When a region in the plane is revolved around an axis, it forms a three‑dimensional solid. The Disk Method and Washer Method use definite integrals to compute the volume of such solids. The idea: slice the solid perpendicular to the axis of revolution to get circular disks or washers, then integrate their areas.
🎯 In this section you will learn
- The Disk Method: $V = \pi \int_a^b [R(x)]^2 dx$ (solid with no hole)
- The Washer Method: $V = \pi \int_a^b ([R(x)]^2 - [r(x)]^2) dx$ (solid with a hole)
- How to choose the radius functions $R(x)$ and $r(x)$
- Revolving around the x‑axis, y‑axis, and other horizontal/vertical lines
📌 The Disk Method (Solid with No Hole)
$$ V = \pi \int_a^b [R(x)]^2 \, dx $$
where $R(x)$ is the distance from the axis of revolution to the curve
💡 Visualizing the Disk Method
Imagine slicing the solid perpendicular to the axis of revolution. Each slice is a thin disk of radius $R(x)$ and thickness $dx$. Volume of a disk = $\pi R^2 \cdot dx$. Summing (integrating) gives the total volume.
Find the volume of the solid obtained by revolving the region under $y = \sqrt{x}$ from $x=0$ to $x=4$ about the x‑axis.
①
Identify $R(x)$
$R(x) = \sqrt{x}$ (distance from x‑axis to the curve)
②
Set up the integral
$V = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx$
③
Evaluate
$V = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi (8 - 0) = 8\pi$
Find the volume of the solid obtained by revolving the region under $y = \sqrt{x}$ from $x=0$ to $x=4$ about the y‑axis.
①
Express $x$ in terms of $y$
$y = \sqrt{x}$ ⇒ $x = y^2$
②
Limits in $y$
When $x=0$, $y=0$; when $x=4$, $y=2$
③
Set up the integral (disk method in $y$)
$R(y) = y^2$, $V = \pi \int_0^2 (y^2)^2 dy = \pi \int_0^2 y^4 dy$
④
Evaluate
$V = \pi \left[\frac{y^5}{5}\right]_0^2 = \pi \cdot \frac{32}{5} = \frac{32\pi}{5}$
📌 The Washer Method (Solid with a Hole)
$$ V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx $$
where $R(x)$ is the outer radius, $r(x)$ is the inner radius
💡 Visualizing the Washer Method
When the solid has a hole (like a pipe), each slice is a washer — a disk with a smaller disk removed. The area of a washer is $\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$.
Find the volume of the solid obtained by revolving the region between $y = x^2$ and $y = x$ around the x‑axis.
①
Find intersection points
$x^2 = x$ ⇒ $x=0, 1$
②
Identify $R(x)$ and $r(x)$
Outer radius: distance to the top curve ($y=x$): $R(x)=x$
Inner radius: distance to the bottom curve ($y=x^2$): $r(x)=x^2$
③
Set up the integral
$V = \pi \int_0^1 [x^2 - (x^2)^2] dx = \pi \int_0^1 (x^2 - x^4) dx$
④
Evaluate
$V = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi (\frac{1}{3} - \frac{1}{5}) = \pi (\frac{5}{15} - \frac{3}{15}) = \frac{2\pi}{15}$
Find the volume when the region bounded by $y = x^2$ and $y = 4$ is revolved around $y = 4$.
①
Find intersection points
$x^2 = 4$ ⇒ $x = \pm 2$
②
Identify $R(x)$
Distance from the axis $y=4$ to the curve $y=x^2$ is $4 - x^2$ (disk method, no hole)
③
Set up the integral
$V = \pi \int_{-2}^2 (4 - x^2)^2 dx = 2\pi \int_0^2 (16 - 8x^2 + x^4) dx$ (by symmetry)
④
Evaluate
$2\pi \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = 2\pi (32 - \frac{64}{3} + \frac{32}{5}) = 2\pi (\frac{480}{15} - \frac{320}{15} + \frac{96}{15}) = 2\pi \cdot \frac{256}{15} = \frac{512\pi}{15}$
Find the volume when the region bounded by $y = x^2$ and $y = 2x$ is revolved around the y‑axis.
①
Find intersection points
$x^2 = 2x$ ⇒ $x=0, 2$; corresponding $y=0, 4$
②
Express $x$ in terms of $y$
$y = x^2$ ⇒ $x = \sqrt{y}$ (right side)
$y = 2x$ ⇒ $x = y/2$
③
Identify $R(y)$ and $r(y)$ (outer and inner radii from y‑axis)
For $y$ from $0$ to $4$, the farther curve from the y‑axis is $x = \sqrt{y}$ (outer), the closer is $x = y/2$ (inner)
④
Set up and evaluate
$V = \pi \int_0^4 [(\sqrt{y})^2 - (y/2)^2] dy = \pi \int_0^4 (y - \frac{y^2}{4}) dy = \pi \left[\frac{y^2}{2} - \frac{y^3}{12}\right]_0^4 = \pi (8 - \frac{64}{12}) = \pi (8 - \frac{16}{3}) = \pi (\frac{24}{3} - \frac{16}{3}) = \frac{8\pi}{3}$
⚠️ Common Mistakes
- Confusing radius: $R(x)$ is the distance from the axis of revolution to the curve, not the $y$-value unless revolving around the x‑axis.
- Forgetting to square: The formula uses $[R(x)]^2$, not $R(x)$.
- Washer order: It's $R^2 - r^2$, not $(R - r)^2$.
- Incorrect limits: Make sure the limits correspond to the variable of integration.
🔍 Key Takeaways
- Disk Method: $V = \pi \int [R(x)]^2 dx$ (solid with no hole).
- Washer Method: $V = \pi \int ([R(x)]^2 - [r(x)]^2) dx$ (solid with a hole).
- Integrate with respect to $x$ when the axis is horizontal, with respect to $y$ when the axis is vertical.
- Radius = distance from the axis of revolution to the curve.
- Always draw a picture to identify the correct radii.
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