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Shells and Integral Applications · Tutorial 695 of 999

Shells About Shifted Lines of Rotation

Use the distance from each slice to a shifted axis to build a correct cylindrical-shell integral.

Advanced 10 min read

What You'll Learn

  • Find vertical-shell radii as distances from a line \(x=a\).
  • Find horizontal-shell radii as distances from a line \(y=b\).
  • Match vertical slices with vertical axes and horizontal slices with horizontal axes.
  • Set up and evaluate shell integrals for regions on one side of a shifted axis.
  • Recognize why integrating absolute distances across both sides of an axis can double-count volume.

Why a Shifted Axis Changes the Radius

The shell method builds a solid from thin slices parallel to the axis of rotation. When the axis is shifted away from the coordinate axis, the shell's radius is no longer simply \(x\) or \(y\). It is the distance from the slice to the actual line of rotation.

For a vertical axis \(x=a\), use vertical slices. Each slice rotates into a cylindrical shell whose radius is the horizontal distance from \(x\) to \(a\). For a horizontal axis \(y=b\), use horizontal slices, and the radius is the vertical distance from \(y\) to \(b\).

Key idea: A shell radius is a distance to the axis of rotation. A shell height is the length of the slice being rotated. For a direct shell integral over the full slice interval, the region must lie on one side of the axis; otherwise, shells from opposite sides may occupy the same radii and be counted more than once.

Shell Formulas for Shifted Lines

Suppose a region lies between \(x=p\) and \(x=q\), on one side of the vertical line \(x=a\). A vertical slice at \(x\) has height \(h(x)\) and thickness \(dx\). Its shell radius is the distance \(|x-a|\). The shell method gives:

$$ V=2\pi\int_p^q |x-a|\,h(x)\,dx. $$

This form applies directly when every vertical slice in the interval lies on the same side of \(x=a\). On the side \(x<a\), the radius is \(a-x\); on the side \(x>a\), it is \(x-a\). In either case, the radius is positive or zero.

Similarly, suppose a region lies between \(y=r\) and \(y=s\), on one side of the horizontal line \(y=b\). A horizontal slice at \(y\) has length \(w(y)\) and thickness \(dy\). Its shell radius is \(|y-b|\), so:

$$ V=2\pi\int_r^s |y-b|\,w(y)\,dy. $$

These formulas match slice direction to the axis: vertical slices for a vertical axis, and horizontal slices for a horizontal axis. The slice direction is parallel to the axis, so its rotation creates a shell. The radius is measured perpendicular to the axis.

Why the Shell Formula Works

A thin slice parallel to the axis, when rotated, forms a thin cylindrical shell. If its radius is \(R\), height is \(H\), and thickness is \(\Delta\), its volume is approximately circumference times height times thickness:

$$ \Delta V\approx 2\pi R H\Delta. $$

For a vertical axis, the slice thickness is \(\Delta x\), the radius is the distance from \(x\) to \(a\), and the height is \(h(x)\). Adding the volumes of many thin shells gives a sum that approaches an integral as the thicknesses shrink. This gives the vertical-shell formula. For a horizontal axis, the same reasoning uses thickness \(\Delta y\), radius \(|y-b|\), and slice length \(w(y)\).

Derivation: Partition the interval \([p,q]\) into small subintervals of widths \(\Delta x_i\), and choose a sample point \(x_i^*\) in each one. The \(i\)th shell has radius \(|x_i^*-a|\) and height approximately \(h(x_i^*)\), so its volume is approximately \(2\pi|x_i^*-a|h(x_i^*)\Delta x_i\). Adding the shells gives a Riemann sum. In the limit, provided the slice-height function is integrable, the sum becomes \(2\pi\int_p^q |x-a|h(x)\,dx\). The horizontal formula follows by partitioning the \(y\)-interval and using slice length \(w(y)\).

Worked Examples

Worked Example: A Rectangle Rotated About a Shifted Vertical Line

The rectangle is \(0\le x\le2\), \(0\le y\le3\). Rotate it about \(x=4\). Find the volume using shells.

The axis is to the right of the entire rectangle, so vertical slices produce shells counted once. The shell height is \(3-0=3\), and the radius at \(x\) is \(4-x\). The bounds are \(x=0\) and \(x=2\):

$$ \begin{aligned} V &=2\pi\int_0^2 (4-x)(3)\,dx\\ &=6\pi\int_0^2 (4-x)\,dx\\ &=6\pi\left[4x-\frac{x^2}{2}\right]_0^2\\ &=6\pi\left((8-2)-(0-0)\right)\\ &=36\pi. \end{aligned} $$

The antiderivative checks because \(\frac{d}{dx}(4x-x^2/2)=4-x\). As a geometric check, the rotated rectangle forms a hollow cylinder of height \(3\), outer radius \(4\), and inner radius \(2\). Its volume is \(\pi(4^2-2^2)(3)=\pi(16-4)(3)=36\pi\), agreeing with the shell integral.

Worked Example: A Region Under a Curve About a Shifted Vertical Line

The region is bounded by \(y=x^2\), \(y=0\), \(x=0\), and \(x=2\). Rotate it about \(x=-1\). Set up and evaluate the shell integral.

