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Washer Radii About Shifted Axes

Find washer radii by measuring perpendicular distances from the region’s boundaries to a shifted axis, then use those distances to calculate volume.

Intermediate 9 min read

What You'll Learn

  • Express radii as distances from a shifted horizontal or vertical axis
  • Decide which boundary gives the outer radius and which gives the inner radius
  • Match slice direction and integration variable to the axis of rotation
  • Build washer-area integrals using shifted radii
  • Check radius ordering and verify exact volume calculations

Why a Shifted Axis Changes the Radius

A washer’s radii are not automatically the \(x\)- or \(y\)-coordinates of a region’s boundaries. Each radius is a distance from the axis of rotation. When the axis is shifted away from the coordinate axes, that distance must include the shift.

For example, a horizontal axis might have equation \(y=-1\), and a vertical axis might have equation \(x=3\). The distance from a point with vertical coordinate \(y\) to \(y=-1\) is \(|y+1|\); its distance from \(y=0\) would instead be \(|y|\). The axis location is part of the radius calculation.

In this tutorial, assume the region lies entirely on one side of the axis over the interval being integrated. If the axis passes through the region, the cross-section geometry requires additional care. The central task here is to calculate the two radii from the correct shifted line.

Definition: Washer Radii as Distances

Definition: A washer cross-section has an outer radius \(R\), the distance from the axis of rotation to the farther boundary of the region, and an inner radius \(r\), the distance to the nearer boundary. Both radii are nonnegative, and \(R\ge r\). Its area is \(A=\pi(R^2-r^2)\).

Distances to horizontal lines are vertical distances. If a point has vertical coordinate \(y\), its distance to the horizontal line \(y=c\) is \(|y-c|\). Distances to vertical lines are horizontal distances: a point with horizontal coordinate \(x\) is \(|x-h|\) units from the vertical line \(x=h\).

For rotation around a horizontal line \(y=c\), slices perpendicular to the axis are vertical. Their boundaries have \(y\)-coordinates, so the radii are measured vertically and the integral is with respect to \(x\). For rotation around a vertical line \(x=h\), perpendicular slices are horizontal, so the radii are measured horizontally and the integral is with respect to \(y\).

$$ \begin{aligned} \text{Rotation about }y=c:\quad R(x),r(x)&=\text{distances }|y-c|,\qquad V=\pi\int_a^b\bigl(R(x)^2-r(x)^2\bigr)\,dx,\\[4pt] \text{Rotation about }x=h:\quad R(y),r(y)&=\text{distances }|x-h|,\qquad V=\pi\int_m^n\bigl(R(y)^2-r(y)^2\bigr)\,dy. \end{aligned} $$

These formulas apply when each perpendicular slice produces a washer and the radii are identified correctly throughout the interval. The shifted line determines the distances; the farther boundary determines \(R\), regardless of whether that boundary is called “upper,” “lower,” “left,” or “right.”

How to Identify the Radii

Suppose the axis is horizontal, \(y=c\), and the region’s vertical boundaries at a fixed \(x\) have coordinates \(y_{\text{low}}\) and \(y_{\text{high}}\). Measure the distance from the axis to each boundary. If the axis lies below the region, the higher boundary is farther away. If the axis lies above it, the lower boundary is farther away. The same distance-based reasoning applies to a vertical axis and the left and right boundaries.

1
Identify the axis and slice direction.
Use slices perpendicular to the axis: vertical slices for a horizontal axis, horizontal slices for a vertical axis.
2
Measure both distances.
Subtract the axis coordinate from each boundary coordinate, using absolute values to represent distance.
3
Assign outer and inner.
The greater distance is \(R\); the lesser distance is \(r\). Check that \(R\ge r\) on the interval.
4
Build the volume integral.
Use \(\pi(R^2-r^2)\), and integrate with respect to the coordinate along the axis of the slices.

Why the Washer Formula Uses These Distances

A perpendicular slice rotated around the axis sweeps out a circular region with a circular hole. The outer circular disk has radius \(R\), and the hole has radius \(r\). A circle of radius \(s\) has area \(\pi s^2\), so subtracting the hole’s area from the full disk’s area gives the washer’s area.

$$ A=\pi R^2-\pi r^2=\pi(R^2-r^2). $$

This argument explains why radii must be measured from the axis itself. The circular disk’s center lies on the axis of rotation, not necessarily on either coordinate axis. A boundary’s coordinate alone is not its radius unless the axis happens to be at coordinate zero.

Worked Examples

Worked Example: A Horizontal Axis Below the Region

The region is bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=2\). It is rotated around \(y=-1\). Find the volume.

Use vertical slices because the axis is horizontal; integrate with respect to \(x\) from \(0\) to \(2\). At each \(x\), the lower boundary is \(y=0\) and the upper boundary is \(y=x\). Their distances from \(y=-1\) are:

$$ |0-(-1)|=1, \qquad |x-(-1)|=x+1. $$

Since \(x\ge0\), \(x+1\ge1\). Thus the farther boundary gives \(R(x)=x+1\), and the nearer boundary gives \(r(x)=1\). The washer integral is

$$ \begin{aligned} V &=\pi\int_0^2\bigl((x+1)^2-1^2\bigr)\,dx\\ &=\pi\int_0^2(x^2+2x)\,dx\\ &=\pi\left[\frac{x^3}{3}+x^2\right]_0^2\\ &=\pi\left(\frac{8}{3}+4-0\right) =\frac{20\pi}{3}. \end{aligned} $$

Check: Differentiating the antiderivative gives \(\frac{d}{dx}(x^3/3+x^2)=x^2+2x\), the integrand. At \(x=0\), both radii are \(1\), so the washer area is zero; at \(x=2\), the radii are \(3\) and \(1\), in the stated outer-to-inner order.

