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Volumes by Cylindrical Shells · Tutorial 396 of 998

Diagnosing Radius, Height, and Differential Errors

Use a short, systematic check to identify and repair the most common radius, height, differential, and limit errors in cylindrical-shell integrals.

Intermediate 10 min read

What You'll Learn

  • Identify shell radius as the perpendicular distance from a slice to the axis of rotation.
  • Choose height by subtracting the correct pair of boundaries in the correct order.
  • Match vertical shells with dx and horizontal shells with dy.
  • Check that integration limits are expressed in the variable of integration.
  • Verify a corrected shell integral by differentiating its antiderivative.

Why Shell Setups Go Wrong

A cylindrical-shell integral can be incorrect even when its overall form looks familiar. A radius may be measured from the wrong line, a shell height may be negative because the boundaries were subtracted in the wrong order, or the differential and limits may refer to a different variable from the one in the integrand.

These errors are easier to catch when each part of the integral is checked against the geometry. The radius comes from distance to the axis of rotation. The height comes from the length of a slice parallel to that axis. The differential and limits describe the variable that locates those slices.

Diagnostic principle: Before evaluating a shell integral, identify the axis, describe one representative slice, and verify that its radius, height, differential, and bounds all refer to that same slice.

The Shell Formula and Its Parts

For a region rotated about a vertical line, vertical slices are parallel to the axis and generate cylindrical shells. If the slices are indexed by \(x\), their radius and height are functions of \(x\), and the differential is \(dx\). For a region rotated about a horizontal line, horizontal slices generate shells; their radius and height are functions of \(y\), and the differential is \(dy\).

Shell formula: When the slices generate non-overlapping shells that describe the solid, the volume is \(V=2\pi\int r h\,dx\) for vertical slices, or \(V=2\pi\int r h\,dy\) for horizontal slices. Here \(r\) is the distance from a slice to the axis, and \(h\) is the length of that slice across the region.

The radius is a distance, so it cannot be negative. For an axis \(x=a\) and a vertical slice at \(x\), the radius is \(|x-a|\). For an axis \(y=b\) and a horizontal slice at \(y\), it is \(|y-b|\). When the region lies entirely on one side of the axis, the absolute value can be written as a simple positive difference. If a region crosses the axis, additional care is needed to avoid counting parts of the solid twice.

The height is also a length. A vertical slice runs from a lower \(y\)-boundary to an upper \(y\)-boundary, so its height is upper minus lower. A horizontal slice runs from a left \(x\)-boundary to a right \(x\)-boundary, so its height is right minus left. Each difference should be nonnegative throughout the interval of integration.

Why Radius Times Height Appears

A thin slice parallel to the axis sweeps out a thin cylindrical shell. If its radius is approximately \(r\), its circumference is approximately \(2\pi r\). If the slice has length \(h\), the shell’s curved surface area is approximately \(2\pi r h\). Giving the slice a small thickness \(\Delta x\) or \(\Delta y\) gives an approximate shell volume of \(2\pi r h\) times that thickness.

Adding the thin shell volumes across the region gives a Riemann sum. In the limit as the thicknesses approach zero, that sum becomes the integral in the shell formula. This explains both the factor \(2\pi\) and why the differential must match the coordinate used to locate the slices. The formula applies when the chosen slices really do generate the solid without unintended overlap, and the radius and height correctly describe each slice over the stated bounds.

A Four-Part Setup Check

1
Name the axis.
Record its equation and whether it is vertical or horizontal. The radius is a perpendicular distance to this line, not automatically the coordinate itself.
2
Match the slice orientation.
For shells, slices are parallel to the axis: vertical slices use \(dx\), and horizontal slices use \(dy\).
3
Build the height as a length.
Use upper minus lower for a vertical slice, or right minus left for a horizontal slice. Check that the result is nonnegative.
4
Match the limits to the differential.
Bounds for \(dx\) must be \(x\)-values; bounds for \(dy\) must be \(y\)-values. The limits should cover the entire region’s range in that variable.

Worked Example: Radius and Boundary Order for Vertical Shells

Worked Example: Radius and Boundary Order for Vertical Shells

Consider the region between \(y=0\) and \(y=4-x^2\), for \(0\leq x\leq 2\). Rotate it about the vertical line \(x=-1\). Find the shell integral and its value.

Step 1: Choose the slice and differential. A vertical axis calls for vertical slices to form shells, so the differential is \(dx\), and the bounds are \(0\leq x\leq 2\).

Step 2: Find the radius. The distance from \(x\) to the axis \(x=-1\) is \(x-(-1)=x+1\). Since \(0\leq x\leq2\), this is positive throughout.

Step 3: Find the height. The upper curve is \(4-x^2\) and the lower curve is \(0\), so \(h=(4-x^2)-0=4-x^2\). This is nonnegative on the interval.

The shell integral is therefore

$$V=2\pi\int_0^2 (x+1)(4-x^2)\,dx.$$

Expand the integrand:

$$ (x+1)(4-x^2)=4x+4-x^3-x^2=-x^3-x^2+4x+4.$$

An antiderivative is \(-\frac{x^4}{4}-\frac{x^3}{3}+2x^2+4x\). Evaluating at the bounds gives

$$\begin{aligned} V&=2\pi\left[-\frac{x^4}{4}-\frac{x^3}{3}+2x^2+4x\right]_0^2\\ &=2\pi\left(-4-\frac{8}{3}+8+8\right)\\ &=2\pi\left(12-\frac{8}{3}\right) =2\pi\cdot\frac{28}{3} =\frac{56\pi}{3}. \end{aligned}$$

Check. Differentiating the antiderivative gives \(-x^3-x^2+4x+4\), which is the expanded shell integrand. At the endpoints, the antiderivative is \(28/3\) at \(x=2\) and \(0\) at \(x=0\), confirming the evaluated integral. The resulting volume is positive.

