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Volumes by Cylindrical Shells · Tutorial 389 of 998

Shell Heights for Regions with Holes

Learn to describe shell heights around a hole as separate intervals and use them to compute volume without counting empty space.

Intermediate 10 min read

What You'll Learn

  • Express a slice through a region with a hole as a union of material intervals.
  • Add the lengths of the material intervals to obtain the shell height.
  • Split the shell integral wherever the cross-section or its boundaries change.
  • Explain why an empty gap contributes no volume.
  • Check a piecewise shell setup by comparing it with an outer region minus its hole.

Why a Hole Changes the Shell Height

The cylindrical-shell method uses the material present in each thin slice of a region. If a slice passes through a hole, the slice is not one unbroken segment: it consists of separate pieces of material with empty space between them. A shell made from that slice therefore has several vertical pieces, not one piece spanning across the gap.

For shells about the \(y\)-axis, a vertical slice at position \(x\) has radius \(x\) when the region lies in \(x\ge 0\). Its shell height is the total length of the material in that slice. If the slice has two or more separate material intervals, add their lengths and do not include the gap.

Definition: Suppose the vertical slice at \(x\) meets a region in disjoint intervals \([a_1(x),b_1(x)],\ldots,[a_n(x),b_n(x)]\). Its shell height is \(H(x)=\sum_{j=1}^{n}(b_j(x)-a_j(x))\). The intervals describe material only; an interval of empty space between them is not part of \(H(x)\).

For a region in \(x\ge 0\) revolved about the \(y\)-axis, the shell-method volume is \(V=\int 2\pi xH(x)\,dx\), with the integral taken over the relevant \(x\)-values. If the number or endpoints of the material intervals change, write the integral piecewise. Individual endpoints do not affect the integral, so it is harmless if adjacent pieces share a boundary value.

Why the Gap Contributes No Material

A thin vertical slice of thickness \(\Delta x\) produces a shell whose circumference is approximately \(2\pi x\). Each material interval in the slice contributes a strip of shell with that circumference and with height equal to the interval’s length. If the slice has several material intervals, the strips together have total height \(H(x)\). The gap between them contains no points of the original region, so rotating it cannot produce part of the solid.

As the slice thickness becomes arbitrarily small, adding these shell volumes gives the integral of circumference times total material height. This explains why the height is a sum of interval lengths, rather than the distance from the lowest point of the slice to its highest point.

Takeaway: Find the material intervals first. Add their lengths to get shell height, then multiply by the shell circumference. Never count an empty gap as part of a shell.

Worked Example: A Rectangular Region with a Rectangular Hole

Worked Example: A Rectangular Region with a Rectangular Hole

Take the rectangle \(1\le x\le 4,\ 0\le y\le 4\), and remove the rectangular hole \(2\le x\le 3,\ 1\le y\le 3\). Revolve the remaining region about the \(y\)-axis. Find its volume using shells.

For \(1\le x<2\) and \(3<x\le 4\), a vertical slice is one interval from \(y=0\) to \(y=4\), so \(H(x)=4\). For \(2<x<3\), the hole removes the interval from \(y=1\) to \(y=3\). The material intervals are \([0,1]\) and \([3,4]\), each of length \(1\), so \(H(x)=1+1=2\).

The volume is therefore

$$ \begin{aligned} V &=2\pi\left(\int_1^2 4x\,dx+\int_2^3 2x\,dx+\int_3^4 4x\,dx\right)\\ &=2\pi\left(4\left[\frac{x^2}{2}\right]_1^2 +2\left[\frac{x^2}{2}\right]_2^3 +4\left[\frac{x^2}{2}\right]_3^4\right)\\ &=2\pi\left(4\cdot\frac{4-1}{2} +2\cdot\frac{9-4}{2} +4\cdot\frac{16-9}{2}\right)\\ &=2\pi(6+5+14)=50\pi. \end{aligned} $$

As a check, the whole rectangle without a hole has shell height \(4\) from \(x=1\) to \(x=4\), giving \(2\pi\int_1^4 4x\,dx=60\pi\). The removed hole has height \(2\) from \(x=2\) to \(x=3\), giving \(2\pi\int_2^3 2x\,dx=10\pi\). Subtracting gives \(60\pi-10\pi=50\pi\), the same result.

Piecewise Heights When a Hole Changes Shape

A hole need not have constant height or occupy the same \(x\)-interval as the rest of the region. The procedure is unchanged: identify each interval of material at a given \(x\), add the lengths, and split the integral wherever those intervals change.

Worked Example: A Hole with a Sloping Boundary

Let the outer region be the rectangle \(0\le x\le 3,\ 0\le y\le 4\). Remove the region \(0\le x\le 2,\ 1\le y\le x+2\), and revolve what remains about the \(y\)-axis.