Worked Example: A Horizontal Axis Above the Region

The region is the rectangle \(0\le x\le1\), \(1\le y\le2\). Rotate it around \(y=4\). Find the volume.

Use vertical slices, so integrate with respect to \(x\) from \(0\) to \(1\). For every slice, the lower boundary \(y=1\) is farther from the axis than the upper boundary \(y=2\):

$$ |1-4|=3, \qquad |2-4|=2. $$

Therefore \(R(x)=3\) and \(r(x)=2\), even though \(y=1\) is the lower boundary. The washer area is constant across the rectangle:

$$ \begin{aligned} V &=\pi\int_0^1(3^2-2^2)\,dx\\ &=\pi\int_0^1 5\,dx\\ &=5\pi[x]_0^1\\ &=5\pi. \end{aligned} $$

Check: The antiderivative \(5x\) has derivative \(5\), matching the integrand. At every \(x\), \(3\ge2\), and the area \(5\pi\) multiplied by the rectangle’s width \(1\) gives the same volume.

Worked Example: A Vertical Axis to the Right of the Region

The region is bounded by \(x=0\), \(x=y\), \(y=0\), and \(y=2\). Rotate it around \(x=3\). Find the volume.

The axis is vertical, so use horizontal slices and integrate with respect to \(y\) from \(0\) to \(2\). At a fixed \(y\), the left boundary is \(x=0\) and the right boundary is \(x=y\). Their distances to \(x=3\) are

$$ |0-3|=3, \qquad |y-3|=3-y, $$

because \(0\le y\le2\). Since \(3\ge3-y\), the left boundary gives the outer radius \(R(y)=3\), while the right boundary gives the inner radius \(r(y)=3-y\). Therefore

$$ \begin{aligned} V &=\pi\int_0^2\bigl(3^2-(3-y)^2\bigr)\,dy\\ &=\pi\int_0^2\bigl(9-(9-6y+y^2)\bigr)\,dy\\ &=\pi\int_0^2(6y-y^2)\,dy\\ &=\pi\left[3y^2-\frac{y^3}{3}\right]_0^2\\ &=\pi\left(12-\frac{8}{3}\right) =\frac{28\pi}{3}. \end{aligned} $$

Check: The derivative of \(3y^2-y^3/3\) is \(6y-y^2\), as required. At \(y=0\), both radii equal \(3\), so the washer has zero area. At \(y=2\), \(R=3\) and \(r=1\), confirming the radius order and the integrand’s nonnegative value.

A Common Mistake: Using Coordinates Instead of Distances

For the first example, a tempting but incorrect setup is to use \(R=x\) and \(r=0\), because the region’s boundaries are \(y=x\) and \(y=0\). Those are distances from the \(x\)-axis, not from the actual axis \(y=-1\). The incorrect calculation would be

$$ V_{\text{wrong}} =\pi\int_0^2(x^2-0^2)\,dx =\frac{8\pi}{3}. $$

But each boundary is one unit farther from the axis \(y=-1\) than its distance from \(y=0\). The correct distances are \(x+1\) and \(1\), giving \(20\pi/3\), as computed above. The correction is to subtract the axis coordinate from each boundary coordinate and use the resulting absolute distances before deciding which is outer.

Why the Distance Check Matters

Shifted-axis problems are easier to organize when you keep three questions separate:

  • Where is the axis? Record its equation, such as \(y=c\) or \(x=h\).
  • What are the boundary distances? Compute \(|y-c|\) or \(|x-h|\) for each boundary.
  • Which distance is larger? Label the larger one \(R\) and the smaller one \(r\).

This method avoids a common confusion: “upper” does not always mean “outer,” and “right” does not always mean “outer.” In the rectangle example, the lower boundary is outer because the axis lies above the region. The distance comparison, not the boundary’s directional name, determines the radius.

Key takeaway: For a shifted axis, first calculate each boundary’s perpendicular distance to that line. Then identify the larger distance as \(R\), the smaller as \(r\), and integrate \(\pi(R^2-r^2)\) using the slice coordinate.

Check Your Understanding

For each question, identify the relevant distances before assigning radii.

  1. A region has vertical boundaries \(y=1\) and \(y=4\) at a fixed \(x\). Rotating about \(y=-2\), what are the inner and outer radii?
  2. A region has horizontal boundaries \(x=0\) and \(x=2\) at a fixed \(y\). Rotating about \(x=5\), which boundary gives the outer radius?
  3. Why is a washer integral about \(y=c\) written with respect to \(x\) when using slices perpendicular to the axis?
  4. For rotation about \(x=1\), a horizontal slice has boundaries \(x=-2\) and \(x=0\). Find \(R\) and \(r\).
  5. State the washer area formula and explain what each radius measures.

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