Why Separate Components Must Be Counted Separately
A horizontal strip through a region does not always form one continuous interval. It may meet the region in two or more separated pieces, with gaps between them. When the region is rotated about the x-axis, each piece of the strip contributes a cylindrical shell.
At a fixed height, all these shells have the same radius, but their lengths can differ. The volume calculation must include every piece. It must also exclude the gaps: the distance from the leftmost edge to the rightmost edge is not necessarily the total width of the region at that height.
This tutorial considers regions lying on or above the x-axis and rotated about that axis. Keeping the region on one side of the axis lets us focus on adding contributions from separate horizontal components.
Horizontal Components and Their Widths
The intervals are written from left to right within each component, so \(a_i(y)\) is that component’s left endpoint and \(b_i(y)\) is its right endpoint. The gap between components is not part of any width.
For rotation about the x-axis, a horizontal strip at height \(y\) has shell radius \(y\). Its height along the axis is the total horizontal width of the slice. Thus, if there are \(n\) components, the shell-method setup is:
Here \(\alpha\) and \(\beta\) are the lower and upper heights of the region, and the component intervals must describe the slice throughout the interval of integration. If the number of components or their boundary formulas change at some height, the integral must be split at that height.
Why the Widths Add
Consider a thin horizontal strip between heights \(y\) and \(y+\Delta y\), where \(\Delta y\) is small and positive. Each component of this strip rotates into a thin shell. At height \(y\), each shell has circumference approximately \(2\pi y\). The shell formed by component \(i\) has axial length \(b_i(y)-a_i(y)\), so its approximate volume is circumference times length times thickness.
Because the components are separated along the axis of rotation, their thin shell pieces occupy separate positions. Their volumes add. The gaps contribute no volume, since the strip contains no region there. Adding the contributions gives an approximate volume for the strip:
Summing these contributions over thin strips and taking the limit as their thicknesses approach zero gives the integral formula. The reason the total width appears inside the integrand is precisely that each component contributes its own shell volume at the same radius.
Worked Example: Two Components of Constant Width
Worked Example: Two Components of Constant Width
Let a region have horizontal slices, for \(0\le y\le2\), given by \([y,y+1]\) and \([y+3,y+4]\). Rotate the region about the x-axis. Find the volume.
First find each component’s width. The left component has width \((y+1)-y=1\). The right component has width \((y+4)-(y+3)=1\). Therefore the total width is \(1+1=2\). The interval from the far-left edge to the far-right edge has length \((y+4)-y=4\), but two of those units are a gap and must not be counted.
The shell radius is \(y\), so:
Thus the volume is \(8\pi\) cubic units. Check the antiderivative used in the integral: the derivative of \(y^2\) is \(2y\), as required. The two widths are both \(1\) at every height, confirming that the integrand uses total width \(2\), not the full span \(4\).
Worked Example: Widths That Vary with Height
Worked Example: Widths That Vary with Height
For \(0\le y\le2\), suppose the horizontal slice consists of \([0,y+1]\) and \([y+3,5]\). Rotate this region about the x-axis and find its volume.
The first width is \((y+1)-0=y+1\). The second width is \(5-(y+3)=2-y\). For \(0\le y\le2\), both widths are nonnegative; at the top endpoint the second width is zero. The total width is:
Substitute this total width into the shell formula:
The volume is \(12\pi\) cubic units. To check the calculation, the derivative of \(\frac{y^2}{2}\) is \(y\), so the antiderivative correctly integrates \(3y\). Also, the component widths sum to \(3\) at both \(y=0\) and \(y=2\): they are \(1\) and \(2\) at the bottom, and \(3\) and \(0\) at the top.
Worked Example: Curved Boundaries
Worked Example: Curved Boundaries
For \(0\le y\le2\), let the horizontal slice consist of \([y^2,y^2+1]\) and \([y^2+3,2y+6]\). These intervals are separated: the right endpoint of the first is \(y^2+1\), while the left endpoint of the second is \(y^2+3\), leaving a gap of \(2\). Rotate the region about the x-axis and find its volume.
The first width is \((y^2+1)-y^2=1\). The second width is:
This second width is positive on \(0\le y\le2\): it equals \((3-y)(y+1)\), and both factors are nonnegative on that interval. The total width is \(1+(-y^2+2y+3)=-y^2+2y+4\). Therefore:
The volume is \(\frac{56\pi}{3}\) cubic units. To check the integration, differentiate the antiderivative: \(-y^3+2y^2+4y\) results, which is the expanded integrand. At \(y=0\), the component widths are \(1\) and \(3\); at \(y=2\), they are \(1\) and \(3\). These endpoint checks agree with the stated width formulas.
A Common Mistake: Counting the Gaps
A frequent error is to use the distance from the slice’s leftmost edge to its rightmost edge as the shell height. That distance includes any gaps, so it overstates the volume whenever the slice has separated components.
For the first example, the leftmost endpoint is \(y\) and the rightmost endpoint is \(y+4\). A wrong setup uses the span \((y+4)-y=4\):
This doubles the correct result because the actual slice contains two intervals of width \(1\), not one interval of width \(4\). The missing distinction is that the two-unit gap between the components contributes no shell. The correction is to calculate the widths separately and add them: \(1+1=2\), giving \(V=8\pi\).
Another way to make the same mistake is to include only one interval. In the first example, that would give \(2\pi\int_0^2 y\,dy=4\pi\), omitting the shell contribution from the second component. A reliable setup records every interval in each slice before forming the integral.
Setting Up the Integral Reliably
When a horizontal slice has multiple components, use the following sequence. It is especially useful when the endpoints are given by curves or when their formulas depend on height.
Find the lowest and highest \(y\)-values occupied by the region. Check whether the component boundaries change within that range.
Write each separate interval from its left endpoint to its right endpoint. Keep gaps outside all the intervals.
For each interval, subtract left endpoint from right endpoint. Add the resulting widths to obtain the total width at height \(y\).
For rotation about the x-axis, use radius \(y\). Multiply \(2\pi y\) by the total width and integrate over the height range.
Confirm that every width is nonnegative on the stated range, gaps have not been counted, and the antiderivative differentiates to the integrand.
If the axis of rotation is a horizontal line \(y=c\) and the entire region lies on one side of it, the same width calculation applies, with shell radius \(|y-c|\). The radius must be the distance to the axis. The case where the axis crosses the region requires additional care and is treated separately.
Check Your Understanding
For each question, write down the component widths before forming or interpreting the shell integral.
- At a fixed height, a slice consists of \([1,3]\) and \([6,9]\). What are the component widths and the total width? What is the full left-to-right span, and why is it different?
- For \(0\le y\le1\), a slice consists of \([0,2-y]\) and \([4,5+y]\). Write the total width as a function of \(y\).
- For a region above the x-axis, what is the shell radius of a horizontal strip at height \(y\)?
- Why does the shell-method integral add component widths rather than use just the width of the longest component?
- For \(0\le y\le2\), a slice consists of \([y,y+2]\) and \([y+5,y+6]\). Write the volume integral for rotation about the x-axis. Do not evaluate it.
