Did you know that the washer method is an extension of the disk method for finding the volume of a solid of revolution to cover solids with a hole?
It’s true!
Let’s jump in and find out more!
Background
To understand how-to, let’s remind ourselves how we would calculate a shaded region area as we did in geometry.
Suppose we are asked to find the area of a rectangle with a triangle missing from the middle.
What would we do?
First, we would see the area of the rectangle and the area of the triangle separately.
Then we would subtract these two values to find the remaining area as seen below.
Well, we can do the same thing for finding solids of revolution. We are going to take a disk and remove a portion.
Assume we have a rectangle that is perpendicular to the axis of revolution, but the rectangle is not directly touching the axis of revolution.
How would we calculate the area of this rectangle? See below.
This means, when we revolve the rectangle about the axis of revolution, we will be finding the volume of the outer radius (R) minus the inner radius (r).
\begin{equation}
V=\pi R^{2} w-\pi r^{2} w=\pi\left(R^{2}-r^{2}\right) w
\end{equation}
Consequently, if we apply this technique for an infinite number of rectangles, we can find the volume of the solid formed by revolving a bounded region about an axis using the following formula.
\begin{equation}
V=\pi \int_{a}^{b}\left(R^{2}-r^{2}\right) d x
\end{equation}
Awesome!
The Washer Method (Step-by-Step)
So, let’s look at an example and see the washer method for solids of revolution in action.
Find the volume of the solid formed by revolving the region bounded by the graphs about the x-axis.
\begin{equation}
y=x^{2} \text { and } y=\sqrt{x}
\end{equation}
Step 1:
First, we will graph our bounded region.
Step 2:
Next, we will identify our rotation axis and create our vertical, rectangular slice perpendicular to the axis of rotation (i.e., the x-axis). In doing so, we determine our thickness to be dx.
Step 3:
Now we must determine our outer radius, R, and our inner radius, r.
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}^{1}(\sqrt{x})^{2}-\left(x^{2}\right)^{2} d x \\
V=\pi \int_{0}^{1}\left(x-x^{4}\right) d x=\pi\left(\frac{x^{2}}{2}-\frac{x^{5}}{5}\right]_{0}^{1}=\frac{3 \pi}{10}
\end{array}
\end{equation}
Wow! We just found that the volume of the bounded region when rotated about the x-axis!
See, not so bad!
Summary
Together, we will work through an abundance of questions in detail to find the volume of a solid generated about the x-axis, y-axis, or any horizontal or vertical line, whose cross-sections are washers.
It’s going to be fun, so let’s get to it!
Video Tutorial w/ Full Lesson & Detailed Examples (Video)
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