So what’s the disk method?
And how does it compare the shell or washer method?
All great questions, that we’re going to answer in today’s lesson.
Let’s jump in!
Background
What if I said that a cookie is a perfect way to visualize how to find the volume of a solid of revolution?
Yep!
Our goal is to take a two-dimensional bounded region and rotate it about a given axis to create a solid of revolution.
But how does this relate to a cookie?
First, we start with a rectangle because it’s a really easy object to find the area.
Then, we spin this rectangle around a line and notice that it creates a circular shape.
Next, we realize that this rectangle creates more than just a circle because it has a thickness (width). And if this circle has a thickness, it makes something that looks like a disk.
And a disk looks just like a cookie! Gosh, math is delicious!
Disk Method Equations
Okay, now here’s the cool part.
We find the volume of this disk (ahem, cookie) using our formula from geometry:
\begin{equation}
\begin{array}{l}
V=(\text { area of base })(\text { width }) \\
V=\left(\pi R^{2}\right)(w)
\end{array}
\end{equation}
But this will only give us the volume of one disk (cookie), so we’ll use integration to find the volume of an infinite number of circular cross-sections of the 3D solid.
Cool!
Important Notes
Now here some critical things to note.
- First, our bounded region must be entirely flush against the axis of rotation to ensure that we will create a disk when rotated.
- Secondly, our green rectangle represents either a vertical or horizontal slice and will always be perpendicular to the axis of rotation.
- And lastly, our R-value represents the height of our rectangle, which is precisely the same as finding the area between two curves:
\begin{equation}
R=(\text { top function })-(\text { bottom function })
\end{equation}
Disk Method Example
Alright, so now let’s walk through an example to help us make sense of everything.
Suppose we are asked to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis:
\begin{equation}
y=x, y=0, x=0 \text { and } x=4
\end{equation}
Step 1:
First, we will graph our bounded region.
Step 2:
Next, we will identify our axis of rotation and create our vertical, rectangular slice perpendicular to the axis of rotation (i.e., the x-axis).
Step 3:
Now we must determine our bounds, or width of the bounded region, which will become our bounds of integration as well as our top and bottom function.
\begin{aligned}
&\text { Limits of Integration: } \underbrace{0}_{a} \leq x \leq \underbrace{4}_{b}\\
&R=\underbrace{(\text { top function })}_{y=x}-\underbrace{(\text { bottom function })}_{y=0}\\
&R=(x)-(0)
\end{aligned}
\end{equation}
Step 4:
Lastly, we plug everything into our formula and integrate it to find the volume of the resulting solid of revolution.
\begin{equation}
\begin{array}{l}
V=\pi \int_{a}^{b}\left(R^{2}\right) d x=\pi \int_{0}^{4}(x-0)^{2} d x \\
V=\pi \int_{0}^{4} x^{2} d x=\pi\left(\frac{x^{3}}{3}\right]_{0}^{4}=\pi\left(\frac{(4)^{3}}{3}-\frac{(0)^{3}}{3}\right)=\frac{64 \pi}{3}
\end{array}
\end{equation}
Wow! We just found that the volume of the bounded region when rotated about the x-axis!
Summary
Together, we will work through numerous questions, step by step, to find the volume of a solid generated about the x-axis, y-axis, or any horizontal or vertical line, whose cross-sections are disks.
Now it’s time to grab a cookie (disk) and click on the video!
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