Tutorials › Calculus II › Choosing Slice Orientation for a Shell Integral

Volumes by Cylindrical Shells · Tutorial 395 of 998

Choosing Slice Orientation for a Shell Integral

Choose shell strips that run parallel to the axis of rotation, then organize the region so each shell’s radius and height are defined without overlap.

Intermediate 11 min read

What You'll Learn

  • Match vertical strips to a vertical axis and horizontal strips to a horizontal axis.
  • Identify whether the region’s boundaries are single-valued in the chosen slice direction.
  • Write shell heights from the top and bottom or right and left boundaries.
  • Check where boundary order changes and split the integral at those points.
  • Apply the shell formula only when generated shells do not count the same part of the solid more than once.

Why Slice Orientation Matters

A cylindrical-shell integral begins with thin strips of a planar region. Rotating a strip about an axis parallel to it sweeps out a thin cylindrical shell. The strip’s orientation determines whether its distance from the axis and its length can be described simply.

The axis determines the basic choice: vertical strips run parallel to a vertical axis, and horizontal strips run parallel to a horizontal axis. The region then determines how convenient those strips are to describe. A good setup has boundaries that are single-valued in the slice direction, a simple expression for the strip length, and as few interval splits as the geometry allows.

Key idea: For the shell method, choose strips parallel to the axis of rotation. Vertical strips are used for a vertical axis; horizontal strips are used for a horizontal axis. Then check that the region’s boundaries and shell heights are correctly described over each interval.

Match the Strips to the Axis

For rotation about the vertical line \(x=c\), a vertical strip at \(x\) runs parallel to the axis. Its shell radius is the horizontal distance \(|x-c|\), its height is the strip’s vertical length, and its thickness is \(dx\). If the region is between a top curve \(y=f(x)\) and a bottom curve \(y=g(x)\), the height is \(f(x)-g(x)\), provided \(f(x)\ge g(x)\) throughout the interval.

$$ V=2\pi\int_a^b |x-c|\,h(x)\,dx. $$

For rotation about the horizontal line \(y=d\), use horizontal strips. If a strip at \(y\) has length \(h(y)\) from its left boundary to its right boundary, its radius is \(|y-d|\), and its thickness is \(dy\). Thus, on an interval where the strip length is given by one expression, the shell integral is

$$ V=2\pi\int_p^q |y-d|\,h(y)\,dy. $$

In either orientation, the shell height is a length: top minus bottom for a vertical strip, or right minus left for a horizontal strip. If the region has several separate pieces in a strip, their lengths must be accounted for separately rather than treating empty space as part of the height.

Non-overlap condition: The shell formula can be applied directly over a strip interval when the region lies on one side of the axis, apart from possibly touching it, so different strips do not generate the same shells. If the region crosses the axis, using the distance to the axis on both sides can count overlapping shells twice. In that situation, decompose the region or otherwise account for each generated shell only once.

Check How the Region Is Described

Once the strip direction is fixed by the axis, inspect the boundary functions. Vertical strips work naturally when the region can be described using \(x\)-intervals and, for each \(x\), a single bottom and top boundary. Horizontal strips work naturally when the region can be described using \(y\)-intervals and, for each \(y\), a single left and right boundary. These descriptions are often called vertically simple and horizontally simple, respectively.

“Single-valued” refers to the boundary as a function of the strip variable. For example, \(y=f(x)\) gives a single \(y\)-value for each allowed \(x\). A curve such as \(x=y^2\) gives \(x\) as a function of \(y\), which is directly useful for horizontal strips. If the region’s boundary has multiple branches, identify which branch forms each edge of the strip and restrict the interval accordingly.

The top and bottom boundaries, or the left and right boundaries, must also stay in the same order over each interval of integration. If they exchange roles at an intersection, split the integral there and write the height separately on each side. A formula that uses “top minus bottom” is not valid on an interval where those labels have switched.

