From a Planar Region to Washers
A solid formed by rotating a plane region around a horizontal axis can be understood by examining thin vertical slices of the region. Each slice is perpendicular to the axis of rotation. When a slice is revolved, it sweeps out a disk or a washer, depending on whether the region reaches the axis.
This tutorial focuses on rotation around the x-axis, the horizontal line \(y=0\). For each value of \(x\), the radii are the vertical distances from that axis to the top and bottom of the slice. The main task is to identify those distances correctly and use the larger one as the outer radius.
Vertical Slices and Radius Distances
Suppose a region extends from \(x=a\) to \(x=b\). At each \(x\) in that interval, assume its vertical slice runs from a lower boundary \(y=g(x)\) to an upper boundary \(y=f(x)\), with
These inequalities say that the slice is on or above the x-axis and that the upper boundary is at least as high as the lower one. Rotating the slice around the x-axis produces a washer. The distance from the axis to the upper boundary is \(f(x)\), and the distance to the lower boundary is \(g(x)\). Thus the outer radius is \(R(x)=f(x)\), and the inner radius is \(r(x)=g(x)\).
A washer’s cross-sectional area is the area of its outer disk minus the area of its inner disk. Therefore its area at position \(x\) is \(\pi(R(x)^2-r(x)^2)\). Adding these areas along the interval gives the volume integral.
Why the Radii Are Vertical Distances
For a fixed \(x\), the slice lies along a vertical line. Its points have heights between \(g(x)\) and \(f(x)\), while the axis of rotation has height zero. The distance from a point at height \(y\) to the x-axis is therefore \(y\) when \(y\geq0\). In particular, the farthest point of the slice from the axis is at \(f(x)\), and the nearest point is at \(g(x)\).
When the slice rotates, the farthest point traces the outer circular edge and the nearest point traces the inner edge. The cross section is an outer disk of radius \(f(x)\) with a disk of radius \(g(x)\) removed. Since a disk of radius \(s\) has area \(\pi s^2\), the washer area is
This explains both the squared radii and their order in the integrand. The integral adds the cross-sectional areas across the full interval from \(a\) to \(b\). The conditions matter: the stated setup assumes the slice lies above the axis and the boundaries are correctly ordered throughout the interval.
Worked Example: A Constant Top Boundary
Worked Example: Rotate a Region Between a Line and a Horizontal Boundary
The region is bounded by \(y=x+1\), \(y=5\), \(x=0\), and \(x=2\). It is rotated around the x-axis. Set up and evaluate the washer integral.
For \(0\leq x\leq2\), the lower boundary is \(x+1\) and the upper boundary is \(5\). They are both nonnegative, and \(x+1\leq3<5\) on this interval. The vertical distances from the x-axis are therefore
Substitute the radii and bounds into the washer formula:
An antiderivative of \(24-2x-x^2\) is \(24x-x^2-\frac{x^3}{3}\). Evaluating at the bounds gives
Check the cross-sectional areas at the endpoints. At \(x=0\), the radii are \(5\) and \(1\), giving area \(\pi(25-1)=24\pi\). At \(x=2\), the lower boundary is \(y=x+1=3\), so the radii are \(5\) and \(3\), giving area \(\pi(25-9)=16\pi\). The integrand gives the same endpoint areas: \(25-(0+1)^2=24\) and \(25-(2+1)^2=16\). Finally, differentiating \(24x-x^2-\frac{x^3}{3}\) gives \(24-2x-x^2\), confirming the antiderivative used.
Worked Example: A Curved Upper Boundary
Worked Example: Rotate a Region Between a Curve and a Line
The region is bounded by \(y=1\), \(y=\sqrt{x+1}\), \(x=0\), and \(x=3\). It is rotated around the x-axis. Find its volume.
On \(0\leq x\leq3\), \(\sqrt{x+1}\geq1\), so the curve is the upper boundary and \(y=1\) is the lower boundary. Both boundaries are above the axis. Their vertical distances from the axis are
Squaring the radii in the washer formula gives
At \(x=0\), both radii are \(1\), so the cross-sectional area is zero; the integrand is also \(0\). At \(x=3\), the radii are \(2\) and \(1\), so the cross-sectional area is \(3\pi\); the integrand is \(3\), giving the same area after multiplication by \(\pi\). Differentiating \(\frac{x^2}{2}\) gives \(x\), verifying the antiderivative.
