Why Horizontal Slices Produce Disks
When a plane region is rotated about the y-axis, a horizontal segment running from the y-axis to the region's boundary sweeps out a disk. Its length is the disk's radius. Adding the volumes of these thin disks gives a definite integral for the solid's volume.
The essential choices are therefore geometric: use slices perpendicular to the axis of rotation, identify the radius at each height, and integrate over the region's vertical extent. The radius must be expressed in terms of \(y\), because each disk corresponds to a horizontal slice at a particular \(y\)-value.
Definition and Disk Formula
Suppose a region has the form \(0\le x\le g(y)\) for \(a\le y\le b\), where \(g\) is continuous on \([a,b]\) and \(g(y)\ge 0\) throughout that interval. Thus, at height \(y\), the region contains the horizontal segment from the y-axis, \(x=0\), to the curve \(x=g(y)\). Rotating that segment about the y-axis produces a disk of radius \(r(y)=g(y)\).
The bounds \(a\) and \(b\) are the lowest and highest \(y\)-values in the region. The radius is a horizontal distance, so it is found by taking the \(x\)-coordinate of the curve at a given \(y\), or by subtracting the left boundary's \(x\)-coordinate from the right boundary's. For this disk setup, one boundary is the y-axis.
If the curve is initially given as \(y=f(x)\), it may be necessary to solve for \(x\) in terms of \(y\) on the branch that bounds the region. The goal is to find the relevant nonnegative radius \(r(y)\). A curve's equation alone does not specify the radius unless the correct branch and the region's bounds are also identified.
Why the Integral Gives the Volume
At height \(y\), a horizontal slice of the rotated solid is a disk with area \(A(y)=\pi[g(y)]^2\). Divide the interval \([a,b]\) into thin subintervals. For a subinterval of width \(\Delta y_i\), the corresponding slice is approximately a short cylinder with volume \(A(y_i^*)\Delta y_i\), where \(y_i^*\) is a height in that subinterval. Summing these approximate volumes gives a Riemann sum.
Because \(g\) is continuous, \(A(y)=\pi[g(y)]^2\) is continuous on the closed interval and hence Riemann integrable. As the widths of the subintervals approach zero, the Riemann sums converge to the integral of \(A\). This gives the disk formula:
More generally, the same formula applies when the cross-sectional area function \(\pi[r(y)]^2\) is Riemann integrable on \([a,b]\). Continuity of the radius is a convenient sufficient condition. Some regularity condition matters: nonnegativity by itself does not guarantee that the area function has a Riemann integral, so it would not justify this integral formula.
A Reliable Setup
Identify the lowest and highest \(y\)-values of the region. These become the integration bounds.
At a typical height \(y\), identify the segment in the region from the y-axis to its boundary.
Express the horizontal distance from the y-axis to the boundary as \(r(y)\ge 0\). If needed, solve the curve equation for \(x\) on the appropriate branch.
Use \(V=\pi\int_a^b[r(y)]^2\,dy\), then evaluate the definite integral.
Worked Examples
Worked Example: A Line Through the Origin
Rotate the region \(0\le y\le 2\), \(0\le x\le y\), about the y-axis. Find its volume.
At height \(y\), the horizontal segment extends from \(x=0\) to \(x=y\). Its length is the radius, so \(r(y)=y\). The bounds are \(0\) and \(2\). The disk formula gives
Check: differentiating the antiderivative \(y^3/3\) gives \(y^2\), the integrand. The result is positive, as a volume must be. At every height, the disk radius is at most \(2\), and the height of the solid is \(2\), so its volume cannot exceed the volume \(8\pi\) of a cylinder of radius \(2\) and height \(2\). Since \(8\pi/3<8\pi\), the answer is consistent with this bound.
Worked Example: Rewriting a Parabola as a Radius
Rotate the region bounded by \(x=0\), \(y=x^2\), and \(y=4\) about the y-axis. Find its volume.
The region extends vertically from \(y=0\) to \(y=4\). The curve is given as \(y=x^2\), but a horizontal slice requires its \(x\)-coordinate as a function of \(y\). Since this region lies to the right of the y-axis, solve for the nonnegative branch:
Thus, at height \(y\), the segment runs from \(x=0\) to \(x=\sqrt y\), so \(r(y)=\sqrt y\). The volume is
Check: the derivative of \(y^2/2\) is \(y\), as required. At the top height \(y=4\), the radius is \(\sqrt4=2\); at the bottom, it is \(0\). The cross-sectional radius increases across the region, matching the stated bounds and curve.
Worked Example: A Decreasing Radius
Rotate the region \(0\le y\le 1\), \(0\le x\le 2-y\), about the y-axis. Find its volume.
At each height, the right boundary has \(x\)-coordinate \(2-y\). On \(0\le y\le1\), this radius is nonnegative: it decreases from \(2\) to \(1\). Therefore \(r(y)=2-y\), and
Check: differentiating \(4y-2y^2+y^3/3\) gives \(4-4y+y^2=(2-y)^2\). The endpoint radii are \(r(0)=2\) and \(r(1)=1\), so both are positive and agree with the geometry. The computed volume is therefore consistent with a solid of height \(1\) whose disks have radii between \(1\) and \(2\).
Common Mistake: Forgetting to Square the Radius
In the first example, the radius at height \(y\) is \(y\). A disk's area is \(\pi r^2\), not \(\pi r\). A mistaken setup would be
This calculation integrates \(\pi\) times the radius, not the area of a disk, so it does not represent the solid's volume. The correction is to square the radius before integrating:
A separate setup check is to match the slice orientation to the axis: disks perpendicular to the y-axis come from horizontal slices and use \(dy\). A vertical slice is parallel to that axis, so it does not sweep out one of these disk cross-sections.
What to Check Before Evaluating
Most errors in a disk integral happen before the antiderivative is found. Confirm that the radius is a horizontal distance, that it is expressed in terms of \(y\), and that the bounds describe the region's vertical extent. If the curve is given as \(y=f(x)\), check that the chosen solution for \(x\) is the branch actually bordering the region. Finally, confirm that the radius stays nonnegative and has enough regularity for the area function to be integrable.
Once the setup is correct, evaluate the definite integral and check the result by differentiating the antiderivative. Also compare the endpoint radii and the general size of the solid with the geometry. These checks cannot replace a correct setup, but they can reveal many mistakes.
Check Your Understanding
For each question, identify the horizontal radius and the bounds before evaluating or describing the integral.
- A region is given by \(0\le y\le 3\) and \(0\le x\le 2y\). What disk integral represents the volume after rotation about the y-axis?
- A region is bounded by \(x=0\), \(y=x^2\), and \(y=9\). What is the radius as a function of \(y\), and what are the bounds?
- For a region with \(r(y)=3-y\) on \(0\le y\le2\), write the disk integral. Is the radius nonnegative throughout the interval?
- Why does a disk integral for rotation about the y-axis use \(dy\) rather than \(dx\)?
- What regularity assumption in the theorem ensures that the disk-area function has a Riemann integral?
