From a Planar Region to Circular Slices
A planar region can generate a three-dimensional solid when it is rotated around a line in its plane. To find the volume by cross-sections, imagine slicing the solid with planes perpendicular to the rotation axis. When the original region reaches the axis, each such slice is a filled circle, or disk.
The two measurements that determine the volume integral are the radius of each disk and the thickness between neighboring slices. The radius is a distance measured perpendicular to the axis. The slice thickness is measured parallel to the axis. Keeping these directions distinct is essential: rotating a region around a horizontal axis calls for horizontal thickness, while rotating around a vertical axis calls for vertical thickness.
This tutorial considers regions that extend from the axis of rotation to a boundary curve, so their cross-sections are disks rather than circles with holes. The volume principle for cross-sections says to integrate the area of each cross-section over the coordinate along the solid.
Why a Slice Becomes a Disk
Take a thin segment of the region that is perpendicular to the axis of rotation and has one endpoint on that axis. Its length is the distance from the axis to the region’s boundary. When the segment rotates around the axis, its points sweep out all distances from zero to that length. The result is a filled disk: the segment’s length becomes the disk’s radius.
For a horizontal axis, the perpendicular direction is vertical, so the radius is a vertical distance and the slices advance horizontally. For a vertical axis, the perpendicular direction is horizontal, so the radius is a horizontal distance and the slices advance vertically.
| Axis orientation | Slice thickness | Radius is measured | Typical volume integral |
|---|---|---|---|
| Horizontal | \(dx\) | Vertically | \(\int_a^b \pi[r(x)]^2\,dx\) |
| Vertical | \(dy\) | Horizontally | \(\int_c^d \pi[r(y)]^2\,dy\) |
These expressions use the general cross-section principle: volume is the integral of cross-sectional area with respect to distance along the solid. The integrand is the disk area at each coordinate, and the differential records the thickness in the direction the slices advance.
Use the first form when the axis is horizontal and the radius is expressed as a function of \(x\). Use the second when the axis is vertical and the radius is expressed as a function of \(y\). The bounds must cover the full extent of the solid in the direction parallel to its axis.
Choosing the Radius and Thickness
A radius is not automatically the value of a coordinate or the distance from the origin. It is the distance from the axis of rotation to the boundary reached by the rotating segment. If the axis is \(y=0\), a point with vertical coordinate \(y\) is a distance \(|y|\) from the axis. If the axis is \(y=k\), its distance is \(|y-k|\). Likewise, distances to a vertical axis \(x=h\) are measured horizontally.
In a disk-slice setup, the region must run from the axis to its outer boundary at each slice coordinate. Then the segment rotating at that coordinate sweeps out a disk. If the region does not touch the axis along a slice, that segment does not sweep out a filled disk; the geometry needs a different cross-section description.
Before writing an integral, identify three things:
- The axis of rotation, including its location.
- The direction of the slices and the coordinate interval they cover.
- The distance from the axis to the region’s boundary at a typical slice.
Worked Examples
Worked Example: Rotating Around the X-Axis
Rotate the region bounded by \(y=0\), \(y=\sqrt{x}\), \(x=0\), and \(x=2\) around the \(x\)-axis. Set up and evaluate the volume integral.
The axis is horizontal, so take vertical slices. These are perpendicular to the axis, and their thickness is \(dx\). At a fixed \(x\), the region extends from the axis \(y=0\) to \(y=\sqrt{x}\). Thus the disk radius is the vertical distance \(r(x)=\sqrt{x}-0=\sqrt{x}\). The \(x\)-coordinate runs from \(0\) to \(2\).
An antiderivative of \(x\) is \(x^2/2\), so
Check the evaluation: \(\frac{d}{dx}(x^2/2)=x\), the integrand after removing the constant \(\pi\). The slice area at \(x=2\) is \(\pi(\sqrt{2})^2=2\pi\), consistent with the radius used. The volume is \(2\pi\) cubic units.
Worked Example: Rotating Around the Y-Axis
Rotate the triangular region bounded by \(x=0\), \(y=4\), and \(x=y/2\) around the \(y\)-axis. Set up and evaluate the volume integral.
