore, the centroid of this region between these two given functions is going to be $(0, \ \frac{8}{5})$ .

Again, the center of mass equations are on my calculus two study guide. You can click here to check that out and get your copy today!

How to find the work required to pump the water out of the tank

In this post, I am going to be showing you how to find the work required to pump the water out of the spout. Work is a common topic in calculus 2, and there are a lot of different applications for it. But finding the work required to pump the water out of a tank is one of the most common applications.

It is also important to keep in mind that the process is slightly different depending on whether your tank is measured in feet or inches versus meters. So, I will show you how both work out. But keep in mind that the process is extremely similar except for one main difference, which I will talk more about more soon, and the units of course.

Find the work required to pump the water out of the spout – Meters

If you are given some sort of a tank like in this case, a spherical tank that we can see has a radius of 3 meters. And then there is the spout sticking out the top of it, which is 1 meter long.

This is going to use one of the formulas mentioned on my calculus 2 study guide. You can click here to check that out, but you can go download that right away. It is available for instant download; it will be a PDF that you can print or save. It is only a couple bucks, so it is very affordable, but this is one of the formulas on there.

The General Strategy

When it comes to figuring out the work required to pump water out of a tank, really what that comes down to is setting up an integral which represents the work required to do this. But when you are setting that integral up, you have to go about it step-by-step and work your way through the different steps. We are not going to go straight to the formula for work quite yet. What we want to do first is imagine, as water is being pumped out of this tank, what is going to be happening is, you can imagine a bunch of little disks. You can see one such disk in the image below.

Each little disk is basically an infinitely thin cylinder, which represents all the water that sits in that layer of our tank. So, you can kind of imagine, instead of thinking of this as a sphere, think of it as a bunch of really thin cylinders stacked on top of each other that make the shape of a sphere. And the reason why we want to think of it like this is each disk of water in this tank, is going to require a certain amount of work since all the water in that layer must be pumped up the same distance. All the water on this layer is going to be the same distance from the top of the spout where we are trying to get that water to. So basically, what we are going to be doing, is figuring out an equation which represents the amount of work required to pump a specific disk out of this tank. And then we are going to sum up all those works that it takes to get each disk, and that be the work that it takes to get all the disks of water out of the tank.

Draw a Sketch and Label Your Variables

To do this, what we want to do, is come up with a new variable, which represents the distance that any given layer of water has to be pumped out. That distance is going to be the distance from whatever layer of water we are looking at up to the top of the spout, which is the distance labeled $x_i^*$ .

And you can imagine, these different disks as we go throughout this whole thing, are all going to have a different distance that they need to be lifted. The disk down at the bottom of the tank is going to have to be pumped further than the disk up at the top of the tank. Once we have this variable that represents the distance that the i^th layer has to be pumped up, what you want to do is figure out an equation for the volume of that layer.

Volume of the i^th Layer

We want to come up with an equation for the i^th layer of the liquid. This is going to be based on $x_i^*$ because, you can see as you go throughout this sphere, the radius of the individual disk changes. The distance from the center of the disk to the edge of the disk is much shorter at the top and the bottom of this sphere, and much longer in the middle of the sphere. In the middle, we know it is 3 meters, but closer to the top or bottom of the tank it is going to be shorter. The volume of each disk is going to be different depending on how far down into the tank you are.

Because the volume of each individual disk is just going to come from the volume of a cylinder equation, we can start with that as a template and adjust from there.

$$V=\pi r^2 h$$

The radius is going to depend on how far down into the tank we are. So, what we need to do is come up with an equation which depends on $x_i^*$ , and represents the radius of that specific disk sitting at the depth of $x_i^*$ . To do that we can graph this circular shape and spout on an x-y-axis. We will say that the top of the spout is at the origin and will graph the depth from the top of the spout ( $x_i^*$ ) on the x-axis, and the radius of the tank at that depth (r) on the y-axis.

