From a Region Between Curves to a Volume
When a region between two curves is rotated about a line, each slice perpendicular to the axis of rotation sweeps out a disk or a washer. The curves determine the boundaries of the region, but they do not automatically give the radii: each radius is a distance from the axis of rotation to a boundary of the region.
A reliable setup therefore requires three decisions: find the region’s bounds, use slices perpendicular to the axis, and measure both radii from that axis. Getting any one of these wrong can produce an integral that looks plausible but represents a different solid.
Finding Bounds and Choosing Slices
The intersection points of the boundary curves often determine the endpoints of the region. If the curves are given as \(y=f(x)\) and \(y=g(x)\), find their intersections by solving \(f(x)=g(x)\). The resulting \(x\)-values give the limits for vertical slices when the region is described using \(x\). For horizontal slices, describe the left and right boundaries as functions of \(y\), and find the corresponding \(y\)-limits.
The slices used for disks and washers must be perpendicular to the axis of rotation. Thus, rotation about a horizontal line uses vertical slices, while rotation about a vertical line uses horizontal slices. If the boundary curves are initially written as \(y=f(x)\), horizontal slicing may require solving for \(x\) in terms of \(y\).
| Axis of rotation | Slice direction | Radius measured horizontally or vertically? |
|---|---|---|
| Horizontal line \(y=c\) | Vertical slices, thickness \(dx\) | Vertical distances from \(y=c\) |
| Vertical line \(x=c\) | Horizontal slices, thickness \(dy\) | Horizontal distances from \(x=c\) |
Each radius is a nonnegative distance. For example, the distance from a point with vertical coordinate \(y\) to the horizontal line \(y=c\) is \(|y-c|\). When the entire region lies on one side of the axis, the outer radius is the greater distance to the region’s boundary and the inner radius is the lesser. Do not assume that the curve called “upper” or “lower” always determines the outer or inner radius if the axis lies between the curves; in that situation, the distance from the axis matters. The earlier material on washers when the axis crosses the region addresses that additional case.
Why the Washer Integral Gives Volume
Consider a small slice of a solid formed by rotation. If its cross section is a washer with outer radius \(R\) and inner radius \(r\), the area of that slice is the area of the outer disk minus the area of the hole:
If the slice has thickness \(\Delta x\), its approximate volume is \(A\Delta x\); for horizontal slices, the thickness is \(\Delta y\). Adding the volumes of thin slices approximates the whole solid. In the limit as the slices become arbitrarily thin, the sums become integrals.
This argument requires that the chosen slices describe the region across the stated interval and that the radii be correctly identified there. If the boundary order or radius order changes, the interval may need to be split so that each formula applies on its own part.
Worked Example: Between Two Curves Rotated About the \(y\)-Axis
Worked Example: Between Two Curves Rotated About the \(y\)-Axis
Find the volume when the region bounded by \(y=x\) and \(y=x^2\) is rotated about the \(y\)-axis.
First find the intersections:
On \(0\le x\le1\), \(x\ge x^2\), so these curves enclose the region between them. The axis is vertical, so use horizontal slices and express both boundaries as \(x\) in terms of \(y\). From \(y=x\), we get \(x=y\). From \(y=x^2\), the relevant branch on this region is \(x=\sqrt{y}\). The \(y\)-values range from \(0\) to \(1\).
For \(0\le y\le1\), \(y\le\sqrt{y}\). The distances to the \(y\)-axis are therefore \(r(y)=y\) and \(R(y)=\sqrt{y}\). The volume is
Check the antiderivative: \(\frac{d}{dy}(\frac{y^2}{2}-\frac{y^3}{3})=y-y^2\), which is the integrand. At \(y=0\), both radii are \(0\); at \(y=1\), both are \(1\), so the washer area is zero at each end, consistent with the boundaries meeting there. The volume is \(\frac{\pi}{6}\) cubic units.
