Why Shell Height Comes from a Difference
A cylindrical shell is formed by rotating a thin slice of a region around an axis parallel to that slice. The shell’s height is the length of the slice within the region. For a region bounded by two curves, finding that length means subtracting one boundary function from the other in the correct order.
The order depends on the slice orientation. A vertical slice extends from a lower curve to an upper curve, so its length is upper minus lower. A horizontal slice extends from a left boundary to a right boundary, so its length is right minus left. The shell radius is a separate measurement: it is the distance from the slice to the axis of rotation.
Match the Slice to the Rotation Axis
For the shell method, slices are parallel to the axis of rotation. A vertical axis, such as the \(y\)-axis or \(x=a\), calls for vertical slices. Their thickness is \(dx\), and their height is measured vertically between the region’s top and bottom boundaries. A horizontal axis, such as the \(x\)-axis or \(y=b\), calls for horizontal slices. Their thickness is \(dy\), and their height is measured horizontally between the region’s left and right boundaries.
| Axis orientation | Slice orientation | Shell height | Thickness |
|---|---|---|---|
| Vertical | Vertical | Upper \(y\)-value minus lower \(y\)-value | \(dx\) |
| Horizontal | Horizontal | Right \(x\)-value minus left \(x\)-value | \(dy\) |
Once the height is identified, the shell-volume formula is \(dV=2\pi rh\,dt\), where \(r\) is the shell radius, \(h\) is its height, and \(dt\) is the slice thickness. Thus, for vertical slices, the volume integral uses \(2\pi r(x)h(x)\,dx\); for horizontal slices, it uses \(2\pi r(y)h(y)\,dy\). The radius is the distance from the slice to the axis, not the distance between the two curve boundaries.
Why the Subtraction Order Works
Consider a fixed vertical coordinate \(x\) at which a region runs continuously from \(y=g(x)\) to \(y=f(x)\), with \(f(x)\ge g(x)\). The segment begins at the lower endpoint and ends at the upper endpoint. Its length is the difference between the endpoint coordinates: \(f(x)-g(x)\). A horizontal segment from \(x=L(y)\) to \(x=R(y)\), with \(R(y)\ge L(y)\), has length \(R(y)-L(y)\) for the same reason.
These lengths become shell heights when the slices are rotated parallel to the axis. The shell formula uses circumference times height times thickness. Adding the contributions of thinner and thinner slices gives the integral for the volume. The sign of a length matters: it must be nonnegative, so identify which boundary is actually higher or farther right before subtracting.
Use slices parallel to the rotation axis: vertical slices for a vertical axis and horizontal slices for a horizontal axis.
For a vertical slice, identify its upper and lower \(y\)-coordinates. For a horizontal slice, identify its right and left \(x\)-coordinates.
Calculate upper minus lower or right minus left, and check that the result is nonnegative over the interval.
Find the distance from the slice to the axis for the radius; do not use the boundary difference as the radius.
Worked Examples
Worked Example: Vertical Shells Between a Line and a Parabola
The region is bounded by \(y=2x\), \(y=x^2\), \(x=0\), and \(x=2\). It is rotated around the \(y\)-axis. Set up and evaluate the shell integral.
The axis is vertical, so use vertical slices. On \(0\le x\le2\), the line is above the parabola because \(2x-x^2=x(2-x)\ge0\). Thus the shell height is
The shell radius is the distance from \(x\) to the \(y\)-axis, so \(r(x)=x\). The thickness is \(dx\). Therefore,
Check the setup: at \(x=1\), the slice runs from \(y=1\) to \(y=2\), so its height is \(2-1=1\), matching \(2(1)-(1)^2=1\). Also, differentiating the antiderivative \(2x^3/3-x^4/4\) gives \(2x^2-x^3=x(2x-x^2)\), the integrand before the constant \(2\pi\). The calculated volume is positive, as it must be.
Worked Example: Horizontal Shells Between a Parabola and a Vertical Line
Consider the region described by \(0\le y\le2\) and \(y^2\le x\le4\). Rotate it around the \(x\)-axis. Find the volume using shells.