https://www.wolframalpha.com/input/?i=y%5E2%2B%28x-4%29%5E2%3D9

What we want to do is figure out an equation for that circle. Generally, the equation for a circle is going to be $(y-y_0)^2+(x-x_0)^2=r^2$ . Since the center of this circle is at (4, 0) and the radius of the tank is 3 m, we can plug in these values to see that the equation of this circle is

$$(y-0)^2+(x-4)^2=3^2$$

Then we can simplify this equation and solve for y, leaving us with

$$y=\pm \sqrt{9-(x-4)^2}$$

So if we want a formula for the radius of our disk we can just take the positive square root piece, since that represents the top half of the circle. Therefore, the radius is going to be $y=\sqrt{9-(x-4)^2}$ . But we’ll want to change the x to $x_i^*$ . So the volume of our i^th disk is just going to be

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ h$$

Now we need to figure out the height. Well, the height of each of these disks will simply be $\Delta x$ . All that means is the distance that you go from one disk, or one layer, of your water to the next. If we make this change, that tells us

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ \Delta x$$

And we can simplify this a bit.

$$V_i = \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Once we have the volume of the i^thdisk, what we needed to do is figure out the mass of that disk.

Mass of the i^th Layer

Once we have found an equation for the volume of the i^th layer of water in the tank, finding the mass of the i^th layer is fairly straight forward. The reason for this is that the mass of the layer will just be the volume of the layer times the mass of water. Since we know the mass of water to be $1000 \frac{kg}{m^3}$ , this tells us that

$$m_i = 1000 * V_i$$

$$m_i = 1000 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Force Acting on the i^th Layer

Again, finding the force acting on the i^th layer is pretty simple once we know the mass of the i^th layer. This is because the force is mass times gravity. Or in other words

$$ F_i = g * m_i $$

Since we know the gravitational constant for the strength of gravity on Earth is $9.8 \frac{m}{s^2}$ , we just have to multiply the mass of the i^th layer by 9.8 to get the force acting on the i^th layer.

$$F_i = 9800 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Work Required to Pump the i^th Layer Out of the Spout

Now that we know the force acting on the i^th layer, we can use this to find the work required to pump the i^th layer of water up and out of the spout. This is where the formula on my calculus 2 study guide comes in. That formula says that work is force times distance.

Since we already know the force of the i^th layer, the only thing left to figure out is the distance that the i^th layer needs to be pumped. Well, if we look back up to the drawing where we named our variables, you will see that $x_i^*$ represents the distance between the i^th layer of water and the top of the spout. So $x_i^*$ is the distance that the i^th layer needs to be pumped. So if we know that

$$ Work \ = \ Force \ * \ Distance$$

And that $x_i^*$ is the distance that the i^th layer needs to be pumped, then

$$W_i = F_i \ * \ x_i^*$$

$$W_i = 9800 x_i^* \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Total Work

At this point we only know the work required to lift the i^th layer, not the work that is required to lift all the water out. What we need to do now is use this to set up an integral which would give us the total amount of work to lift all the i layers, which is all the water. To do that we can set up an integral. And we are going to integrate this work that it takes to lift the i^th layer. Basically, what the integral is doing is summing up all the amounts of work that it takes to lift each individual layer. It is adding all those up together to give us one total amount of work that it takes to pump all the water out of the spout.

When you put it in terms of an integral, this $\Delta x$ changes to a dx. We would also need to change the $x_i^*$ to x. This change of variable just changes us from being in the context of a sum to the context of an integral, which is just a special type of an infinite sum itself.

$$\int 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

But what about the bounds?

Now what we need to figure out is the bounds of our integral. The bounds are going to represent the range of these x values, that we want to integrate over. To do this we want to think about where the x came from. Again, this was the distance that we must pump our water. What we need to figure out is: what is the range that we have to pump each layer of water? What are the total range of distances?

Well, we can see looking back up to our drawing, that the top layer of water is already at the bottom of the spout. It just has to be pumped up 1 meter to get to the top of the spout. So, the shortest distance we are going to have to pump is 1 meter. That tells us that the lower bound of our integral is going to be 1.

