Piecewise Shell Setups for Changing Boundaries

Why a Shell Setup May Need More Than One Integral

A region can have a simple overall shape while its boundary is described by different curves on different parts of the coordinate interval. A shell height that is correct on one part may become incorrect after two boundary curves meet and exchange which one forms the edge of the region.

The remedy is to partition the interval wherever the boundary changes, write the shell height separately on each piece, and add the resulting integrals. The crucial bookkeeping is geometric: the pieces must cover the region’s full coordinate range, and the formulas on each piece must describe its actual boundary.

This tutorial considers regions for which a slice at each coordinate in the interval is a single segment. The task is to handle changes in the curves that form the segment’s edges, not to handle disconnected slices.

Piecewise Shell Heights

For vertical shells about the $y$-axis, use vertical slices. At a given $x$, the shell height is the upper $y$-boundary minus the lower $y$-boundary. If the radius is $x$, the shell method gives the contribution $2\pi x h(x)\,dx$, where $h(x)$ is that height.

For horizontal shells about the $x$-axis, use horizontal slices. At a given $y$, the shell height is the right $x$-boundary minus the left $x$-boundary. If the radius is $y$, the contribution is $2\pi y h(y)\,dy$. In either orientation, a boundary change can make the height function piecewise defined.

Definition: A piecewise shell setup partitions the coordinate interval into adjacent subintervals, writes the shell height using the correct boundary functions on each subinterval, and sums the corresponding shell integrals. The partition points include every point where the boundary formula changes.

To locate a change, find where candidate boundary curves meet, then check which curve bounds the region on either side. Solving an intersection equation identifies possible change points; it does not, by itself, establish which curve is the relevant boundary on each interval. The earlier material on Finding Shell Bounds from Intersections addresses finding such points.

A Reliable Partitioning Procedure

Choose the slice coordinate.
For vertical shells, track how the top or bottom boundary varies with $x$. For horizontal shells, track how the right or left boundary varies with $y$.

Mark every boundary change.
Find the relevant intersections and endpoints. Order their coordinate values to create adjacent subintervals.

Determine the actual boundary on each interval.
Use the region’s geometry or test values to decide which curve is upper and lower, or left and right.

Write and add the shell integrals.
On each subinterval, form height as upper minus lower or right minus left, multiply by $2\pi$ times the radius, and sum.

If the coordinate interval is $[a,c]$ and the boundary changes at $b$, the vertical-shell integral has the form $2\pi\int_a^b xh_1(x)\,dx+2\pi\int_b^c xh_2(x)\,dx$. The exact expressions for $h_1$ and $h_2$ depend on the boundary curves. At the shared endpoint $b$, both pieces meet; counting that single endpoint in both definite integrals does not create a positive-volume overlap.

Worked Examples

Worked Example: An Upper Boundary Changes at an Intersection

Consider the region for $0\le x\le2$ between $y=0$ and an upper boundary that is $y=2x$ for $0\le x\le1$, then $y=4-2x$ for $1\le x\le2$. Find the volume when the region is revolved about the $y$-axis.

The upper curves meet where $2x=4-2x$, so $4x=4$ and $x=1$. On the left interval, the shell height is $2x-0=2x$. On the right interval, it is $(4-2x)-0=4-2x$. The radius is $x$.

\begin{aligned} V &=2\pi\int_0^1 x(2x)\,dx +2\pi\int_1^2 x(4-2x)\,dx\\ &=2\pi\int_0^1 2x^2\,dx +2\pi\int_1^2(4x-2x^2)\,dx. \end{aligned}

The first integral is $\left[\frac{2}{3}x^3\right]_0^1=\frac23$. An antiderivative for the second integrand is $2x^2-\frac23x^3$, so its value is $\left(8-\frac{16}{3}\right)-\left(2-\frac23\right)=\frac83-\frac43=\frac43$. Therefore, $V=2\pi(\frac23+\frac43)=4\pi$ cubic units.

Check: differentiating $\frac23x^3$ gives $2x^2$, and differentiating $2x^2-\frac23x^3$ gives $4x-2x^2$, the two integrands used. The boundary height is positive on both intervals and is $2$ at the joining point from either formula.

Worked Example: A Lower Boundary Changes

Let $0\le x\le2$, with upper boundary $y=3$ and lower boundary $y=0$ for $0\le x\le1$, then $y=x-1$ for $1\le x\le2$. Revolve the region about the $y$-axis.

The lower-boundary formulas agree at $x=1$: both give $y=0$. The shell radius is $x$. The heights are $3-0=3$ on the first interval and $3-(x-1)=4-x$ on the second. In particular, $4-x$ stays positive from $x=1$ through $x=2$.

\begin{aligned} V &=2\pi\int_0^1 3x\,dx +2\pi\int_1^2 x(4-x)\,dx\\ &=2\pi\left( \left[\frac32x^2\right]_0^1 +\left[2x^2-\frac13x^3\right]_1^2 \right). \end{aligned}

The first integral is $\frac32$. The second is $(8-\frac83)-(2-\frac13)=\frac{16}{3}-\frac53=\frac{11}{3}$. Thus $V=2\pi(\frac32+\frac{11}{3})=2\pi(\frac{31}{6})=\frac{31\pi}{3}$ cubic units.