A Derivation of the Shell Setup

Consider a region lying to one side of the vertical axis \(x=c\), with one vertical strip at each \(x\in[a,b]\). Suppose the strip has continuous height \(h(x)\), and the region does not cross the axis. For a small width \(dx\), rotating the strip through a full turn produces a thin shell with radius \(r(x)=|x-c|\), height \(h(x)\), and thickness \(dx\).

A cylinder of radius \(r\) and height \(h\) has volume \(\pi r^2h\). A shell of thickness \(\Delta x\) can be viewed as the difference between two nearby cylinders. If its inner and outer radii are \(r\) and \(r+\Delta x\), respectively, its volume is

$$ \pi(r+\Delta x)^2h-\pi r^2h = \pi\bigl(2r\Delta x+(\Delta x)^2\bigr)h. $$

After dividing by \(\Delta x\), the first-order contribution per unit thickness is \(2\pi rh\); the term \(\pi(\Delta x)^2h\) becomes negligible as the thickness tends to zero. Summing these shell contributions over a partition of \([a,b]\) and taking the limit gives the integral.

$$ V=\int_a^b 2\pi r(x)h(x)\,dx =2\pi\int_a^b |x-c|h(x)\,dx. $$

This reasoning requires that the shells represent the solid without counting any part more than once. It also assumes the radii and heights are integrable over the interval; continuous, piecewise-defined boundaries satisfy this condition. The horizontal-strip formula follows by the same argument, using \(dy\) for thickness and distance \(|y-d|\) for radius.

Worked Example: Vertical Shells About the \(y\)-Axis

Worked Example: Vertical Shells About the \(y\)-Axis

Find the shell-method volume when the region bounded by \(y=x^2\), \(y=4\), and \(x=0\) is rotated about the \(y\)-axis.

The axis is vertical, so use vertical strips. The region runs from \(x=0\) to the intersection of \(y=x^2\) and \(y=4\). Solve \(x^2=4\), giving \(x=2\) on the region with \(x\ge0\). The strip’s top is \(4\) and its bottom is \(x^2\), so its height is \(4-x^2\). The region lies to the right of the axis, so each strip generates a shell with radius \(x\), without overlap.

$$ V=2\pi\int_0^2 x(4-x^2)\,dx =2\pi\int_0^2(4x-x^3)\,dx. $$

An antiderivative is \(2x^2-\frac{x^4}{4}\). Evaluating at the bounds gives

$$ V=2\pi\left[2x^2-\frac{x^4}{4}\right]_0^2 =2\pi\left((8-4)-0\right) =8\pi. $$

As a check on the setup, at \(x=0\) the shell radius is zero, and at \(x=2\) the shell height is \(4-4=0\), matching the boundaries of the region. The computed volume is \(8\pi\) cubic units.

Worked Example: Horizontal Shells About the \(x\)-Axis

Worked Example: Horizontal Shells About the \(x\)-Axis

Find the shell-method volume when the region bounded by \(x=y^2\), \(x=4\), and \(y=0\) is rotated about the \(x\)-axis.

The axis is horizontal, so use horizontal strips. The region has \(0\le y\le2\), since \(y^2=4\) gives \(y=2\) on the part above the \(x\)-axis. At each such \(y\), the strip extends from \(x=y^2\) to \(x=4\). Its length is therefore \(4-y^2\). The radius is \(y\), and the region lies above the axis, so these strips generate distinct shells.

$$ V=2\pi\int_0^2 y(4-y^2)\,dy =2\pi\int_0^2(4y-y^3)\,dy. $$

An antiderivative is \(2y^2-\frac{y^4}{4}\). Therefore,

$$ V=2\pi\left[2y^2-\frac{y^4}{4}\right]_0^2 =2\pi\left((8-4)-0\right) =8\pi. $$

At \(y=0\), the shell radius is zero; at \(y=2\), its height is \(4-4=0\). These checks agree with the region’s boundary. The volume is \(8\pi\) cubic units.

Worked Example: Split Where the Boundaries Switch

Worked Example: Split Where the Boundaries Switch

Let the region consist of the points between \(y=x\) and \(y=4-x\), for \(0\le x\le3\). Find the volume when this region is rotated about the \(y\)-axis.