The very, very bottom bit of water down at the bottom of our tank is going to have to be pumped all the way up through the tank and then all the way up through the spout. If the radius of our sphere is 3 meters, it is going to have to go 3 meters to get up to the center of our sphere, another 3 meters to get up to the top of the sphere, and then another 1 meter to get to the top of the spout. So, that is going to be $3 + 3 + 1 = 7$ total meters. That tells us that the upper bound is going to be 7.

$$\int_1^7 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

So now we can integrate this and that would give us the total work that it would take to pump all the water out of the spout.

$$9800 \pi \int_1^7 x \Big( 9-(x-4)^2 \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x-4)(x-4) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x^2-8x+16) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( -x^2+8x-7) \Big) \ dx$$

$$9800 \pi \int_1^7 -x^3+8x^2-7x \ dx$$

This can be done using the power rule, then evaluating over our bounds of 1 to 7, we get $1,411,200 \pi$ . So, we know that it would take $1,411,200 \pi$ Joules worth of work to pump all of the water out of the spherical tank.

So, like I said, this is one of the formulas on the Jake’s Math Lessons calculus 2 study guide, click here to go check that out and download your copy today.

Find the work required to pump the water out of the spout – Feet and Pounds

How to Find the Integral of e^x+x^e

$$\int e^x + x^e \ dx$$

Finding the integral of $\mathbf{f(x)=e^x+x^e}$ can be tricky at first glance. I’m sure you’re probably familiar with how to take the integral of the $\mathbf{e^x}$ part of it. In general, you’ll just want to remember that $$\int e^x \ dx = e^x$$

However, the $\mathbf{x^e}$ piece looks a little weird. At first glance it may even look completely foreign. What have you seen that looks like this piece of the function?

Let’s think about what’s going on here

If we look back at our original integral, you’ll want to notice the dx at the end of it. Remember we had $$\int e^x + x^e \ dx$$

The dx is important because it indicates what letter is our variable that we will be integrating with respect to. Since it’s dx, that means we will be integrating with respect to x. Therefore, e will need to be treated as a constant.

In fact, e is a known constant with a specific value. It’s kind of like $\mathbf{\pi}$ . We know that $\mathbf{e \approx 2.71}$ . So when we are trying to integrate the $\mathbf{x^e}$ part of this function, what we are integrating just comes down to “a variable raised up to a constant power.”

When you are trying to integrate x raised up to some constant power, you would want to use the power rule. The power rule for integrating tells us that if n is some constant other than n = -1, then $$\int{x^n} \ dx \ = \ \frac{x^{n+1}}{n+1} \ = \ \frac{1}{n+1} \cdot x^{n+1}$$

So since e is a constant, we can basically just replace the n in the above formula with e to find the integral of $\mathbf{x^e}$ . Using this, we can see that $$\int{x^e} \ dx \ = \ \frac{x^{e+1}}{e+1} \ = \ \frac{1}{e+1} \cdot x^{e+1}$$

Putting these integrals together

Now that we have figured out how to integrate the $\mathbf{x^e}$ part of our function, let’s go back to the original integral. As we do this, let’s also think about one of the basic integral properties. Specifically, the one that tells us what to do about integrating a function that is a sum of two simpler functions. $$\int f(x)+g(x) \ dx \ = \int f(x) \ dx + \int g(x) \ dx$$

So as a result of this, we can break down our problem into two simpler integrals that we already know. $$\int e^x + x^e \ dx \ = \int e^x \ dx + \int x^e \ dx$$

And since we already found both of these integrals earlier, we know $$\int e^x + x^e \ dx \ = \ e^x + \frac{x^{e+1}}{e+1} + C$$

Cylinder/Shell Method – Rotate around a horizontal line

Before reading through this problem, I’d recommend checking out my lesson on finding volumes of rotation using the cylinder shell method. I’m not going to go into quite as much detail here as I did in that lesson. It might help you make more sense of what’s going on if your start there.

Other than that there isn’t much else to add so let’s jump into an example!

Example 1

Find the area of the solid created by rotating the area bounded between $y= (x-1)^3-3$ , $y=-x-2$ , and $y=-2$ about the line $y=-1$ .

Just as before I’ll use the same 4 step process as in the cylinder method lesson.