Check: the derivative of $\frac32x^2$ is $3x$, and the derivative of $2x^2-\frac13x^3$ is $4x-x^2=x(4-x)$. At $x=1$, the two height formulas both give $3$, as required for adjoining pieces.

Worked Example: A Right Boundary Changes for Horizontal Shells

Consider the region with $0\le y\le2$, left boundary $x=0$, and right boundary $x=y$ for $0\le y\le1$, then $x=2-y$ for $1\le y\le2$. Revolve it about the $x$-axis.

For horizontal shells, the radius is $y$, and the height is right minus left. Thus the heights are $y-0=y$ and $(2-y)-0=2-y$, respectively. The boundary changes at $y=1$, where both right-boundary expressions equal $1$.

\begin{aligned} V &=2\pi\int_0^1 y(y)\,dy +2\pi\int_1^2 y(2-y)\,dy\\ &=2\pi\left( \left[\frac13y^3\right]_0^1 +\left[y^2-\frac13y^3\right]_1^2 \right). \end{aligned}

The first integral is $\frac13$. The second is $(4-\frac83)-(1-\frac13)=\frac43-\frac23=\frac23$. Hence $V=2\pi(\frac13+\frac23)=2\pi$ cubic units.

Check: the derivatives of $\frac13y^3$ and $y^2-\frac13y^3$ are $y^2$ and $2y-y^2=y(2-y)$. Both shell heights are nonnegative on their assigned intervals; at $y=1$ they agree.

Why Adding the Pieces Is Valid

The sum of the piecewise integrals is justified by the additivity of definite integrals over adjacent intervals. If $a<b<c$ and the shell integrand is $f$, then $\int_a^c f(t)\,dt=\int_a^b f(t)\,dt+\int_b^c f(t)\,dt$, provided the integrals exist. The same fact applies when a boundary change makes the formula for the integrand differ on the two intervals: integrate the correct formula on each interval and add.

Geometrically, each interval contributes the shells generated by slices whose coordinates lie in that interval. Adjacent intervals share only the slice at their common endpoint. That shared boundary has no volume, so it does not create double counting. If the intervals are adjacent and together span the entire coordinate range, they also leave no gap.

Theorem: Suppose the shell radius is $r(t)$, and the shell height is given by $h_i(t)$ on each of adjacent intervals $[a_{i-1},a_i]$, for $i=1,\ldots,n$. If these intervals cover the full coordinate range and the shell formula applies throughout, then the volume is $$ V=2\pi\sum_{i=1}^{n}\int_{a_{i-1}}^{a_i} r(t)h_i(t)\,dt. $$ Each $h_i$ must describe the region’s actual shell height on its own interval.

A Common Mistake: Keeping One Boundary Formula Too Long

In the first example, the upper boundary $y=2x$ is valid only through $x=1$. A wrong setup applies it all the way to $x=2$, giving $2\pi\int_0^2x(2x)\,dx=2\pi[\frac23x^3]_0^2=\frac{32\pi}{3}$. This is incorrect because for $1<x\le2$, the actual upper boundary is $4-2x$, not $2x$.

The correction is to split at $x=1$, use each actual height on its interval, and obtain $4\pi$, as computed above. The wrong integral can also be diagnosed geometrically: it assigns positive height $2x$ near $x=2$, even though the actual region’s height tends to $0$ there. A quick endpoint and sign check can reveal a boundary formula that has been extended beyond its valid interval.

Checking a Piecewise Setup

Before evaluating, verify the structure of the setup. List the coordinate endpoints in increasing order, and confirm that consecutive intervals meet with no gaps. On each interval, check that the stated top is at least the bottom, or that the stated right boundary is at least the left. At each change point, confirm that the adjoining boundary descriptions meet if the region’s edge is continuous there. Finally, check that the shell radius is measured from the axis of rotation and that every interval contributes exactly once.

These checks are especially useful when several curves appear in the description of the region. An intersection is a possible place to split; the geometry determines whether the boundary actually changes there. A correct split is not merely an algebraic convenience: it is what keeps every shell height tied to the region it represents.

Check Your Understanding

For each question, identify the interval structure and explain how you would check the resulting shell setup.

For vertical shells about the $y$-axis, a region’s upper boundary is $y=3x$ up to $x=2$, then $y=12-3x$ from $x=2$ to $x=4$, while its lower boundary is $y=0$. What are the two shell heights?
For horizontal shells, the left boundary changes at $y=1$, while the right boundary stays the same. Which boundary difference defines the shell height on each interval?
Why does counting a shared endpoint in two adjacent definite integrals not count a positive volume twice?
What should be checked after solving an equation where two candidate boundary curves intersect?
A proposed shell height is negative on part of its interval. What geometric feature of the setup should be re-examined?