Worked Example: Between Two Curves Rotated About the \(x\)-Axis
Worked Example: Between Two Curves Rotated About the \(x\)-Axis
Find the volume when the region between \(y=4-x^2\) and \(y=x^2\) is rotated about the \(x\)-axis.
Find the intersection points by equating the curves:
At \(x=0\), the upper curve has value \(4\), and the lower curve has value \(0\). Over the interval \([-\sqrt{2},\sqrt{2}]\), the region lies on or above the \(x\)-axis. Its outer radius is \(R(x)=4-x^2\), and its inner radius is \(r(x)=x^2\).
For the check, the derivative of \(16x-\frac{8x^3}{3}\) is \(16-8x^2\), the simplified integrand. At \(x=0\), the radii are \(R=4\) and \(r=0\), as expected for a full disk. At either intersection, \(4-x^2=x^2=2\), so the radii match and the washer area is zero. The volume is \(\frac{64\pi\sqrt{2}}{3}\) cubic units.
Worked Example: Measuring Radii from a Shifted Axis
Worked Example: Measuring Radii from a Shifted Axis
Find the volume when the region bounded by \(y=x^2\) and \(y=2\) is rotated about the line \(y=3\).
The curves meet where
The region has lower boundary \(y=x^2\) and upper boundary \(y=2\). It lies below the axis \(y=3\). For vertical slices, the distance from \(y=3\) to the lower boundary is \(3-x^2\), and the distance from \(y=3\) to the upper boundary is \(3-2=1\). Thus \(R(x)=3-x^2\) and \(r(x)=1\). The outer radius is the distance to the lower curve because that curve is farther from the axis.
Check the antiderivative: its derivative is \(8-6x^2+x^4\). At \(x=0\), the radii are \(R=3\) and \(r=1\); at \(x=\pm\sqrt{2}\), the two boundaries meet at \(y=2\), so both radii are \(1\) and the washer area is zero. The volume is \(\frac{48\pi\sqrt{2}}{5}\) cubic units.
Common Mistake: Using the Wrong Slice Direction
For the region between \(y=x\) and \(y=x^2\) rotated about the \(y\)-axis, a tempting but incorrect setup is
This treats \(x\) as the slice variable and uses the curves’ \(y\)-values as washer radii. But a vertical slice is parallel to the \(y\)-axis, not perpendicular to it, so it does not sweep out a washer. It also measures distances from \(y=0\), rather than horizontal distances from the \(y\)-axis. For washers about the \(y\)-axis, use horizontal slices, write the boundaries as \(x=y\) and \(x=\sqrt{y}\), and measure radii from \(x=0\). The correct washer setup is
Before integrating, check that the slice is perpendicular to the axis and that every radius is a distance from that axis.
A Practical Setup Checklist
Solve for curve intersections and identify which curve forms each boundary on the interval.
Use vertical slices for a horizontal axis and horizontal slices for a vertical axis.
Express both radii as nonnegative distances from the stated line of rotation.
Confirm \(R\ge r\ge0\), use the slice coordinate for integration, and split the interval if the relevant boundary or radius ordering changes.
Use \(\pi(R^2-r^2)\) for each washer, or \(\pi R^2\) when the inner radius is zero.
The curves locate the region’s boundaries; the axis determines the radii. Keeping those roles separate makes it easier to construct correct volume integrals, including when the axis is shifted away from the coordinate axes.
Check Your Understanding
For each question, identify the bounds, slice direction, and radii before calculating any integral.
- Where do \(y=2x\) and \(y=x^2\) intersect, and which curve is above the other between the intersection points?
- For a region rotated about the vertical line \(x=2\), should a washer setup use vertical or horizontal slices?
- The region between \(y=x^2\) and \(y=4\) is rotated about the \(x\)-axis. What are the intersection limits and the outer and inner radii?
- For the region between \(y=x^2\) and \(y=2\) rotated about \(y=3\), why is the outer radius \(3-x^2\), not \(3-2\)?
- What two checks should you make before integrating a washer formula?
