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A Piecewise Region Shell Setup Workshop

Build a piecewise cylindrical-shell integral by finding where the boundary curves switch and writing the correct radius and height on each interval.

Intermediate 10 min read

What You'll Learn

  • Identify the interval over which a region extends and the points where its boundary curves intersect.
  • Decide which curve supplies the upper and lower boundaries on each subinterval.
  • Express a changing shell height as a piecewise function.
  • Assemble the shell integral with the correct radius, height, and differential.
  • Check a piecewise setup at endpoints and representative points.

Why a Region Sometimes Needs More Than One Shell Integral

A shell integral is built from the radius and height of a typical thin slice. When the same two curves bound the region across the entire interval, the height has one formula. But sometimes the upper or lower boundary changes at an intersection. Then one formula for the height cannot describe the whole region: the interval must be split, and the shell integral must be written as a sum.

This tutorial focuses on that setup. The essential task is not merely to solve equations for intersections. It is to use those intersections to determine where the boundary changes, identify the correct height on each resulting interval, and include every piece exactly once.

The Shell Setup for Vertical Slices

For shells formed by rotating vertical slices about the \(y\)-axis, the slice at \(x\) has radius \(x\) when the region lies to the right of the axis. Its height is the vertical distance between the upper and lower edges of the region. If those edges are \(y=U(x)\) and \(y=L(x)\), respectively, the height is \(U(x)-L(x)\).

Shell formula: For a region to the right of the \(y\)-axis with vertical-slice height \(h(x)\) on \(a\le x\le b\), its shell-method volume is \(V=2\pi\int_a^b xh(x)\,dx\). If the region lies to the left of the \(y\)-axis, the radius is \(|x|=-x\), not \(x\). If a region crosses the axis, using \(|x|\) without first checking for overlap can count parts of the solid more than once.

In a piecewise setup, the shell heights may be \(h_1(x)\), \(h_2(x)\), and so on, on adjacent intervals. The integral is then the sum of the shell integrals over those intervals. The radius is still measured from the axis, and the height is still upper boundary minus lower boundary on each piece.

Why the Shell Integral Has This Form

A thin vertical slice of width \(\Delta x\), radius approximately \(x\), and height approximately \(h(x)\) sweeps out a thin cylindrical shell. Its circumference is approximately \(2\pi x\), so its volume is approximately circumference times height times thickness:

$$ \Delta V\approx 2\pi xh(x)\,\Delta x. $$

Adding the shell volumes across the interval gives a Riemann sum for the volume. As the slice widths shrink, that sum approaches the definite integral of \(2\pi xh(x)\). If the boundary changes at finitely many intersections, the same reasoning applies on each interval where the height formula is fixed, and the total volume is the sum of those integrals. The split is necessary because a single formula for the height may not remain valid across a boundary switch.

A Reliable Setup Procedure

First determine the full \(x\)-interval occupied by the region. Then find the intersections of the curves that might supply its boundaries, keeping only intersection \(x\)-values inside that interval. Sort those values together with the interval endpoints. On each resulting subinterval, compare the curves to decide which one is upper and which is lower. Finally, write the height and shell integral separately for each piece.

1
Set the domain.
Identify the leftmost and rightmost \(x\)-values of the region, including any vertical boundaries.
2
Find relevant intersections.
Solve equations between curves that could exchange roles as an upper or lower boundary. Retain only solutions in the region's \(x\)-interval.
3
Test each interval.
Use a point strictly inside each interval to determine the curve order. A single test point is enough when the curves do not intersect inside that interval.
4
Write height and radius.
On each piece, subtract lower boundary from upper boundary and measure the radius from the axis of rotation.
5
Add the pieces.
Write one shell integral for each interval, using matching bounds, height, radius, and \(dx\).

Worked Example: Two Lines Form the Upper Boundary

Worked Example: Two Lines Form the Upper Boundary

Consider the region bounded by \(x=0\), \(y=0\), \(y=x\), and \(y=4-x\). Set up its volume when rotated about the \(y\)-axis.

The two sloping curves meet where \(x=4-x\). Solving gives \(2x=4\), so \(x=2\), and the intersection is \((2,2)\). The region extends horizontally from \(x=0\) to \(x=4\): the line \(y=x\) meets \(y=0\) at \(x=0\), and \(y=4-x\) meets \(y=0\) at \(x=4\).

For \(0<x<2\), test \(x=1\). The line values are \(y=x=1\) and \(y=4-x=3\), so \(y=x\) is the upper curve there. For \(2<x<4\), test \(x=3\). The values are \(y=x=3\) and \(y=4-x=1\), so \(y=4-x\) is upper on that interval. The lower boundary is \(y=0\) throughout. Therefore, the height is \(x\) on \([0,2]\) and \(4-x\) on \([2,4]\).

$$ \begin{aligned} V &=2\pi\int_0^2 x(x-0)\,dx +2\pi\int_2^4 x\bigl((4-x)-0\bigr)\,dx\\ &=2\pi\int_0^2 x^2\,dx +2\pi\int_2^4 x(4-x)\,dx. \end{aligned} $$

Check: At the switch \(x=2\), both candidate heights equal \(2\). At the left endpoint \(x=0\), the first height is \(0\), and at the right endpoint \(x=4\), the second height is \(0\). The test points also confirm the selected upper boundary on each side. The radius \(x\) is nonnegative throughout, as required for this form of the shell formula.

