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A Shifted-Axis Shell Computation Workshop

Use the shell method when the axis is parallel to the slices, measuring each radius from the shifted axis.

Intermediate 10 min read

What You'll Learn

  • Identify whether vertical or horizontal slices are parallel to a shifted axis of rotation.
  • Measure shell radii as distances from the axis, not from a coordinate axis.
  • Determine shell heights from the region’s upper and lower or right and left boundaries.
  • Set up shell integrals with the appropriate variable and bounds.
  • Evaluate shifted-axis shell integrals and check the dimensions and geometry of the result.

Why a Shifted Axis Changes the Shell Calculation

A cylindrical shell is formed by rotating a thin strip that is parallel to the axis of rotation. Its volume depends on three quantities: the shell’s radius, its height, and the strip’s thickness. When the axis is shifted away from a coordinate axis, the height and thickness may be familiar, but the radius must be measured from the new axis.

For a vertical axis \(x=a\), use vertical strips, whose thickness is \(dx\). The radius is the horizontal distance from \(x\) to \(a\). For a horizontal axis \(y=b\), use horizontal strips, whose thickness is \(dy\), and measure the vertical distance from \(y\) to \(b\). The region’s coordinate boundaries determine the shell height; the shifted axis determines the shell radius.

Definition: If the axis is \(x=a\) and the region lies entirely on one side of it, a vertical strip at \(x\) has shell radius \(r(x)=|x-a|\). Its shell height is the vertical length of the strip within the region. If the axis is \(y=b\), a horizontal strip at \(y\) has radius \(r(y)=|y-b|\) and height equal to the horizontal length of that strip within the region.

The absolute value expresses the geometric meaning of radius: a distance cannot be negative. When the region lies entirely to the right of \(x=a\), the radius is \(x-a\); when it lies entirely to the left, it is \(a-x\). Similarly, above \(y=b\) the radius is \(y-b\), and below it the radius is \(b-y\).

Why the Shell Formula Works

Consider a vertical strip of height \(h\) and width \(\Delta x\), with its nearer edge at radius \(r\) from a vertical axis. Rotating the strip makes a thin cylindrical layer. Its exact volume is the difference between two cylinders, both with height \(h\): the outer cylinder has radius \(r+\Delta x\), and the inner cylinder has radius \(r\). Thus

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

The leading term is circumference times height times thickness: \(2\pi r h\,\Delta x\). As a region is divided into thinner strips, the extra term involving \((\Delta x)^2\) contributes no volume in the limit. Adding the leading terms and taking that limit gives the shell integral. The same reasoning applies to horizontal strips around a horizontal axis.

Theorem: For a region with a single continuous shell height on the stated interval, rotated about a parallel axis that does not cross the region, the shell-method volume is $$ V=2\pi\int r\,h\,d(\text{slice variable}). $$ For a vertical axis and vertical strips, this is \(V=2\pi\int r(x)h(x)\,dx\). For a horizontal axis and horizontal strips, it is \(V=2\pi\int r(y)h(y)\,dy\). If the boundary changes, split the integral where the shell height changes.

This formula is a limit of the thin-shell sums, not a claim that every finite strip makes a shell with exactly constant radius. The factor \(2\pi r\) is the circumference at the strip’s representative radius, and the integral accounts for the changing radius and height across the region.

Worked Example: A Vertical Axis to the Left

Worked Example: A Vertical Axis to the Left

Rotate the triangular region bounded by \(y=0\), \(y=x\), and \(x=0\) to \(x=2\) about the vertical line \(x=-1\). Find the volume using shells.

First sketch the region in the coordinate plane. It lies above \(y=0\), below \(y=x\), and between \(x=0\) and \(x=2\). Draw the axis \(x=-1\), one unit left of the vertical boundary \(x=0\). A vertical slice at \(x\) rotates around that axis. Its radius is the horizontal distance \(x-(-1)=x+1\), and its height is the distance from the lower curve to the upper curve, \(x-0=x\).

The bounds are \(0\le x\le 2\), so

$$ \begin{aligned} V &=2\pi\int_0^2 (x+1)x\,dx\\ &=2\pi\int_0^2 (x^2+x)\,dx\\ &=2\pi\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^2\\ &=2\pi\left(\frac{8}{3}+2-0\right)\\ &=2\pi\cdot\frac{14}{3} =\frac{28\pi}{3}. \end{aligned} $$

Check the antiderivative: \(\frac{d}{dx}\left(\frac{x^3}{3}+\frac{x^2}{2}\right)=x^2+x\), the integrand. A coordinate-sketch check also confirms the factors: at \(x=0\), the radius is \(1\) and the height is \(0\); at \(x=2\), the radius is \(3\) and the height is \(2\). The shell radii are measured from \(x=-1\), not from the \(y\)-axis. The volume is \(\frac{28\pi}{3}\) cubic units.

Worked Example: A Vertical Axis to the Right

Worked Example: A Vertical Axis to the Right

Rotate the region bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=2\) about \(x=3\). Find the volume with shells.