The axis is horizontal, so use horizontal slices. At each \(y\), the left boundary is \(x=y^2\) and the right boundary is \(x=4\). The shell height is therefore right minus left:
The region lies above the \(x\)-axis, so a slice at \(y\) has radius \(r(y)=y\). Its thickness is \(dy\). The shell integral is
Check the height at \(y=1\): the slice runs horizontally from \(x=1\) to \(x=4\), so its length is \(4-1=3\), which agrees with \(4-(1)^2=3\). The derivative of \(2y^2-y^4/4\) is \(4y-y^3\), the integrand before \(2\pi\). The radius \(y\) is distinct from the horizontal length \(4-y^2\).
Worked Example: A Shifted Horizontal Axis
Use the same region \(0\le y\le2\), \(y^2\le x\le4\), but rotate it around \(y=-1\). Find the shell volume.
The axis is horizontal, so the slices remain horizontal and the height is still the distance from the left boundary to the right boundary:
Changing the axis changes the radius, not the horizontal length of the slice. Since the region lies above \(y=-1\), the radius is \(r(y)=y-(-1)=y+1\). Hence
Check the geometry at \(y=1\): the shell height is \(4-1=3\), and its radius is \(1-(-1)=2\). The integrand before \(2\pi\) is then \(2\cdot3=6\), matching \((y+1)(4-y^2)=2\cdot3=6\). Differentiating the antiderivative gives \(4y+4-y^3-y^2=(y+1)(4-y^2)\), confirming the integration.
When the Boundaries Are Given as Functions of \(x\)
Vertical shell heights are often easy to read when the curves are already written as \(y=f(x)\) and \(y=g(x)\). Horizontal shell heights require horizontal endpoints, usually expressed as \(x=R(y)\) and \(x=L(y)\). If a boundary is initially given as \(y=f(x)\), finding its horizontal-slice endpoint may require solving for \(x\) in terms of \(y\). Use only the inverse branch that describes the boundary segment of the region.
The integral limits must also match the slice variable. Vertical slices use \(x\)-limits and \(dx\); horizontal slices use \(y\)-limits and \(dy\). Before integrating, check a representative slice against the graph or boundary description. This catches errors such as reversing the subtraction or using the wrong branch.
Common Mistake: Reversing the Height
For the region \(x^2\le y\le4\), \(0\le x\le2\), a common error is to write the vertical shell height as \(x^2-4\). At \(x=1\), this gives \(1-4=-3\), even though the slice clearly runs upward from \(y=1\) to \(y=4\) and has length \(3\). The correct subtraction is upper minus lower:
Reversing the subtraction can make a volume integral negative. A negative result is a signal to recheck the boundary order; volume is nonnegative. Do not repair the error by inserting an absolute value without checking the geometry. Identify the upper and lower, or right and left, boundaries first.
A Caution About Regions Crossing the Axis
The simple shell integral assumes that the shells generated by the chosen slices do not count the same part of the solid more than once. In particular, if a region crosses the axis of rotation, slices on opposite sides can sweep out shells at the same radius. Integrating \(2\pi\) times an absolute radius times the slice height over the entire region may then double-count volume.
For example, rotating the rectangle \(0\le x\le1\), \(-1\le y\le1\) around the \(x\)-axis produces a cylinder of radius \(1\) and length \(1\), with volume \(\pi\). A naive shell integral over all horizontal slices gives \(2\pi\int_{-1}^{1}|y|\,dy=2\pi\), counting shells from both sides of the axis at the same radii. In such a case, the shell setup must be reformulated to count each shell once; the absolute value alone does not resolve the overlap.
Check Your Understanding
For each question, identify the slice orientation and write the shell height before evaluating any volume integral.
- A region has upper boundary \(y=5\) and lower boundary \(y=x^2\) on \(0\le x\le2\). What is the vertical shell height?
- A horizontal slice runs from \(x=y^2\) to \(x=9\). What is its shell height?
- For rotation around a vertical line, which slice orientation produces shells, and which differential represents thickness?
- For rotation around \(y=-2\), what is the radius of a horizontal slice at \(y=3\)? Explain why this does not change its shell height.
- Why can integrating absolute radius over a region that crosses the axis count some shells more than once?