The region lies to the right of \(x=-1\). A vertical slice at \(x\) has height \(x^2-0=x^2\), and its radius is \(x-(-1)=x+1\). Thus:

$$ \begin{aligned} V &=2\pi\int_0^2 (x+1)x^2\,dx\\ &=2\pi\int_0^2 (x^3+x^2)\,dx\\ &=2\pi\left[\frac{x^4}{4}+\frac{x^3}{3}\right]_0^2\\ &=2\pi\left(\frac{16}{4}+\frac{8}{3}-0\right)\\ &=2\pi\left(4+\frac{8}{3}\right)\\ &=2\pi\left(\frac{20}{3}\right) =\frac{40\pi}{3}. \end{aligned} $$

The antiderivative checks because \(\frac{d}{dx}(x^4/4+x^3/3)=x^3+x^2\). The radius factor is \(x+1\), not \(x\), because the axis is one unit to the left of the \(y\)-axis.

Worked Example: A Rectangle Rotated About a Shifted Horizontal Line

The rectangle is \(0\le x\le2\), \(1\le y\le3\). Rotate it about \(y=-1\). Find the volume using horizontal shells.

The axis lies below the entire rectangle. A horizontal slice has length \(2-0=2\), and its radius is \(y-(-1)=y+1\). The bounds are \(y=1\) and \(y=3\):

$$ \begin{aligned} V &=2\pi\int_1^3 (y+1)(2)\,dy\\ &=4\pi\int_1^3 (y+1)\,dy\\ &=4\pi\left[\frac{y^2}{2}+y\right]_1^3\\ &=4\pi\left(\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}+1\right)\right)\\ &=4\pi(6)\\ &=24\pi. \end{aligned} $$

The antiderivative checks because \(\frac{d}{dy}(y^2/2+y)=y+1\). Independently, the solid is a hollow cylinder with length \(2\), outer radius \(4\), and inner radius \(2\). Its volume is \(\pi(4^2-2^2)(2)=24\pi\).

Worked Example: Horizontal Shells with a Variable Slice Length

The region is bounded by \(x=0\), \(x=y\), \(y=1\), and \(y=3\). Rotate it about \(y=-1\). Set up and evaluate the shell integral.

The region lies above the axis. At a given \(y\), the horizontal slice runs from \(x=0\) to \(x=y\), so its length is \(w(y)=y\). Its shell radius is \(y+1\). Therefore:

$$ \begin{aligned} V &=2\pi\int_1^3 (y+1)y\,dy\\ &=2\pi\int_1^3 (y^2+y)\,dy\\ &=2\pi\left[\frac{y^3}{3}+\frac{y^2}{2}\right]_1^3\\ &=2\pi\left(\left(9+\frac{9}{2}\right)-\left(\frac{1}{3}+\frac{1}{2}\right)\right)\\ &=2\pi\left(\frac{81}{6}-\frac{5}{6}\right)\\ &=2\pi\left(\frac{38}{3}\right) =\frac{76\pi}{3}. \end{aligned} $$

The antiderivative checks because \(\frac{d}{dy}(y^3/3+y^2/2)=y^2+y\). The changing slice length \(y\) belongs in the height factor; the shifted-axis distance \(y+1\) belongs in the radius factor.

A Crucial Caution: Do Not Count the Same Radius Twice

The distance expressions \(|x-a|\) and \(|y-b|\) correctly measure radii, but that does not mean they can always be integrated over a region that crosses the axis. Slices on opposite sides of an axis can rotate into the same shells. Summing both sets of slice volumes would count some of the solid more than once.

For example, take the rectangle \(-1\le x\le1\), \(0\le y\le1\), and rotate it about \(x=0\). The solid is a cylinder of radius \(1\) and height \(1\), so its volume is \(\pi\). The tempting calculation

$$ 2\pi\int_{-1}^{1}|x|\,dx =2\pi\left(\int_{-1}^{0}(-x)\,dx+\int_0^1x\,dx\right) =2\pi\left(\frac{1}{2}+\frac{1}{2}\right) =2\pi $$

is wrong: it counts the shells generated by the left half and the right half separately, even though those halves produce the same radii. Since this particular region is symmetric, using just the right half counts every shell once:

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

For a region that crosses an axis, do not apply the one-side formula over all slices without checking whether shells repeat. The calculation must be split or reformulated so each resulting shell is counted once. The absolute value supplies a nonnegative distance; it does not prevent double-counting.

A Reliable Setup Routine

1
Identify the axis.
Record its equation as \(x=a\) or \(y=b\); the constant determines the distance measurement.
2
Choose slices parallel to the axis.
Use vertical slices for \(x=a\), and horizontal slices for \(y=b\).
3
Find the radius and slice length separately.
The radius is a distance to the axis; the height or width is the length of the slice in the region.
4
Check which side contains the region.
If all slices lie on one side, the direct formula applies. If the region crosses the axis, check for repeated radii before integrating.
5
Set bounds in the slice variable.
Use \(dx\) with vertical slices and \(dy\) with horizontal slices, then evaluate the integral.

Check Your Understanding

For each setup, identify the shell radius and slice length before evaluating any integral.

  1. A region lies between \(x=1\) and \(x=4\), with vertical slice height \(h(x)\), and is rotated about \(x=6\). What is the shell radius at \(x\), and what differential belongs in the integral?
  2. A region lies between \(y=2\) and \(y=5\), with horizontal slice length \(w(y)\), and is rotated about \(y=-1\). State the shell radius and the integral setup.
  3. Why does the direct integral \(2\pi\int_p^q |x-a|h(x)\,dx\) require a check that the region lies on one side of \(x=a\)?
  4. For the rectangle \(0\le x\le3\), \(0\le y\le2\), rotated about \(x=5\), write the shell integral without evaluating it.
  5. What geometric error occurs if slices from both sides of an axis generate the same radii but are all included in a shell integral?

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