Rotating functions around an axis to create a 3-D shape then finding its volume is one of the more common applications of integrals. This is commonly referred to as **finding a volume using the disk method**. It seems like a complicated type of problem, but if you think about what you are actually measuring it isn’t so bad.

Let’s think about a specific example. Imagine taking the function between and and rotating it around the x-axis then finding the volume of this solid using the disk method.

## 1. Graph the 2-D function

The first thing I would recommend doing with a problem like this is to graph the function that’s given to you. Here we are graphing within our given domain of . Using Wolfram Alpha we can see this graph below.

## 2. Rotate the 2-D function around the given axis

Once you graph the function on the 2-D x-y-plane we need to imagine rotating it around the axis given in the problem. This will result in creating a 3-D figure whose volume we need to find.

In this case we need to rotate this portion of our function around the x-axis. Another way to say this is that we are rotating around the line . Again, using Wolfram Alpha we can see what this figure would look like.

I would always recommend drawing out the 2-D graph and 3-D rotation anytime you need to find the volume of a solid like this. It helps to visualize the solid whose volume you are trying to measure and it makes it much easier to make sure you are setting up your problem correctly.

## 3. Setting up the integral

As with any problem where we need to find a volume using the disk method, what we want to imagine here is having infinitely thin cylinders stacking up to create our 3-D figure. **When we add up the volumes of all of these infinitely thin cylinders, get the volume of the entire figure.** This is all the integral is doing.

You can see in this drawing our function has been rotated around the x-axis to create a round cone-like 3-D figure. The green cylinder in the figure represents one of the infinitely thin disks that we are slicing the figure into.

Since we are trying to make this integral represent the sum of all of these disks, we need to think about the volume of each disk in particular. Clearly each disk is a very thin cylinder. So in order to find their volumes, we should start with the volume of a cylinder, which is

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

Thinking back to our example of rotating around the x-axis, let’s determine the radius and height of our cylinders.

### How do we find the radius?

Considering that each cylinder will be a different size, it seems clear that the radius of each cylinder will depend on which cylinder we’re considering. In fact, if you look at our drawing, you can see that the radius of each cylinder will simply depend on the *x* value where it’s sitting.

You can see in the drawing above that I drew a copy of the disk we are considering down below the function. Imagine we are trying to find the distance between the two points we labeled. The center point of our disk is labeled *(x, 0)*. **We know that the center of our disk will always have a y-coordinate of ****0**** because we rotated our function around the line** . And we will leave the x-coordinate as the variable *x* because we are trying to find the volume of any disk along this figure with all different *x *values, not just the one disk drawn above.

Now consider the upper point. This point is labeled *(x, y)*. This is just meant to be **any ****(x, y)**** combination that sits on our function** . This point will always have the same x-value as our other labeled point, so **the distance between these two points will simply be the distance between their y-coordinates**, which are *y* and *0*. To find the distance between these two values, we just need to do the larger value minus the smaller one, or , which is just *y*.

But we need this to be in terms of *x*. Remember we realized earlier that the radius of our disk will depend on *x*, so we want everything in terms of *x*. We know that our point *(x, y)* we were looking at is some point that lies on our function. **So we know** . Therefore, if the radius of our disk is *y*, we can also say that the radius is . So,

$$r=x^2.$$

### How do we find the height?

The height of our infinitely thin cylinders is actually quite simple. Just like when we integrate a 2-D function to find the area under the curve, our slices here are all the same width. We don’t have to worry about each disk, or cylinder, having a different height.

The height of each cylinder will just be how far we always move over before taking another slice. Since we are moving over in the *x* direction as we imagine the next slice, this can simply be our **change in ****x**** between the slices. Change in ****x ****is always represented as ****dx****.** So we can simply say the height of each cylinder is

$$h=dx.$$

If we were instead rotating our function around a vertical line (like ) the height of our disks would be *dy*. **This is only if we are using the disk method and would not necessarily be the case when using the shell method.**

### Back to the integral

Now we got our height and radius of each disk we get from slicing this figure. Putting this together, we can say that the volume of a single disk can be represented as

$$V=\pi \big( x^2 \big)^2dx.$$

But this of course is just one disk. We need to add up the volume of all of the disks to get the volume of the full figure. This is exactly what the integral accomplishes. Before doing this, remember we are only rotating this function between and . So these will be the bounds of our integral. Therefore, the volume of our entire figure can be found with the following integral.

$$\int_0^2 \pi \big( x^2 \big)^2dx$$

## 4. Solve the integral

We made it through the hard part! Now all we need to do is solve this integral and we will have the volume of our figure. First let’s simplify this integral a little, then we can integrate using the power rule and evaluate at the given bounds.

$$\int_0^2 \pi \big( x^2 \big)^2dx$$

$$\pi \int_0^2 x^4 dx$$

$$\pi \Bigg[ \frac{1}{5} x^5 \Bigg]^2_0$$

$$\frac{\pi}{5} \Big[ (2)^5 – (0)^5 \Big]$$

$$\frac{\pi}{5} \Big[ 32 \Big]$$

$$\frac{32 \pi}{5}$$

So we found that the volume of the solid is ! Hopefully this has helped you with the disk method, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions about integrals. Once you know and understand the disk method, another good application of integrals to check out would be the washer method. If you can’t find the topic or question you’re looking for just let me know by emailing me at **jakesmathlessons@gmail.com**!

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