1. Graph the 2-D functions

As I always say, I suggest starting any problem possible by drawing what is being described to you. Go ahead and start with graphing all of the functions described in the problem. I’ll do this using Desmos. You should end up with something like the graph below. I also went ahead and shaded the bounded region gray to make it a little easier to see (this was not done in Desmos).

$y= (x-1)^3-3$ , $y=-x-2$ , $y=-2$ , and $y=-1$

$y= (x-1)^3-3$ , $y=-x-2$ , $y=-2$ , and $y=-1$

2. Rotate the 2-D area around the given axis

Again, we want to visualize what the question is asking us to find. We will need to take the shaded region in the above graph and rotate it around the line $y=-1$ . Doing this would create a 3-D figure whose volume we’ll need to find. But first let’s draw it.

To do this, imagine the 2-D gray region coming off the paper or screen and rotating around the axis of rotation. Doing this would give us something like the figure below.

Figure resulting from rotating the area around a horizontal. — Result of rotating the region about the line $y=-1$ .

Result of rotating the region about the line $y=-1$ .

3. Setting up the integral

I’m not going to go into as much detail to explain where this integral comes from as I did in the cylinder method lesson, but if the following integral confuses you I’d recommend checking that lesson out by clicking on the above link.

Long story short, we want to imagine our 3-D figure is made up of several infinitely thin cylindrical shells. Adding up the volume of all of these shells would result in an integral like this: $$\int 2 \pi r h \ dr.$$

In order to help with coming up with each of these pieces, we need to relate them back to our figure and the functions that created it. In order to visualize this, let’s draw our figure with one of these infinitely thin shells that make up the entire figure. We can consider this one shell and how to represent these dimensions in terms of the given functions.

You can see one of these cylindrical shells represented in the drawing below with a labeled version of the cylinder draw in the upper-right hand corner.

3-D figure with a cylindrical shell — 3-D figure with a sample cylindrical shell shown in green.

As with all cylinder shell method problems, we need to imagine integrating from the center of the cylinder out to the outer edge. Since our cylinder is laying horizontally, moving from its center to its edge moves up and down. This means we are moving in the y direction. Therefore, we need to integrate in the y direction and represent our integral only in terms of y (we shouldn’t have any x‘s).

So let’s think about each of the three pieces that make up our integral one at a time.

Finding r

The radius of this cylinder would simply be the distance between the center of the cylinder and the edge. You can see in the smaller version of the cylinder drawn off to the side that the radius is represented by the red line measuring between the points labeled $(x_2, \ -1)$ and $(x_2, \ y)$ .

Since these two points have the same x value, we can find the distance between them by simply finding the distance between their y values. To do this we just need to take the larger value and subtract the smaller one from it. $$r=-1-y$$

Finding h

The height of a cylinder will always be measured as the distance between the two flat, parallel faces. Usually they would be the top and bottom, but since our cylinder is sideways, we need the distance between the left side and right side.

Looking at the smaller cylindrical shell off to the side in the drawing above, you can see the height of this cylinder is represented by the red line measuring the distance between the points $(x_1, \ y)$ and $(x_2, \ y)$ .

Similar to what we did before, these two points have the same y value. As a result, the distance between them would be the same as the distance between their x values. So we just need to take the larger x value and subtract away the smaller one. $$h=x_2-x_1$$

But remember earlier I said we need everything just in terms of y?

So we need to think about how we can rewrite $x_1$ and $x_2$ in terms of y.

Finding $\mathbf{x_1}$

We know that $x_1$ lies on the function $y=-x-2$ so we know that the relationship between $x_1$ and y can be described in the same way $$y=-x_1-2.$$ If we rearrange this to solve for $x_1$ instead of y, we can use this to replace the $\mathbf{x_1}$ in our equation for h. $$y=-x_1-2$$ $$y+x_1=-2$$ $$x_1=-y-2$$

We can use this to rewrite h but replace the $x_1$ with $(-y-2)$ since we know they are equal. $$h= \ x_2- (-y-2)$$ Now we need to do the same thing with $x_2$ .