Worked Example: Horizontal Shell Height and dy

Worked Example: Horizontal Shell Height and dy

Let a region be described by \(0\leq y\leq2\) and \(y\leq x\leq3\). Rotate it about the horizontal line \(y=-1\). Set up and evaluate the shell integral.

Step 1: Match slices to the axis. The axis is horizontal, so horizontal slices form shells. Their variable is \(y\), so use \(dy\) and the limits \(0\leq y\leq2\).

Step 2: Find the radius. A horizontal slice at \(y\) is \(y-(-1)=y+1\) units from the axis.

Step 3: Find the height. The slice runs from the left boundary \(x=y\) to the right boundary \(x=3\). Thus \(h=3-y\), which is nonnegative for \(0\leq y\leq2\).

The integral is

$$V=2\pi\int_0^2 (y+1)(3-y)\,dy.$$

First expand:

$$ (y+1)(3-y)=3y-y^2+3-y=3+2y-y^2.$$

An antiderivative is \(3y+y^2-\frac{y^3}{3}\). Therefore,

$$\begin{aligned} V&=2\pi\left[3y+y^2-\frac{y^3}{3}\right]_0^2\\ &=2\pi\left(6+4-\frac{8}{3}\right)\\ &=2\pi\cdot\frac{22}{3} =\frac{44\pi}{3}. \end{aligned}$$

Check. The derivative of \(3y+y^2-\frac{y^3}{3}\) is \(3+2y-y^2\), exactly the expanded integrand. Its endpoint difference is \(22/3-0=22/3\), so the volume is \(\frac{44\pi}{3}\), a positive result.

Worked Example: Repairing a Differential and Limits

Worked Example: Repairing a Differential and Limits

Consider the region \(0\leq y\leq2\), \(0\leq x\leq3-y\), rotated about the \(x\)-axis, \(y=0\). A proposed setup is \(2\pi\int_0^3 y(3-y)\,dx\). Diagnose and correct it, then evaluate the volume.

The proposed setup mixes variables. Its limits \(0\) and \(3\) suggest an \(x\)-integral, but the integrand contains \(y\), the coordinate that locates the horizontal slices. Moreover, the horizontal slices are parallel to the \(x\)-axis, so they—not vertical slices—generate shells here. The differential must be \(dy\).

For a horizontal slice at height \(y\), the radius is its distance to \(y=0\), namely \(y\). The slice extends from \(x=0\) to \(x=3-y\), so its height is \(3-y\). The region’s stated vertical range supplies the correct limits \(0\leq y\leq2\). Thus the repaired setup is

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

Expand the integrand and integrate:

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

Check. Differentiating \(\frac{3y^2}{2}-\frac{y^3}{3}\) gives \(3y-y^2\), the expanded integrand. The antiderivative is \(10/3\) at \(y=2\) and \(0\) at \(y=0\), confirming the calculation and the positive volume.

Common Mistakes and Their Corrections

Using a coordinate instead of a distance. If the axis is \(x=-1\), writing the radius as \(x\) ignores the shift. For example, at \(x=1\), the distance to the axis is \(1-(-1)=2\), not \(1\). Compute the perpendicular distance from a representative slice to the actual axis.

Subtracting boundaries in the wrong order. In the first example, writing the height as \(0-(4-x^2)\) gives \(x^2-4\), which is negative on \(0\leq x<2\). The resulting shell integral would be negative, which cannot be a volume. Correct it by taking upper minus lower: \((4-x^2)-0\).

Pairing a differential with the wrong bounds. A \(dy\) integral must use \(y\)-bounds, and a \(dx\) integral must use \(x\)-bounds. When changing which coordinate describes the slices, re-express both the boundaries and the limits in that coordinate; do not simply change \(dx\) to \(dy\) while leaving the rest unchanged.

Assuming the radius is always \(x\) or \(y\). That shortcut works only for an axis at the corresponding zero coordinate and a region on one side of it. For a shifted axis, subtract its coordinate; in all cases, interpret the result as a nonnegative distance.

Why These Checks Matter

A shell integral combines geometric information with calculus. An integral can be evaluated perfectly and still give the wrong volume if its radius, height, or bounds do not describe the intended shells. Checking those pieces before integrating is therefore not just a way to catch algebra errors; it verifies that the model represents the solid.

A useful final scan is to read the integral aloud: “twice pi times distance to the axis times slice length, integrated across the range of those slices.” If the differential or limits do not fit that description, repair the setup before calculating. If the radius or height becomes negative, revisit the distance or the boundary order.

Takeaway: A valid shell setup keeps four choices consistent: the axis determines the radius, the slice determines the height, the slice orientation determines \(dx\) or \(dy\), and that variable determines the limits.

Check Your Understanding

Use the geometry of the slices to answer each question before evaluating any integral.

  1. For shells about \(x=2\), what expression gives the radius of a vertical slice at \(x\), when the region lies entirely to the left of the axis?
  2. A horizontal slice runs from \(x=1\) to \(x=5-y\). Which boundary should be subtracted from which to obtain its height?
  3. What differential and type of bounds belong in an integral over horizontal shells?
  4. Why is \(2\pi\int_0^3 y(3-y)\,dx\) inconsistent when \(y\) indexes the horizontal shells?
  5. If a proposed shell height is negative throughout an interval, which geometric subtraction should you check first?

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