Why Shell Height Matters
A cylindrical-shell calculation depends on two geometric measurements: the distance from a slice to the axis of rotation, and the length of that slice within the region. For a vertical slice, that length is a vertical distance. When the region lies between two graphs, its height is found by subtracting the lower graph from the upper one.
This subtraction is not just a detail in the setup. Reversing the order can make the height negative, while using only one boundary function can omit part of the region. The goal is to represent the actual vertical segment of the region at each position \(x\).
From Two Curves to a Shell Integral
Consider rotating such a region about the vertical line \(x=a\). A vertical slice has thickness \(dx\). When revolved, it sweeps out a thin cylindrical shell. Its radius is the horizontal distance from \(x\) to the axis, namely \(|x-a|\), and its height is \(f(x)-g(x)\).
The lateral surface area of a cylinder is circumference times height. Thus a shell of radius \(r\), height \(h\), and thickness \(dx\) has approximate volume \(2\pi rh\,dx\). When the axis lies entirely to one side of the region, the vertical slices sweep out non-overlapping shells, so adding their volumes gives the integral.
In many problems the axis is the \(y\)-axis, so \(a=0\) and the radius is \(x\) when the region is to its right. The bounds \(p\) and \(q\) describe the horizontal extent of the region. If the two graphs intersect at the endpoints, those intersection \(x\)-coordinates often provide the bounds. Before using one height formula over the entire interval, check that the same graph remains above the other throughout it.
Why the Formula Works
The shell formula follows by approximating the region with narrow vertical strips and adding the volumes of the resulting shells. The argument below makes the role of the height difference explicit.
At a position \(x_i\), a vertical strip of width \(\Delta x\) runs from \(g(x_i)\) to \(f(x_i)\). Its length is \(f(x_i)-g(x_i)\).
The strip sweeps out a shell with radius \(|x_i-a|\), height \(f(x_i)-g(x_i)\), and thickness approximately \(\Delta x\). Its volume is approximately \(2\pi |x_i-a|\bigl(f(x_i)-g(x_i)\bigr)\Delta x\).
Summing these approximations over the interval gives a Riemann sum for the continuous function \(2\pi |x-a|(f(x)-g(x))\). As the strip widths approach zero, the sum approaches its definite integral, which is the stated volume formula.
The argument requires the strips to represent the region once and their revolutions not to overlap in a way that counts volume repeatedly. The condition that the axis lies entirely to one side of the region ensures this for the setup used here.
Worked Examples
Worked Example: A Parabola Above the \(x\)-Axis
Find the volume formed by rotating the region bounded by \(y=4-x^2\), \(y=0\), and \(x=0\) about the \(y\)-axis.
The upper curve is \(4-x^2\), and the lower curve is \(0\). The curves \(y=4-x^2\) and \(y=0\) meet when \(4-x^2=0\), so \(x^2=4\). On the part of the region to the right of the \(y\)-axis, the interval is \(0\le x\le 2\). The shell radius is \(x\), and the height is \((4-x^2)-0=4-x^2\).
An antiderivative of \(4x-x^3\) is \(2x^2-\frac{x^4}{4}\), since
Therefore,
The integral is nonnegative on \([0,2]\), as expected for shell volumes. The result is \(8\pi\) cubic units.
Worked Example: Two Intersecting Curves and a Shifted Axis
The curves \(y=2x\) and \(y=x^2\) intersect at \(x=0\) and \(x=2\), since \(2x=x^2\) gives \(x(x-2)=0\). On \(0\le x\le 2\), the line is above the parabola: \(2x-x^2=x(2-x)\ge 0\). Rotate the enclosed region about \(x=-1\).
The whole region lies to the right of the axis. Thus the shell radius is \(x-(-1)=x+1\), and the height is upper minus lower, \(2x-x^2\).
Expanding the product gives
An antiderivative is \(-\frac{x^4}{4}+\frac{x^3}{3}+x^2\), verified by differentiation:
Evaluate at the bounds:
The height is nonnegative throughout the interval, and the computed volume is positive. The result is \(\frac{16\pi}{3}\) cubic units.
Worked Example: Checking Which Curve Is Above
Rotate the region between \(y=1+x\) and \(y=1+x^2\), from their intersections, about \(x=-2\).
The curves intersect where \(1+x=1+x^2\), or \(x=x^2\). Hence \(x(x-1)=0\), giving the bounds \(0\) and \(1\). For \(0\le x\le1\), \(x\ge x^2\), so \(1+x\) is the upper curve and \(1+x^2\) is the lower curve. The shell height is \((1+x)-(1+x^2)=x-x^2\). Since the region is right of \(x=-2\), the radius is \(x+2\).
An antiderivative is \(x^2-\frac{x^3}{3}-\frac{x^4}{4}\), because its derivative is \(2x-x^2-x^3\). Thus
The height \(x-x^2\) is nonnegative on \([0,1]\), confirming the chosen order of subtraction. The volume is \(\frac{5\pi}{6}\) cubic units.
A Common Mistake: Subtracting in the Wrong Order
Suppose the region between \(y=2x\) and \(y=x^2\) on \([0,2]\) is rotated about \(x=-1\). A common error is to write the height as \(x^2-2x\), perhaps because the curves were listed in that order. But on the interval, \(x^2-2x=x(x-2)\le0\). That is a signed difference in the wrong direction, not a physical shell height.
A negative volume signals that the setup is wrong. Compare the curves on the interval: \(2x-x^2=x(2-x)\ge0\), so \(2x\) is above \(x^2\). The corrected height is \(2x-x^2\), giving the positive volume \(\frac{16\pi}{3}\) cubic units, as calculated in the preceding example.
A Reliable Setup
For a region between two graphs, organize the setup before evaluating any integral. Confirm that the interval describes the region, identify which graph is upper, and only then form the shell height. This keeps the geometric meaning of every factor visible.
- Write the horizontal interval \([p,q]\) for the region under consideration.
- Compare the boundary functions on that interval to identify the upper and lower curves.
- Subtract lower from upper to obtain the shell height.
- Measure the horizontal distance from \(x\) to the vertical axis of rotation to obtain the radius.
- Use \(2\pi\int_p^q (\text{radius})(\text{height})\,dx\), then check the sign and the units.
If the upper and lower curves exchange roles inside the interval, a single expression based on one ordering will not describe the height everywhere. Likewise, if the region is not entirely on one side of the axis, the simple setup may count parts of the solid more than once. Those cases require additional care beyond the one-interval situation considered here.
Why the Height Difference Is Useful
Writing the height as a difference of functions converts the shape of a two-boundary region into a single expression that can be integrated. The upper curve contributes the top of each vertical slice; the lower curve removes the part beneath the region. That difference is exactly the length of the slice, not an approximation to some other geometric quantity.
This perspective also makes the setup easier to audit. A shell integral has a radius, a height, and a thickness. The radius describes where the slice is relative to the axis; the height describes how far the region extends vertically; and \(dx\) records the slice's width. Keeping those roles distinct helps catch errors before they affect the calculation.
Check Your Understanding
Use the definitions and setup principles in this tutorial to answer the following questions.
- On \(0\le x\le 2\), which curve is above: \(y=2x\) or \(y=x^2\)? What is the corresponding shell height?
- A region lies between \(y=f(x)\) above and \(y=g(x)\) below, and is rotated about \(x=3\). What are its shell height and shell radius at \(x\)?
- Why does the shell height use upper minus lower rather than the difference in either arbitrary order?
- For a region on \(0\le x\le1\) between \(y=1+x\) and \(y=1+x^2\), identify the upper function and state the height.
- What geometric issue can arise if the region extends across the axis of rotation?