The axis is vertical, so use vertical strips. Each strip lies entirely to the right of the axis, and its radius is \(x\). The two boundary curves intersect when \(x=4-x\), so \(2x=4\) and \(x=2\). Before \(x=2\), \(4-x\) is above \(x\); after \(x=2\), \(x\) is above \(4-x\). The height must therefore be split into two expressions.

$$ h(x)= \begin{cases} (4-x)-x=4-2x, & 0\le x\le2,\\ x-(4-x)=2x-4, & 2\le x\le3. \end{cases} $$

Applying the shell formula on each interval gives

$$ V=2\pi\left( \int_0^2 x(4-2x)\,dx +\int_2^3 x(2x-4)\,dx \right). $$

For the first integral,

$$ \int_0^2(4x-2x^2)\,dx = \left[2x^2-\frac{2}{3}x^3\right]_0^2 =8-\frac{16}{3} =\frac{8}{3}. $$

For the second,

$$ \int_2^3(2x^2-4x)\,dx = \left[\frac{2}{3}x^3-2x^2\right]_2^3 =0-\left(\frac{16}{3}-8\right) =\frac{8}{3}. $$

Thus

$$ V=2\pi\left(\frac{8}{3}+\frac{8}{3}\right) =\frac{32\pi}{3}. $$

The split at \(x=2\) is essential: using \(4-2x\) as the height throughout \([0,3]\) would make the height negative after the curves cross. The region stays on one side of the axis, so no shell is counted twice. The volume is \(\frac{32\pi}{3}\) cubic units.

A Common Mistake: Counting Shells Twice

Suppose the rectangle \(-1\le x\le1,\ 0\le y\le1\) is rotated about the \(y\)-axis. A tempting but incorrect setup is to use vertical strips across the whole rectangle and write

$$ 2\pi\int_{-1}^{1}|x|\,dx =2\pi\left(\frac12+\frac12\right) =2\pi. $$

This doubles the correct volume. A strip at \(x\) and a strip at \(-x\) sweep out the same cylindrical shell, so the integral counts every shell twice. The solid is a cylinder of radius \(1\) and height \(1\), whose volume is \(\pi\). A non-overlapping shell setup uses only one side of the rectangle:

$$ 2\pi\int_0^1 x(1)\,dx =2\pi\left[\frac{x^2}{2}\right]_0^1 =\pi. $$

The correction is not to ignore the distance from the axis. It is to ensure that each shell is included exactly once. When the region crosses the axis, check explicitly whether strips on opposite sides generate the same shells before setting up an integral.

A Practical Decision Process

1
Identify the axis.
A vertical axis calls for vertical shell strips; a horizontal axis calls for horizontal shell strips.
2
Describe each strip.
For vertical strips, identify top and bottom boundaries as functions of \(x\). For horizontal strips, identify right and left boundaries as functions of \(y\).
3
Find where the description changes.
Check intersections and endpoints. Split the interval whenever the boundary order or the strip components change.
4
Check radii and overlap.
Use the distance to the axis for the radius, and verify that distinct strips do not generate any shell more than once.
5
Write the integral piece by piece.
Multiply circumference \(2\pi r\) by height and thickness, then integrate over each interval where the formulas apply.

This process helps distinguish a genuinely simple shell setup from one that only appears simple when the region’s changing boundaries or possible overlap are overlooked. If the required strip direction produces several branches or many pieces, keep the geometry explicit and divide the integral wherever its description changes.

Check Your Understanding

Use the axis direction, strip boundaries, and non-overlap condition to answer these questions.

  1. For rotation about \(x=3\), which strip orientation produces cylindrical shells, and what is the radius of a strip at \(x\)?
  2. For a horizontal strip extending from \(x=L(y)\) to \(x=R(y)\), how is its shell height expressed?
  3. Why must an integral be split when the top and bottom curves exchange order inside the interval?
  4. When can strips on opposite sides of an axis generate overlapping shells?
  5. For rotation about the \(x\)-axis, what strip orientation is parallel to the axis, and what variable measures strip thickness?

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