For \(0\le x<2\), the material intervals are \([0,1]\) and \([x+2,4]\). Their lengths are \(1\) and \(4-(x+2)=2-x\), so the total shell height is \(H(x)=1+(2-x)=3-x\). For \(2<x\le 3\), the slice misses the removed region and has height \(4\). Thus

$$ \begin{aligned} V &=2\pi\left(\int_0^2 x(3-x)\,dx+\int_2^3 4x\,dx\right)\\ &=2\pi\left(\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^2 +\left[2x^2\right]_2^3\right)\\ &=2\pi\left(6-\frac{8}{3}+18-8\right)\\ &=2\pi\left(\frac{10}{3}+10\right) =\frac{80\pi}{3}. \end{aligned} $$

For an independent check, the outer rectangle gives \(2\pi\int_0^3 4x\,dx=36\pi\). The removed region has vertical height \((x+2)-1=x+1\) for \(0\le x\le 2\), so its shell volume is

$$ 2\pi\int_0^2 x(x+1)\,dx =2\pi\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^2 =2\pi\left(\frac{8}{3}+2\right) =\frac{28\pi}{3}. $$

The difference is \(36\pi-\frac{28\pi}{3}=\frac{80\pi}{3}\), confirming the piecewise setup.

Worked Example: A Circular Region with a Circular Hole

Worked Example: A Circular Region with a Circular Hole

Consider the right half of the disk \(x^2+y^2\le 9\), with the right half of the disk \(x^2+y^2<1\) removed. Revolve the remaining region about the \(y\)-axis.

For \(0\le x<1\), the outer circle has top and bottom values \(y=\pm\sqrt{9-x^2}\), while the hole has top and bottom values \(y=\pm\sqrt{1-x^2}\). The material intervals are \([-\sqrt{9-x^2},-\sqrt{1-x^2}]\) and \([\sqrt{1-x^2},\sqrt{9-x^2}]\). Each has length \(\sqrt{9-x^2}-\sqrt{1-x^2}\). For \(1\le x\le 3\), the slice is one full interval of length \(2\sqrt{9-x^2}\). Therefore

$$ \begin{aligned} V &=2\pi\left( \int_0^1 2x\bigl(\sqrt{9-x^2}-\sqrt{1-x^2}\bigr)\,dx +\int_1^3 2x\sqrt{9-x^2}\,dx\right)\\ &=4\pi\left( \int_0^3 x\sqrt{9-x^2}\,dx -\int_0^1 x\sqrt{1-x^2}\,dx\right). \end{aligned} $$

To evaluate these integrals, an antiderivative of \(x\sqrt{R^2-x^2}\) is \(-\frac{1}{3}(R^2-x^2)^{3/2}\): differentiating gives \(-\frac{1}{3}\cdot\frac{3}{2}(R^2-x^2)^{1/2}(-2x)=x\sqrt{R^2-x^2}\). Consequently,

$$ \begin{aligned} \int_0^3 x\sqrt{9-x^2}\,dx &=\left[-\frac{1}{3}(9-x^2)^{3/2}\right]_0^3 =0-(-9)=9,\\ \int_0^1 x\sqrt{1-x^2}\,dx &=\left[-\frac{1}{3}(1-x^2)^{3/2}\right]_0^1 =0-\left(-\frac{1}{3}\right)=\frac{1}{3}. \end{aligned} $$

Thus \(V=4\pi(9-\frac{1}{3})=\frac{104\pi}{3}\). As a check, the shell contribution from the full outer half-disk is \(4\pi\int_0^3x\sqrt{9-x^2}\,dx=36\pi\). The removed half-disk contributes \(4\pi\int_0^1x\sqrt{1-x^2}\,dx=\frac{4\pi}{3}\). Their difference is \(36\pi-\frac{4\pi}{3}=\frac{104\pi}{3}\).

A Common Mistake: Spanning the Gap

In the rectangular example, a slice with \(2<x<3\) has lowest material point \(y=0\) and highest material point \(y=4\). Using \(4-0=4\) as its shell height is wrong: the interval from \(y=1\) to \(y=3\) is empty. That incorrect height gives \(2\pi\int_1^4 4x\,dx=60\pi\), the volume of the entire outer rectangle after rotation. It counts the hole as if it were material.

The correction is to list the material intervals: \([0,1]\) and \([3,4]\). Their lengths add to \(1+1=2\), not \(4\). The correct volume is \(50\pi\), as confirmed by subtracting the hole’s \(10\pi\) contribution from the outer volume \(60\pi\).

A Reliable Setup

1
Locate the changing slices.
Mark the \(x\)-values where a hole begins or ends, or where its boundary meets another boundary.
2
List material intervals.
At each \(x\)-range, write every interval of \(y\)-values that belongs to the region. Do not include gaps.
3
Find the total height.
Subtract the endpoints of each material interval and add those lengths.
4
Integrate piecewise.
Use \(2\pi xH(x)\) for shells about the \(y\)-axis, splitting the integral wherever the height formula changes.
5
Check the setup.
When convenient, compute the outer region’s contribution and subtract the hole’s contribution. The result should agree with the piecewise shell integral.

This interval-based description is useful whenever a cross-section has an interior gap, whether the boundaries are straight or curved. The essential question is always the same: which parts of this slice contain material? The shell height is the total length of those parts and nothing more.

Check Your Understanding

For each question, identify the material intervals before writing a shell-height formula.

  1. A vertical slice meets a region in \([0,2]\) and \([5,7]\). What is its shell height?
  2. A rectangular region has vertical height \(6\), and a hole removes the interval from \(y=2\) to \(y=4\) for \(1<x<3\). What is the shell height on each side of, and inside, that \(x\)-interval?
  3. Why is the distance from the lowest to the highest point of a slice not necessarily its shell height?
  4. For shells about the \(y\)-axis, what factor multiplies the total shell height in the volume integrand?
  5. How can subtracting the hole’s shell contribution help check a piecewise setup?

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