Why Use Horizontal Shells?
When a region is revolved around the x-axis, horizontal strips are parallel to the axis of rotation. Rotating one such strip produces a thin cylindrical shell. Adding the volumes of these shells gives a way to calculate the solid’s volume without first describing circular cross-sections.
This approach is especially direct when the region is described using \(x=g(y)\). If the region lies between the y-axis and that curve, a horizontal strip at height \(y\) has length \(g(y)\). Its distance from the x-axis is \(y\), so the shell’s height and radius are read directly from the description of the region.
Radius, Height, and Thickness
Imagine drawing a horizontal strip through the region at a particular value of \(y\). The strip begins at \(x=0\) and ends at \(x=g(y)\), so its length is \(g(y)\). When rotated around the x-axis, that strip sweeps out a shell.
- Radius: \(y\), the distance from the strip to the x-axis.
- Height: \(g(y)\), the strip’s horizontal length.
- Thickness: \(dy\), a small change in the vertical coordinate.
A cylindrical shell with radius \(r\), height \(h\), and small thickness \(dy\) has approximate volume equal to its circumference times its height times its thickness: \(2\pi r h\,dy\). In this setting, \(r=y\) and \(h=g(y)\), so the small shell volume is approximately \(2\pi y g(y)\,dy\). Integrating across the y-bounds accumulates the shell volumes.
The bounds \(a\) and \(b\) must include the full vertical extent of the region. The stated formula assumes that the region is above or on the x-axis, so each radius is \(y\), not a distance involving a negative coordinate. It also assumes that the region extends horizontally from the y-axis to the curve and that \(g(y)\ge 0\) throughout the interval.
Why the Shell Integral Works
The formula follows by dividing the region into thin horizontal strips and summing the volumes of the shells they produce. The shell formula gives the volume of each thin layer in the limit as the strips become arbitrarily narrow.
Divide \([a,b]\) into subintervals with widths \(\Delta y_i\), and choose a sample height \(y_i\) in each subinterval.
At height \(y_i\), the shell has approximate radius \(y_i\), height \(g(y_i)\), and thickness \(\Delta y_i\). Its volume is approximately \(2\pi y_i g(y_i)\Delta y_i\).
The total is approximated by the sum \(\sum_i 2\pi y_i g(y_i)\Delta y_i\).
As the largest strip width tends to zero, continuity of \(g\) makes these sums converge to the definite integral \(2\pi\int_a^b y g(y)\,dy\).
In the limiting sum, the circumference factor is \(2\pi y\), and the strip length is \(g(y)\). Their product is the integrand. This also explains why the thickness is \(dy\): the partition is made along the vertical coordinate.
Setting Up the Integral
Before evaluating an integral, describe the region and identify the three pieces of shell information. For this setup, the region is bounded by \(x=0\), \(x=g(y)\), \(y=a\), and \(y=b\). A horizontal slice at any \(y\) between the bounds extends from \(0\) to \(g(y)\). Its radius is measured vertically to the axis, while its height is measured horizontally across the region.
Once those quantities are identified, insert them into the shell formula. The integrand is \(2\pi\) times radius times height. The variable of integration is \(y\) because the strips are horizontal and their thickness is vertical.
Worked Example: A Region Under \(x=\sqrt{y}\)
Consider the region \(0\le x\le\sqrt{y}\), \(0\le y\le4\), revolved about the x-axis. At height \(y\), the radius is \(y\), the shell height is \(\sqrt{y}\), and the thickness is \(dy\). Therefore,
An antiderivative of \(y^{3/2}\) is \(\frac{2}{5}y^{5/2}\), so
Check the integration: \(\frac{d}{dy}\left(\frac{2}{5}y^{5/2}\right)=\frac{2}{5}\cdot\frac{5}{2}y^{3/2}=y^{3/2}\), the integrand after removing the constant factor \(2\pi\). Thus the shell integral gives \(V=\frac{128\pi}{5}\) cubic units.
Further Worked Examples
Worked Example: A Region Under \(x=y^2\)
Let \(0\le x\le y^2\) for \(0\le y\le3\), and revolve the region about the x-axis. Each horizontal shell has radius \(y\) and height \(y^2\), so
Since an antiderivative of \(y^3\) is \(\frac{y^4}{4}\), evaluate at both bounds:
The check is \(\frac{d}{dy}(y^4/4)=y^3\), which matches the integrand inside the constant factor. The volume is \(\frac{81\pi}{2}\) cubic units.
Worked Example: A Decreasing Shell Height
Take the region \(0\le x\le4-y\), \(0\le y\le2\), and revolve it about the x-axis. The shell radius is \(y\), while its height is \(4-y\). Hence
An antiderivative of \(4y-y^2\) is \(2y^2-\frac{y^3}{3}\). Therefore,
Differentiate the antiderivative: \(\frac{d}{dy}(2y^2-y^3/3)=4y-y^2\), as required. Thus the volume is \(\frac{32\pi}{3}\) cubic units.
A Common Setup Error
The radius is the distance from the horizontal strip to the axis of rotation. It is not the strip’s horizontal length. The horizontal length is the shell height. Confusing these measurements can produce a different integral, even when the bounds and curve are correct.
Why This Method Matters
Choosing shells is useful when the region’s horizontal slices have simple lengths and the resulting integral is straightforward. Here, writing the boundary as \(x=g(y)\) makes those lengths easy to read, and the distance to the x-axis is already the integration variable \(y\). A clear geometric setup often prevents more errors than any later algebraic check.
Before integrating, verify that the strip is parallel to the axis of rotation, the radius is measured perpendicular to that axis, and the height is the strip’s length. Then check that the bounds cover the entire region and that the stated assumptions hold. In particular, the formula above uses a region above the x-axis; if a region crosses the axis, applying it without examining how the rotation overlaps can count parts of the solid incorrectly.
Check Your Understanding
For each question, identify the shell dimensions before evaluating any integral.
- For \(0\le x\le y+1\), \(0\le y\le2\), revolved about the x-axis, what are the shell radius and height?
- Set up, but do not evaluate, the shell integral for \(0\le x\le 3-y\), \(0\le y\le1\), revolved about the x-axis.
- For a region \(0\le x\le g(y)\) with \(0\le y\le b\), explain why the shell thickness is \(dy\), not \(dx\).
- In the region \(0\le x\le y^2\), \(0\le y\le3\), which quantity is the radius, and which is the shell height?
- State the assumptions on \(g\) and the y-bounds used in the formula \(V=2\pi\int_a^b y g(y)\,dy\).
