Why Vertical Slices Work About the Y-Axis
When a region is rotated about the y-axis, a vertical slice travels around that axis. Instead of forming a solid disk, the slice sweeps out a thin cylindrical shell. Adding the volumes of these shells gives the volume of the solid.
This method is especially direct when the region is described by its top and bottom boundaries as functions of \(x\). The slice width, and therefore the integration variable, is \(dx\). For a region to the right of the y-axis, the shell radius is the slice’s distance from that axis.
The formula assumes that the region is to the right of the axis and that the stated top and bottom boundaries describe its vertical slices across the interval. If the upper or lower boundary changes, the setup must account for that change rather than use one height formula across the whole interval.
Where the Shell Formula Comes From
Consider a very narrow vertical strip at a position \(x\), with width \(\Delta x\) and height \(h(x)=T(x)-B(x)\). When it rotates around the y-axis, its distance from the axis is about \(x\). The strip sweeps out a thin hollow cylinder: its circumference is about \(2\pi x\), its height is \(h(x)\), and its thickness is \(\Delta x\).
Thus its volume is approximately circumference times height times thickness. The approximation becomes exact in the limit as the strips get arbitrarily narrow. The following derivation makes that limiting step precise.
Divide \([a,b]\) into narrow subintervals of widths \(\Delta x_i\), and choose a sample point \(x_i\) in each one. The corresponding strip has approximate height \(h(x_i)\).
A strip at radius \(x_i\) has approximate volume \(2\pi x_i h(x_i)\Delta x_i\), using circumference times height times thickness.
The sum of these approximations approaches the Riemann integral as the largest strip width approaches zero. Therefore the solid’s volume is \(2\pi\int_a^b xh(x)\,dx\).
For an exact check on the thin-shell approximation, if a strip has constant height \(h\), inner radius \(r\), and width \(\Delta x\), its rotated volume is \(\pi\bigl((r+\Delta x)^2-r^2\bigr)h\). Expanding gives \(2\pi rh\Delta x+\pi h(\Delta x)^2\). The second term is an error of order \((\Delta x)^2\), which vanishes in the limit of increasingly fine partitions. The remaining sum is precisely the Riemann sum for \(2\pi\int_a^b xh(x)\,dx\).
Setting Up a Vertical-Shell Integral
Before integrating, identify three geometric quantities: the radius, the height, and the thickness. For rotation about the y-axis, a vertical strip at \(x\ge0\) has radius \(x\). Its height is the vertical distance from the bottom of the region to the top. The thickness is \(dx\).
- Find the x-interval. These endpoints describe the leftmost and rightmost positions of the region, not its lowest and highest y-values.
- Find the vertical slice’s endpoints. Write the top and bottom boundaries as functions of \(x\).
- Calculate the shell height. Subtract bottom from top: \(h(x)=T(x)-B(x)\).
- Use the shell volume formula. Multiply \(2\pi\), radius, height, and \(dx\), then integrate over the x-interval.
- Evaluate and check. Confirm that the integrand represents a nonnegative volume and, when practical, verify the result using another volume setup.
The bounds are often the source of confusion. Since vertical shells are indexed by \(x\), the integral is with respect to \(x\), and the limits must be x-coordinates. A y-coordinate such as the region’s maximum height is not automatically an integration limit for this method.
Worked Examples
Worked Example: The Region Between \(y=x\) and \(y=x^2\)
Find the volume formed by rotating the region between \(y=x\) and \(y=x^2\), for \(0\le x\le1\), about the y-axis.
On this interval, \(x\ge x^2\), because \(x-x^2=x(1-x)\ge0\). Thus the top curve is \(T(x)=x\), the bottom curve is \(B(x)=x^2\), and the shell height is \(x-x^2\). The radius is \(x\), and the bounds are 0 and 1.
As a check, a horizontal slice of the rotated solid gives a washer with outer radius \(\sqrt y\) and inner radius \(y\), for \(0\le y\le1\). Its volume is \(\pi\int_0^1(y-y^2)\,dy=\pi\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1=\pi\left(\frac12-\frac13\right)=\frac{\pi}{6}\), agreeing with the shell calculation.
Worked Example: The Region Under a Downward-Opening Parabola
Rotate the region bounded by \(y=4-x^2\), the x-axis, and \(0\le x\le2\) about the y-axis.
The region is above the x-axis because \(4-x^2\ge0\) on \([0,2]\). Its top boundary is \(4-x^2\), its bottom boundary is 0, and its shell height is \(4-x^2\). The shell radius is \(x\).
To check, horizontal slices give disks. At height \(y\), the curve satisfies \(x=\sqrt{4-y}\), so the disk radius is \(\sqrt{4-y}\), for \(0\le y\le4\). The washer calculation is \(\pi\int_0^4(\sqrt{4-y})^2\,dy=\pi\int_0^4(4-y)\,dy=\pi[4y-y^2/2]_0^4=\pi(16-8)=8\pi\). This matches the shell result.
Worked Example: A Triangular Region Under a Line
Find the volume obtained by rotating the region under \(y=3-x\), above the x-axis, for \(0\le x\le3\), about the y-axis.
On \([0,3]\), the line is nonnegative. The top boundary is \(3-x\), the bottom boundary is 0, and the shell height is \(3-x\). The radius is \(x\), so the integral is taken with respect to \(x\) from 0 to 3.
For a check, horizontal slices produce disks whose radius at height \(y\) is \(3-y\), for \(0\le y\le3\). Their volume is \(\pi\int_0^3(3-y)^2\,dy=\pi\left[-\frac{(3-y)^3}{3}\right]_0^3=\pi(0-(-9))=9\pi\), agreeing with the shell integral.
A Common Mistake: Reversing the Shell Height
Shell height is a vertical distance, so it must be nonnegative. It is found by subtracting the bottom boundary from the top boundary, not the other way around. In the region between \(y=x\) and \(y=x^2\) on \([0,1]\), \(x\) is the top curve and \(x^2\) is the bottom curve.
A reversed subtraction gives the incorrect calculation
A negative volume signals that the setup is wrong. The correction is to use top minus bottom, \(x-x^2\), giving \(2\pi\int_0^1x(x-x^2)\,dx=\frac{\pi}{6}\). Checking the order of the curves at a typical point in the interval is a reliable way to catch this error before integrating.
What to Remember
For a region to the right of the y-axis, vertical slices create shells whose radius is \(x\), whose height is the top function minus the bottom function, and whose thickness is \(dx\). The volume is the integral of \(2\pi\) times radius times height across the region’s x-interval.
Check Your Understanding
For each question, identify the shell geometry before evaluating any integral.
- For the region between \(y=2x\) and \(y=x^2\) on \(0\le x\le2\), which curve is on top, and what is the shell height?
- What are the radius, height, and thickness for a vertical strip at \(x\) when a region to the right of the y-axis is rotated about that axis?
- Set up, but do not evaluate, the shell integral for the region under \(y=5-x\) and above the x-axis on \(0\le x\le5\), rotated about the y-axis.
- A shell setup produces a negative integral. What geometric quantity should you check first, and how should it be defined?
- Why are the limits in a vertical-shell integral about the y-axis x-coordinates?