The axis is vertical, so use horizontal slices with thickness \(dy\). At a fixed height \(y\), the region extends horizontally from the axis \(x=0\) to the line \(x=y/2\). Rotating this segment creates a disk whose radius is the horizontal distance \(r(y)=y/2-0=y/2\). The height ranges from \(0\) to \(4\).
An antiderivative of \(y^2\) is \(y^3/3\). Therefore,
Check the integration: \(\frac{d}{dy}(y^3/3)=y^2\). At the top boundary \(y=4\), the radius is \(4/2=2\), so the top disk has area \(4\pi\); at \(y=0\), the radius and area are both zero. The volume is \(\frac{16\pi}{3}\) cubic units.
Worked Example: Rotating Around a Shifted Horizontal Axis
Rotate the region between \(y=2\) and \(y=2+x^2\), for \(0\le x\le 1\), around the line \(y=2\). Set up and evaluate the volume integral.
The axis \(y=2\) is horizontal, so vertical slices have thickness \(dx\). At each \(x\), the region runs from the axis \(y=2\) to \(y=2+x^2\). The radius is the vertical distance between these curves, not the height \(2+x^2\) itself:
The slice coordinate runs from \(x=0\) to \(x=1\), so the disk-slice integral is
An antiderivative of \(x^4\) is \(x^5/5\). Evaluating gives
Check the antiderivative: \(\frac{d}{dx}(x^5/5)=x^4\). At \(x=0\), the radius is \(0\); at \(x=1\), it is \(1\), as expected from the boundary separation \((2+1)-2=1\). The volume is \(\frac{\pi}{5}\) cubic units.
Why the Integral Gives the Volume
The cross-section principle applies to solids whose cross-sectional area varies along an interval: integrate that area over the interval to obtain the volume. Here, the geometry of rotation supplies the area at each coordinate. A segment of length \(r\) that starts on the axis sweeps out a disk, and the area of a disk of radius \(r\) is \(\pi r^2\). Substituting this area into the cross-section principle gives the disk-slice integrals.
The result depends on describing the cross-section correctly at every coordinate in the bounds. If the boundary expression changes, the radius may need a different formula on different intervals. The integral must then reflect that piecewise geometry. Likewise, a region that leaves a gap between itself and the axis does not produce a disk on that slice.
Common Mistake: Using the Wrong Distance
Suppose the region between \(y=0\) and \(y=2x\), for \(0\le x\le1\), is rotated around the \(x\)-axis. A common error is to use \(x\) as the radius because \(x\) is the slice coordinate. That would give the incorrect setup \(\pi\int_0^1x^2\,dx=\pi/3\).
But \(x\) measures position along the axis; it is not the perpendicular distance from the region to the axis. The vertical slice at \(x\) runs from \(y=0\) to \(y=2x\), so its radius is \(2x\). The correct calculation is
Checking the integration, \(\frac{d}{dx}(x^3/3)=x^2\). The incorrect result is too small by a factor of \(4\), because squaring a radius also squares any factor in that radius. A reliable check is to describe the perpendicular segment in words before writing its length as a formula.
Setting Up a Disk-Slice Model
Record its orientation and position, such as \(y=0\), \(y=2\), or \(x=0\).
Use vertical slices for a horizontal axis and horizontal slices for a vertical axis.
Find the distance from the axis to the region’s boundary at a typical slice coordinate.
Use the full coordinate interval parallel to the axis, and integrate \(\pi r^2\) with respect to that coordinate.
The resulting integral encodes both the shape and the orientation of the solid. In particular, it makes clear why the radius cannot be chosen independently of the axis and why the differential is tied to the direction in which slices are stacked.
Check Your Understanding
For each question, identify the cross-section geometry and explain how the radius and thickness are chosen.
- A region between \(y=0\) and \(y=x+1\), for \(0\le x\le2\), is rotated around the \(x\)-axis. What are the radius function, bounds, and volume integral?
- A region from \(x=0\) to \(x=3-y\), for \(0\le y\le3\), is rotated around the \(y\)-axis. What are the radius function and slice thickness?
- When a region is rotated around \(y=4\), why is the radius generally not the boundary’s \(y\)-coordinate?
- What geometric feature of a slice must hold for its rotation to create a disk rather than a cross-section with a hole?
- For a horizontal axis, which coordinate measures slice thickness, and in which direction is the radius measured?
