Why the Shifted Axis Changes the Radius
In the shell method, a horizontal strip rotates around a horizontal axis to form a cylindrical shell. The shell's thickness comes from the strip's small vertical thickness, and the shell's height is the horizontal length of that strip. The radius is the distance from the strip to the axis of rotation.
When the axis is the \(x\)-axis, its vertical coordinate is \(0\). A strip at height \(y\) is therefore a distance \(y\) from the axis, provided the region lies above it. If the axis is shifted to \(y=b\), the distance instead depends on the difference between \(y\) and \(b\).
For this tutorial, suppose the region lies entirely on one side of the axis, with vertical bounds \(\alpha\le y\le\beta\). Let \(L(y)\) denote the horizontal length of the strip through the region at height \(y\). The shell height is \(L(y)\). The task here is to use the axis location and the bounds to express the radius correctly; the region's horizontal length is taken as known.
Choosing Bounds and Radius
If the axis lies below the whole region, then \(b\le\alpha\). Every strip has \(y\ge b\), so its distance from the axis is \(y-b\). If the axis lies above the whole region, then \(b\ge\beta\). Every strip has \(y\le b\), so its distance is \(b-y\). In either case, integrating over the region's vertical bounds keeps the radius nonnegative.
The shell-volume formula is circumference times shell height times thickness, accumulated over the strips. With thickness \(dy\), the circumference is \(2\pi r(y)\), and the shell height is \(L(y)\). Thus the volume is the integral of \(2\pi r(y)L(y)\) over the vertical bounds.
The bounds \(\alpha\) and \(\beta\) describe the vertical extent of the region, not the location of the axis. The axis determines which distance expression applies. Keeping these roles separate is a reliable way to avoid a negative radius or an integral with the wrong sign.
Why the Formula Uses a Distance
Consider a thin horizontal strip at height \(y\) with horizontal length \(L(y)\) and vertical thickness \(\Delta y\). When it rotates, it sweeps out a thin cylindrical shell. The circle traced by the strip is centered on the axis \(y=b\), so its radius is the vertical distance between \(y\) and \(b\), namely \(|y-b|\). Its circumference is \(2\pi|y-b|\).
For a sufficiently thin strip, the shell's volume is approximately its circumference multiplied by its height and thickness. This gives \(2\pi|y-b|L(y)\Delta y\). Adding these contributions over a partition of the region approximates the total volume. As the largest strip thickness tends to zero, the sums become the definite integral of \(2\pi|y-b|L(y)\). If the region lies on one side of the axis, the absolute value simplifies to one of the two expressions above throughout the interval.
Worked Examples
Worked Example: A Rectangle Rotated About an Axis Below It
A rectangle has horizontal width \(3\) units and extends vertically from \(y=1\) to \(y=4\). Find the volume when it is rotated about \(y=-1\).
The axis is below the region because \(-1\le1\). At height \(y\), the radius is \(y-(-1)=y+1\). Each horizontal strip has length \(L(y)=3\), and the bounds are \(1\le y\le4\). Therefore,
An antiderivative of \(y+1\) is \(\frac{y^2}{2}+y\). Evaluating at both bounds gives
The volume is \(63\pi\) cubic units. To check the integration, differentiate the antiderivative: \(\frac{d}{dy}\left(\frac{y^2}{2}+y\right)=y+1\). The endpoint values are \(12\) at \(y=4\) and \(\frac32\) at \(y=1\), giving the stated difference of \(\frac{21}{2}\). The radius ranges from \(2\) to \(5\) units, so it is positive throughout.
Worked Example: A Rectangle Rotated About an Axis Above It
Use the same rectangle, with horizontal width \(3\) units and vertical bounds \(1\le y\le4\). Find the volume when it is rotated about \(y=5\).
Now the axis is above the region because \(5\ge4\). The distance from a strip at \(y\) to the axis is \(5-y\), not \(y-5\). Thus
An antiderivative of \(5-y\) is \(5y-\frac{y^2}{2}\). Evaluating gives
The volume is \(45\pi\) cubic units. The derivative of the antiderivative is \(5-y\), as required. Its endpoint values are \(12\) and \(\frac92\), whose difference is \(\frac{15}{2}\). The radius decreases from \(4\) units at \(y=1\) to \(1\) unit at \(y=4\), so the radii are nonnegative throughout the interval.
Worked Example: A Changing Strip Length and a Shifted Axis
Suppose a region extends from \(y=0\) to \(y=2\), and its horizontal strip length is known to be \(L(y)=y+1\) units. Find the volume when it is rotated about \(y=-2\).
The axis is below the region, so the radius is \(y-(-2)=y+2\). The bounds are \(0\le y\le2\). Using the given strip length,
An antiderivative of the integrand is \(\frac{y^3}{3}+\frac{3y^2}{2}+2y\). Therefore,
The volume is \(\frac{76\pi}{3}\) cubic units. Differentiating the antiderivative gives \(y^2+3y+2=(y+2)(y+1)\), matching the integrand. The antiderivative is \(\frac{38}{3}\) at \(y=2\) and \(0\) at \(y=0\). The radius runs from \(2\) to \(4\) units, confirming that the chosen expression stays positive over the bounds.
A Common Sign Mistake
For an axis above the region, it is easy to subtract in the wrong order. In the rectangle example with \(1\le y\le4\) and axis \(y=5\), the expression \(y-5\) is negative on the entire interval. Using it as a radius would produce this incorrect calculation:
A negative volume cannot describe the solid. The error is not in the bounds; it is in treating a signed difference as a distance. Since the axis is above every strip, the distance is \(5-y\). Using that radius gives \(45\pi\), as the second worked example verifies.
Using the Setup Reliably
Before evaluating a horizontal-shell integral about \(y=b\), check three features of the setup: the variable of integration is \(y\), the bounds cover the region's vertical extent, and the radius is the distance \(|y-b|\). If the region lies below the axis, write \(b-y\); if it lies above, write \(y-b\). The remaining factor \(L(y)\) is the horizontal length of the strip.
The one-sided condition matters. The direct formulas \(y-b\) or \(b-y\) apply across the whole interval only when the region is entirely on that side of the axis. If the axis falls within the region's vertical range, a single choice of subtraction order will not describe the distance at every height. Do not force one of the one-sided formulas onto the full interval; that situation requires additional care in selecting the setup.
Check Your Understanding
Use the distance from each strip to the axis, and check that the radius is nonnegative over the stated bounds.
- A region lies between \(y=2\) and \(y=6\). Is the radius \(y-1\) or \(1-y\) for rotation about \(y=1\)?
- A region lies between \(y=-3\) and \(y=0\). Write the radius for rotation about \(y=2\).
- A region has horizontal strip length \(4\) and bounds \(0\le y\le2\). Set up, but do not evaluate, the shell integral for rotation about \(y=-1\).
- Why is \(y-5\) not a valid radius for a strip with \(1\le y\le4\) rotated about \(y=5\)?
- What condition on \(b\) and the bounds ensures that one subtraction order gives a nonnegative radius throughout the interval?