Worked Example: A Slice That Reaches the Axis
Worked Example: Rotate a Region With a Zero Inner Radius at One Endpoint
The region is bounded by \(y=x\), \(y=x+2\), \(x=0\), and \(x=3\). It is rotated around the x-axis. Find its volume.
On \(0\leq x\leq3\), the upper boundary is \(x+2\) and the lower boundary is \(x\). Both are nonnegative. The radii are
At \(x=0\), the inner radius is zero, so that one cross section is a disk rather than a washer with a hole. For other \(x\) in the interval, the inner radius is positive. The same washer formula applies:
For an endpoint check, at \(x=0\) the radii are \(2\) and \(0\), giving area \(4\pi\). At \(x=3\), they are \(5\) and \(3\), giving area \(\pi(25-9)=16\pi\). The integrand \(4x+4\), multiplied by \(\pi\), gives \(4\pi\) and \(16\pi\) at those endpoints. The derivative of \(2x^2+4x\) is \(4x+4\), so the integration is also verified.
A Reliable Setup Process
Before evaluating an integral, make sure the geometry supports the radii you have chosen. A small sketch or a list of boundary values can help establish which curve is above the other and whether the region is on the intended side of the axis.
For rotation around the x-axis, the axis is \(y=0\). Identify the interval \(a\leq x\leq b\) covered by the region.
Identify the lower and upper boundary heights at a typical \(x\) in the interval.
For a slice above the x-axis, each radius is its boundary height because the axis has height zero.
The boundary farther from the axis gives the outer radius; the boundary nearer the axis gives the inner radius.
Use \(V=\pi\int_a^b(R(x)^2-r(x)^2)\,dx\), then evaluate and check the setup at useful points.
A Common Radius Reversal
A frequent mistake is to subtract the outer disk from the inner disk. In the first worked example, that reversal would produce
The integrand is negative throughout the interval because \(x+1<5\). That cannot represent the area of a washer: the inner disk cannot have greater area than the outer disk. The correction is to identify the boundary farther from the axis as the outer radius and subtract the smaller squared radius from the larger one:
Be careful to identify radii as distances from the axis, not simply as whichever formula happens to be written first. In these examples the region lies above the x-axis, so the boundary heights themselves are the distances. If the region is not on that side of the axis, the assumptions behind this particular setup must be checked rather than applied automatically.
Why Endpoint Checks Help
Evaluating the two radii at an endpoint is a quick way to catch a boundary or subtraction error. For example, in the first example the lower boundary at \(x=2\) is \(x+1=3\), not \(5\). The resulting endpoint washer has outer radius \(5\), inner radius \(3\), and area \(16\pi\). Checking the radii in the integrand gives \(\pi(25-9)=16\pi\) as well.
Endpoint checks do not replace checking the full interval. The boundary order and the position of the region relative to the axis must support the chosen radii throughout the interval. But they are useful safeguards: a negative cross-sectional area, or an endpoint radius that does not match the boundary height, signals that the setup needs attention.
Check Your Understanding
For each question, identify the radii from vertical distances to the x-axis before writing an integral.
- A region lies between \(y=2\) and \(y=6\) for \(1\leq x\leq4\). What are the outer and inner radii when it is rotated around the x-axis?
- A region lies between \(y=x+1\) and \(y=4\) for \(0\leq x\leq2\). Write the washer integral, without evaluating it.
- For the region between \(y=1\) and \(y=\sqrt{x+1}\), \(0\leq x\leq3\), what are the two radii at \(x=2\), and what is the cross-sectional area there?
- Why is \(\pi\int_a^b(r(x)^2-R(x)^2)\,dx\) not the correct washer-volume setup when \(R(x)\) is the outer radius?
- For a region with lower boundary \(y=x\) and upper boundary \(y=x+2\), what happens to the cross section at \(x=0\), and why does the washer formula still apply?