On a sketch, the whole region lies left of \(x=3\). A vertical slice at \(x\) therefore has radius \(3-x\), not \(x-3\). Its height is \(x\), since the slice runs from \(y=0\) to \(y=x\), and the bounds remain \(0\le x\le2\). Hence

$$ \begin{aligned} V &=2\pi\int_0^2 (3-x)x\,dx\\ &=2\pi\int_0^2 (3x-x^2)\,dx\\ &=2\pi\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^2\\ &=2\pi\left(6-\frac{8}{3}-0\right)\\ &=2\pi\cdot\frac{10}{3} =\frac{20\pi}{3}. \end{aligned} $$

Differentiate the antiderivative: \(\frac{d}{dx}\left(\frac{3x^2}{2}-\frac{x^3}{3}\right)=3x-x^2\), as required. The sketch gives an independent check on the radius: at the left edge \(x=0\), the distance to \(x=3\) is \(3\); at the right edge \(x=2\), it is \(1\). The shell heights are \(0\) and \(2\) at those same edges. The volume is \(\frac{20\pi}{3}\) cubic units.

Worked Example: A Horizontal Axis Below the Region

Worked Example: A Horizontal Axis Below the Region

Consider the region described by \(0\le y\le2\) and \(0\le x\le y^2\). Rotate it about the horizontal line \(y=-1\). Find the volume using shells.

Sketch the boundary as horizontal slices: at each \(y\) from \(0\) to \(2\), the slice starts at \(x=0\) and ends at \(x=y^2\). Its height, measured horizontally, is therefore \(y^2-0=y^2\). The region is above \(y=-1\), so the shell radius is \(y-(-1)=y+1\). The bounds are \(0\le y\le2\).

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

Check by differentiation: \(\frac{d}{dy}\left(\frac{y^4}{4}+\frac{y^3}{3}\right)=y^3+y^2\). On the sketch, the radius at \(y=0\) is \(1\) and the height is \(0\); at \(y=2\), the radius is \(3\) and the height is \(4\). These are vertical distances to the horizontal axis and horizontal lengths across the region, respectively. The volume is \(\frac{40\pi}{3}\) cubic units.

A Reliable Setup and Sketch Check

Before integrating, identify the axis direction and write down what a representative slice measures. The slice must be parallel to the axis for the shell method. For a vertical axis, use vertical slices and integrate with respect to \(x\); for a horizontal axis, use horizontal slices and integrate with respect to \(y\). Then find the radius as a distance to the shifted axis and the height as the length of the slice inside the region.

1
Locate the axis and region.
Sketch the boundary curves, the shifted line of rotation, and the interval or range over which the region extends.
2
Choose a parallel slice.
Use a vertical strip for a vertical axis, or a horizontal strip for a horizontal axis. Its thickness determines the integration variable.
3
Measure radius and height separately.
Radius is perpendicular distance to the axis. Height is the length of the strip within the region, found from its bounding curves.
4
Build and check the integral.
Use \(2\pi\int rh\,d(\text{slice variable})\). Check that the bounds cover the region and that test slices on the sketch give plausible radii and heights.

If the axis crosses the region, simply applying the one-sided formulas across the entire interval can count the same cylindrical layers more than once. That situation requires separate care. Here, the examples keep the axis outside the region, so each shell radius is measured consistently from the same side.

A Common Radius Error

In the first example, a tempting but incorrect setup is to use \(x\) as the radius because the strip is located at coordinate \(x\). That would give

$$ V_{\mathrm{wrong}} =2\pi\int_0^2 x\cdot x\,dx =2\pi\left[\frac{x^3}{3}\right]_0^2 =\frac{16\pi}{3}. $$

This calculation treats the axis as \(x=0\). But the actual axis is \(x=-1\), so every vertical slice is one unit farther from the axis than its \(x\)-coordinate alone indicates. At \(x=0\), for example, the strip is one unit from \(x=-1\), not zero units away. The correction is \(r=x+1\), which gives \(\frac{28\pi}{3}\), as calculated above.

Key check: Pick a visible slice on the sketch and measure its distance to the rotation axis. If the radius formula does not give that distance, correct the formula before evaluating the integral.

Check Your Understanding

For each question, identify the slice direction and reason from the stated axis before writing or interpreting a shell integral.

  1. A region lies between \(x=1\) and \(x=4\) and is rotated about \(x=6\). For a vertical slice at \(x\), what is its shell radius?
  2. A region lies below \(y=5\) and above \(y=1\), and is rotated about \(y=0\). For a horizontal slice at \(y\), what is its shell radius?
  3. For rotation about \(x=-2\), what variable should a shell integral use, and which direction should its slices run?
  4. A horizontal slice at \(y\) runs from \(x=L(y)\) to \(x=R(y)\). What is its shell height about a horizontal axis?
  5. Why can using the coordinate \(x\) itself as a shell radius be wrong when the axis is \(x=a\) rather than \(x=0\)?

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