Why One Vertical Strip Can Give More Than One Shell Height
A region does not have to be vertically connected. At a particular value of \(x\), a vertical strip may pass through the region in two or more separate intervals. When the region is revolved about a vertical axis, each interval contributes part of the shell at that radius.
The key is to account for all the intervals without treating the empty gaps between them as part of the region. Find the height of each interval, add those heights, and use that total in the shell integral. If the components are present over different ranges of \(x\), include each only over its own range.
Describe Each Vertical Component
Suppose a region lies between vertical lines \(x=a\) and \(x=b\). At each \(x\), describe its vertical intersection as a collection of separate intervals. For instance, if the strip intersects the region in
then the two component heights are \(v_1(x)-u_1(x)\) and \(v_2(x)-u_2(x)\). The total height is their sum. The intervals must represent the actual pieces of the region: the gap between them is not included.
For rotation about the \(y\)-axis, a vertical strip at \(x\ge 0\) sweeps out a shell of radius \(x\), circumference \(2\pi x\), and thickness \(dx\). Its height is the total length of the strip inside the region. Thus, for two components present throughout \([a,b]\),
For rotation about the shifted vertical line \(x=c\), if the region lies to the right of the axis, replace the radius \(x\) by \(x-c\). The component heights are unchanged. If the component intervals exist over different \(x\)-ranges, integrate each component over its own range, or split the integral at the points where the list of active components changes.
Why the Component Heights Add
The shell formula uses the height of the material swept out by a vertical strip. When that strip meets the region in separate intervals, the lengths of those intervals add to give the strip’s total length inside the region. The spaces between intervals contribute no material.
At a chosen \(x\), list the disjoint vertical intervals belonging to the region.
For each interval, subtract its lower \(y\)-boundary from its upper \(y\)-boundary.
Add the interval lengths to get the total shell height, then multiply by the shell circumference and thickness.
This also explains why the component contributions may be integrated separately and then added. For a thin strip of width \(dx\), each interval sweeps out a thin shell section with the same radius and thickness. The sections occupy separate vertical ranges, so their volumes add. Taking the limit of these sums gives the integral of the sum of their heights.
Worked Examples
Worked Example: Two Components Revolved About the \(y\)-Axis
Let \(R\) consist of the points with \(0\le x\le 1\) in either of these intervals: \(0\le y\le x\), or \(2\le y\le 3-x\). Find the volume when \(R\) is revolved about the \(y\)-axis.
At each \(x\in[0,1]\), the lower component has height
The upper component has height
The intervals are separated: the first ends at \(y=x\le1\), while the second starts at \(y=2\). Their total height is \(x+(1-x)=1\). The shell radius is \(x\), so
An antiderivative of \(x\) is \(x^2/2\), since \(\frac{d}{dx}(x^2/2)=x\). Therefore,
Check by calculating each component separately. The lower component contributes \(2\pi\int_0^1 x\cdot x\,dx=2\pi(1/3)=2\pi/3\). The upper contributes \(2\pi\int_0^1 x(1-x)\,dx=2\pi(1/2-1/3)=\pi/3\). Their sum is \(2\pi/3+\pi/3=\pi\), agreeing with the total-height calculation.
Worked Example: Two Components About a Shifted Vertical Axis
For \(0\le x\le2\), let a region consist of the intervals \(0\le y\le x+1\) and \(x+3\le y\le2x+5\). Find the volume when it is revolved about \(x=-1\).
The first component has height \((x+1)-0=x+1\). The second has height \((2x+5)-(x+3)=x+2\). They are disjoint because the first ends at \(x+1\), which is below the second’s lower boundary \(x+3\). The total height is \(2x+3\).
Because the region is to the right of \(x=-1\), the shell radius is \(x-(-1)=x+1\). Hence
Multiplying out gives \((x+1)(2x+3)=2x^2+5x+3\). An antiderivative is \(\frac{2}{3}x^3+\frac{5}{2}x^2+3x\), as differentiation gives \(2x^2+5x+3\). Thus
Check the result by adding the two component volumes. The first is \(2\pi\int_0^2(x+1)^2\,dx=2\pi(8/3+4+2)=52\pi/3\). The second is \(2\pi\int_0^2(x+1)(x+2)\,dx=2\pi(8/3+6+4)=76\pi/3\). Their sum is \(128\pi/3\), as before.
