Why Radius Order Requires Care
A washer has an outer radius and an inner radius. The outer radius is the distance from the axis of rotation to the boundary farther from that axis; the inner radius is the distance to the nearer boundary. These are geometric distances, so they are nonnegative.
This distinction matters when a region lies below the horizontal axis or to the left of the vertical axis. In those locations, a boundary can have a smaller coordinate value while being farther from the axis. Comparing the boundary values alone can therefore reverse the radii and produce an incorrect washer area.
In this tutorial, consider regions lying entirely on one side of the axis of rotation. For each slice, compare the distances from its two endpoints to the axis. The farther endpoint gives the outer radius, and the nearer gives the inner radius.
Radii Are Distances, Not Coordinates
For rotation about the horizontal axis \(y=0\), a vertical slice has endpoints with \(y\)-coordinates. The distance of a point \((x,y)\) from the horizontal axis is \(|y|\). Thus a boundary at \(y=-5\) is five units from the axis, while one at \(y=-2\) is two units away.
For rotation about the vertical axis \(x=0\), a horizontal slice has endpoints with \(x\)-coordinates. The distance of \((x,y)\) from the vertical axis is \(|x|\). For example, \(x=-4\) is four units from the axis, while \(x=-1\) is one unit away.
The order in \(R^2-r^2\) is essential: the outer disk has area \(\pi R^2\), and removing the inner disk leaves the washer. Since \(R\geq r\), the area cannot be negative.
A Reliable Comparison Procedure
For each slice, identify its two boundary coordinates and find each boundary’s distance to the axis. Compare those distances, not merely the coordinates. A useful shortcut applies when both boundary points are on the same side of the axis:
- If both \(y\)-coordinates are positive, the larger \(y\)-coordinate is farther from \(y=0\). If both are negative, the smaller \(y\)-coordinate is farther from \(y=0\).
- If both \(x\)-coordinates are positive, the larger \(x\)-coordinate is farther from \(x=0\). If both are negative, the smaller \(x\)-coordinate is farther from \(x=0\).
These shortcuts follow from absolute value. On the positive side of zero, distance from zero increases with the coordinate; on the negative side, distance increases as the coordinate decreases. If unsure, calculate both absolute values directly.
Use a vertical slice for rotation about a horizontal axis and a horizontal slice for rotation about a vertical axis.
Identify the two values on the slice that bound the region.
For the standard axes, use \(|y|\) for distance to \(y=0\) and \(|x|\) for distance to \(x=0\).
The larger distance is \(R\), the smaller is \(r\), and the washer area is \(\pi(R^2-r^2)\).
Why the Distance Comparison Works
At a fixed slice, imagine a line segment joining the two boundary points. Rotating that segment around the axis sweeps out a circular cross section. The endpoint farther from the axis traces the outer circle, while the nearer endpoint traces the inner circle. The cross section is the disk inside the outer circle with the inner disk removed.
Because a circle’s radius is its distance from its center—in this case, the axis—the radii must be distances rather than signed coordinates. The area calculation follows directly from subtracting the inner disk’s area from the outer disk’s area:
Integrating these cross-sectional areas over the interval of slices gives the volume. The radius comparison determines which expression belongs in each position in the integrand.
Worked Examples
Worked Example: A Region Below the Horizontal Axis
Rotate the region between \(y=-x\) and \(y=-2x\), for \(1\leq x\leq 3\), about the horizontal axis \(y=0\). Find the volume.
Both curves are below the axis. At each \(x\) in the interval, their distances to the axis are
Since \(x>0\), \(2x>x\). Therefore the curve \(y=-2x\) gives the outer radius, even though \(-2x\) is the smaller \(y\)-value. The radii are \(R(x)=2x\) and \(r(x)=x\).
Set up the washer integral and simplify its integrand:
Evaluate:
Check the antiderivative: \(\frac{d}{dx}(x^3)=3x^2\), so the integration is correct. The area integrand \(3\pi x^2\) is positive throughout \(1\leq x\leq3\), as expected for washer areas.
Worked Example: A Region Left of the Vertical Axis
Rotate the rectangle \(-4\leq x\leq-1\), \(0\leq y\leq2\), about the vertical axis \(x=0\). Find the volume using horizontal washers.
Each horizontal slice runs from \(x=-4\) to \(x=-1\). Its endpoint distances from the axis are
The left endpoint is farther from the vertical axis, so \(R(y)=4\). The right endpoint is nearer, so \(r(y)=1\). The radii do not depend on \(y\), and the slice interval is \(0\leq y\leq2\).
Check: differentiating the antiderivative \(15\pi y\) gives the integrand \(15\pi\). Also, each cross section has area \(15\pi\), and the rectangle’s vertical length is \(2\), so the volume \(30\pi\) agrees with area times length.
Worked Example: Radii That Vary Along the Slice Interval
For \(0\leq y\leq2\), a region lies between \(x=-y-2\) on the left and \(x=-1\) on the right. Rotate it about the vertical axis \(x=0\). Find the volume.
At a fixed \(y\), the distances to the axis are
Because \(y\geq0\), \(y+2\geq2>1\). Thus the left boundary is farther from the axis at every slice: \(R(y)=y+2\), and \(r(y)=1\).
Check the antiderivative by differentiating: \(\frac{d}{dy}(\frac{y^3}{3}+2y^2+3y)=y^2+4y+3\), the integrand before multiplying by \(\pi\). The outer radius is at least \(2\) and the inner radius is \(1\), so each washer has positive area, consistent with the positive volume.
Common Mistake: Ordering by Function Values
In the first example, a tempting but incorrect choice is to call \(y=-x\) the outer boundary because \(-x>-2x\) for \(x>0\). That comparison correctly identifies the larger function value, but not the boundary farther from the axis. The relevant comparison is \(|-x|\) versus \(|-2x|\), and \(2x>x\).
If the values are assigned in the wrong order, the calculation becomes
A negative volume signals the reversed radii. Correct the setup by placing the greater distance in \(R\): \(R=2x\), \(r=x\). Then \(R\geq r\), and the washer area is nonnegative.
Check Your Understanding
For each question, identify the farther boundary by comparing distances to the stated axis.
- A slice has endpoints \(y=-3\) and \(y=-7\). Rotated about \(y=0\), which endpoint gives \(R\), and what are \(R\) and \(r\)?
- A slice has endpoints \(x=-2\) and \(x=-6\). Rotated about \(x=0\), which endpoint gives \(R\), and what are the radii?
- A region below \(y=0\) is bounded on a vertical slice by \(y=-x\) and \(y=-4x\), where \(x>0\). State the outer and inner radius expressions.
- A horizontal slice extends from \(x=-y-3\) to \(x=-2\), with \(y\geq0\), and is rotated about \(x=0\). Which boundary gives the outer radius?
- Why does an expression for washer area that is negative indicate that the radii were assigned incorrectly?