Finding $\mathbf{x_2}$

We are going to apply the same idea here as in the previous section. We know that $x_2$ lies on the function $y= (x-1)^3-3$ . Therefore, we can describe the relationship between $x_2$ and y as $$y= (x_2-1)^3-3.$$ Now we can solve this equation for $x_2$ and plug this into our equation for h. $$y \ = \ (x_2-1)^3-3$$ $$y+3 \ = \ (x_2-1)^3$$ $$\sqrt[3] {y+3} \ = \ x_2-1$$ $$\sqrt[3] {y+3} +1 \ = \ x_2$$ Now going back to our equation for h, this tells us $$h \ = \ \sqrt[3] {y+3} +1 – (-y-2).$$ And to simplify a bit: $$h \ = \ \sqrt[3] {y+3} +1 + y+2$$ $$h \ = \ \sqrt[3] {y+3} + y+3.$$ Now that we have h and r, we just need to find dr.

Finding dr

This is actually the simplest part to find. The dr represents the change in the cylinder’s radius as we go from each shell to the next. Since we move in the same direction of the radius as we integrate to find our volume, the change in r should be the same as the change in y between each step. Therefore, we can say that $$dr=dy.$$

Putting it all back into an integral

We already figured out that the volume of our figure can be found by using the integral $$\int 2 \pi rh \ dr.$$ And we just found these three pieces to be $$r=-1-y$$ $$h \ = \ \sqrt[3] {y+3} + y+3$$ $$dr=dy.$$ So we can just plug them into our integral. $$ \int 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy$$

Now we need one last piece. We need to add bounds on the integrals.

Since we are integrating with respect to y, the bounds of our integrals need to be the range of y values that make up our original 2-D area. Looking back at our original graph, we can see that the original area bounded by the given functions spans over all of the y values between $y=-2$ and $y=-3$ . Therefore, we know that the volume of our figure will be $$V \ = \int_{-3}^{-2} 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy.$$

4. Solve the integral

Now all we need to do is solve the integral we just found and that will leave us with our volume. This is actually a pretty complicated integral as is it, so let’s start with simplifying it a bit. We’ll do this by pulling out the constant, distributing out through the parenthesis, and combining like terms.

$$V \ = \int_{-3}^{-2} 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy$$ $$V \ = \ 2 \pi \int_{-3}^{-2} \ – \big(y+3 \big)^{\frac{1}{3}} – y \ – 3 -y \big(y+3 \big)^{\frac{1}{3}} – y^2 – 3y \ \ dy$$ $$V \ = \ 2 \pi \int_{-3}^{-2} \ – \big(y+3 \big)^{\frac{1}{3}} -y \big(y+3 \big)^{\frac{1}{3}} – y^2 – 4y -3 \ \ dy$$

Now that we have it in a form that is simplest to integrate we can go ahead and integrate this function one term at a time. I’m not going to show every step of how to do this, but if you’d like to work it out on your own, I’d suggest using u-substitution on the $-(y+3)^{1/3}$ term and using integration by parts on the $-y(y+3)^{1/3}$ term.

$$V \ = \ 2 \pi \Bigg[ – \frac{3}{14} \big( y+3 \big)^{\frac{4}{3}}\big( 2y-1 \big) – \frac{1}{3}y^3 – 2y^2 – 3y \Bigg]_{-3}^{-2}$$

Again, I’m not going to show every step of this. Instead I used Wolfram Alpha from here, but if you evaluate this expression from $y=-3$ to $y=-2$ , you’ll see that $$V \ = \ 2 \pi \bigg(\frac{73}{42} \bigg)$$ $$V \ = \ \frac{73 \pi}{21}$$

Hopefully all of this helps you gain a bit of a better understanding of this method, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I’ll send you my bonus FREE calc 1 study guide! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about integrals as well.

Rotating Volumes with the Cylinder/Shell Method

Similar to using the disk or washer method, we will use the cylinder method to find the volume of a solid. Specifically, it’s used when we rotate a function or region around an axis of rotation. In fact, most problems that require finding the volume of a solid of rotation can use the disk/washer method or the cylinder method. However, one will usually be significantly easier.

I’ll explain what I mean by this with an example.