Worked Example: The Region Lies Below Both Curves

Worked Example: The Region Lies Below Both Curves

Let a region have \(0\le x\le 3\), lower boundary \(y=0\), and upper boundary constrained by both \(y=x^2\) and \(y=4\). Set up its volume around the \(y\)-axis.

Because the region lies below both curves, its upper edge is the smaller of \(x^2\) and \(4\). Find the switch by solving \(x^2=4\). The solutions are \(x=-2\) and \(x=2\); only \(x=2\) lies in \([0,3]\). For \(0<x<2\), test \(x=1\): \(x^2=1<4\), so the upper boundary is \(y=x^2\). For \(2<x<3\), test \(x=3\): \(x^2=9>4\), so the upper boundary is \(y=4\). The lower boundary is zero on both pieces, giving heights \(x^2\) and \(4\).

$$ \begin{aligned} V &=2\pi\int_0^2 x(x^2-0)\,dx +2\pi\int_2^3 x(4-0)\,dx\\ &=2\pi\int_0^2 x^3\,dx +2\pi\int_2^3 4x\,dx. \end{aligned} $$

Check: At \(x=2\), both proposed upper-boundary formulas give \(4\). At the test points, the chosen heights are \(1\) at \(x=1\) and \(4\) at \(x=3\), matching the smaller of the two curve values. The radius is \(x\) on the full interval \([0,3]\).

Worked Example: The Lower Boundary Changes

Worked Example: The Lower Boundary Changes

Consider the region with \(0\le x\le 4\), upper boundary \(y=5\), and lower boundary formed by whichever is higher of \(y=x\) and \(y=4-x\). Set up its volume around the \(y\)-axis.

The lower-boundary curves intersect where \(x=4-x\), so \(2x=4\) and \(x=2\). On \(0<x<2\), test \(x=1\): \(4-x=3\) is greater than \(x=1\), so \(4-x\) is the lower edge of the region (the higher of the two proposed lower bounds). On \(2<x<4\), test \(x=3\): \(x=3\) is greater than \(4-x=1\), so \(x\) is the lower edge. Subtracting these from the upper boundary \(5\) gives heights \(5-(4-x)=1+x\) and \(5-x\).

$$ \begin{aligned} V &=2\pi\int_0^2 x\bigl(5-(4-x)\bigr)\,dx +2\pi\int_2^4 x(5-x)\,dx\\ &=2\pi\int_0^2 x(1+x)\,dx +2\pi\int_2^4 x(5-x)\,dx. \end{aligned} $$

Check: At \(x=2\), both lower-boundary curves equal \(2\), so both height formulas give \(5-2=3\). At \(x=1\), the selected lower boundary is \(3\), giving height \(2=1+1\). At \(x=3\), the selected lower boundary is \(3\), giving height \(2=5-3\). Each height is nonnegative, and the radius \(x\) is nonnegative across the domain.

A Common Mistake: Choosing the Wrong Curve

In the second example, a tempting but incorrect choice is to use the larger of \(x^2\) and \(4\) as the upper boundary, on the assumption that the larger curve is “on top.” For a region below both curves, the upper edge must satisfy both upper limits. It is therefore the smaller value, not the larger one.

$$ \text{Incorrect on }[0,2]:\quad h(x)=4 \qquad \text{Correct on }[0,2]:\quad h(x)=x^2, $$

For instance, at \(x=1\), the incorrect choice gives height \(4\), even though the region must lie below \(y=x^2=1\). The correct height is \(1\). The correction is to compare the curve values on each interval and then subtract the actual lower edge from the actual upper edge. For a region between curves, “upper minus lower” is the rule; which formula is upper must be established rather than guessed.

What a Complete Setup Must Show

A complete piecewise shell setup makes the geometry visible in the integral. It states the interval, identifies every relevant switch, gives the boundary order on each piece, and uses the corresponding height with the correct radius. If the region lies on the left side of the \(y\)-axis, replace \(x\) in the radius with \(|x|\); do not let a negative coordinate produce a negative volume. If it crosses the axis, first check whether distinct slices generate overlapping shells before summing integrals.

These checks also help distinguish a real boundary change from an irrelevant intersection. An algebraic intersection outside the region's \(x\)-interval does not create a piece. An intersection of curves that are not actually bounding the region on either side may not create a switch either. The partition should reflect changes in the boundary of the region, not merely every solution of every curve equation.

Check Your Understanding

For each question, focus on how the boundary order determines the integral setup.

  1. If a region lies below both \(y=f(x)\) and \(y=g(x)\), which curve gives its upper boundary on an interval where \(f(x)<g(x)\)?
  2. Why should an intersection outside the region's \(x\)-interval not be used as a shell-integral split point?
  3. For the region in the first worked example, what are the shell radii and heights on the interval \(2\le x\le4\)?
  4. A region lies to the left of the \(y\)-axis. What radius should be used for a vertical shell at coordinate \(x<0\)?
  5. Why is it important to check a test point inside each interval created by the intersection values?

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