Vertical Shells About a Shifted Axis

Why a Shifted Axis Changes the Radius

When vertical strips are rotated around the \(y\)-axis, a strip at \(x\) has radius \(x\), provided it lies to the right of that axis. If the axis is instead the vertical line \(x=a\), the strip’s radius is not generally \(x\). It is the horizontal distance from the strip to the line.

Distance is never negative. A strip to the left of the axis and a strip to the right of the axis both generate shells with positive radii. To set up the integral correctly, identify which side of \(x=a\) contains the region and choose bounds accordingly.

The Shell Setup About \(x=a\)

Suppose a region lies between \(x=u\) and \(x=v\), with \(u<v\), and is entirely on one side of the axis \(x=a\). Let \(h(x)\) denote the vertical height of the region at \(x\). For example, if its top and bottom boundaries are \(y=f(x)\) and \(y=g(x)\), then \(h(x)=f(x)-g(x)\) wherever \(f(x)\geq g(x)\).

A vertical strip of thickness \(dx\) rotates into a thin cylindrical shell. Its circumference is \(2\pi r(x)\), its height is \(h(x)\), and its thickness is \(dx\). Thus its approximate volume is circumference times height times thickness. Adding these contributions across the interval gives the shell integral.

The expressions \(x-a\) and \(a-x\) are not interchangeable: the correct one depends on the region’s position relative to the axis. In either case, the radius is nonnegative throughout the chosen interval.

Why the Integral Gives the Volume

The shell formula follows by adding the volumes of increasingly thin shells. The argument requires that the region’s height be integrable on the interval and that the interval lie on one side of the axis, so each vertical strip corresponds to one shell radius without reversing direction.

More explicitly, partition \([u,v]\) into subintervals of width \(\Delta x_i\), and choose a sample point \(x_i^*\) in each subinterval. The corresponding shell-volume sum is

As the largest subinterval width tends to zero, this Riemann sum tends to the integral of the continuous (or, more generally, integrable) shell-volume density. Therefore the volume is

This explains why the shifted-axis formula uses distance rather than a signed coordinate difference. A negative value of \(x-a\) is not a negative physical radius; it only indicates that the strip is left of the axis.

Worked Examples

A Common Radius Mistake

A frequent error is to use \(x-a\) as the radius no matter which side of the axis contains the region. For the second example, this would give \(x-4\), which is negative throughout \(1\leq x\leq 3\). The incorrect calculation would be

A volume cannot be negative. The error is using a signed difference instead of a distance. Since the region is left of \(x=4\), its radius is \(4-x=|x-4|\), not \(x-4\). Correcting the radius changes the answer to \(\frac{44\pi}{3}\) cubic units, as computed above.

Choosing Bounds and Knowing the Setup’s Scope

For this one-sided setup, the axis must not lie between the interval’s endpoints. If the region occupies \(u\leq x\leq v\), then \(a\leq u\) places it to the right of the axis, while \(a\geq v\) places it to the left. These checks tell you which radius expression to use before evaluating any integral.

If an axis passes through a region, simply inserting an absolute value into a single integral over the entire \(x\)-interval can count shells at the same radius from opposite sides separately. That situation needs a different setup; the one-sided formula here is intended for bounds that keep the region on one side of the axis.

The method is useful because a shifted axis changes the radius but does not change the basic shell structure: each contribution is still circumference times height times thickness. Careful choices of bounds make the distance interpretation explicit and keep every shell radius physically meaningful.