Worked Example: Components Present Over Different \(x\)-Ranges
Consider three components: \(0\le x\le1,\ 0\le y\le x\); \(0\le x\le2,\ 2\le y\le3\); and \(1\le x\le2,\ 4\le y\le x+3\). Find the volume obtained by revolving their union about the \(y\)-axis.
The first component has height \(x\) on \([0,1]\). The second has height \(1\) on \([0,2]\). The third has height \((x+3)-4=x-1\) on \([1,2]\). The intervals do not overlap vertically: the first lies at or below \(y=1\), the second lies from \(2\) to \(3\), and the third starts at \(4\).
On \(0\le x\le1\), the active components have total height \(x+1\). On \(1\le x\le2\), the active components have total height \(1+(x-1)=x\). The shell radius is \(x\), so
For the first integral, \(x(x+1)=x^2+x\), with antiderivative \(x^3/3+x^2/2\). Its value is \(1/3+1/2=5/6\). For the second, an antiderivative of \(x^2\) is \(x^3/3\), so its value is \(8/3-1/3=7/3\). Therefore,
Check by integrating each component over its own range. Their respective volumes are \(2\pi\int_0^1x^2\,dx=2\pi/3\), \(2\pi\int_0^2x\,dx=4\pi\), and \(2\pi\int_1^2x(x-1)\,dx\). For the last integral, \(x(x-1)=x^2-x\), with antiderivative \(x^3/3-x^2/2\); the endpoint difference is \((8/3-2)-(1/3-1/2)=2/3+1/6=5/6\). The third volume is therefore \(5\pi/3\). Adding gives \(2\pi/3+4\pi+5\pi/3=19\pi/3\), which verifies the result.
A Common Mistake: Counting the Gap as Part of the Height
In the first example, the vertical intervals are \([0,x]\) and \([2,3-x]\). A tempting but incorrect calculation is to use the distance from the bottom of the lowest interval to the top of the highest interval as the shell height:
That expression includes the empty gap between \(y=x\) and \(y=2\). It is not the length of the region inside the strip. The correct total height is the sum of the two interval lengths:
The incorrect setup would give \(2\pi\int_0^1x(3-x)\,dx=7\pi/3\), not the correct volume \(\pi\). The correction is to list the separate intervals before calculating any height. Only if the vertical intersection is one continuous interval can its total height be found by subtracting the lowest boundary from the highest.
A Reliable Setup
When several vertical components are present, organize the geometry before integrating. For each component, record its \(x\)-range, its lower boundary, and its upper boundary. Then calculate its height and include it only where that component exists. This prevents both missing contributions and accidentally filling gaps.
- Confirm that the intervals at a fixed \(x\) are disjoint components of the region.
- Compute each height as upper boundary minus lower boundary.
- Add the heights of all components active at that \(x\).
- Use the distance from the vertical axis of rotation for the shell radius.
- Split the \(x\)-range when the active components or their boundaries change.
- Check the setup by summing the component integrals separately when practical.
The same principle applies whether the component boundaries are lines, curves, or constants. The geometry determines the height; the axis determines the radius. Keeping those roles separate makes the shell integral easier to construct and audit.
Check Your Understanding
For each question, identify the separate vertical lengths before writing a shell integral.
- A strip intersects a region in \([0,2]\) and \([5,7-x]\). What is its total height, and for which \(x\)-values are both intervals nonempty?
- A region to the right of \(x=1\) has two vertical components of heights \(x\) and \(3-x\). What radius and total height belong in the shell integrand for rotation about \(x=1\)?
- Why is the distance from the lowest point in a strip to its highest point not generally the correct total height?
- If one component exists for \(0\le x\le1\) and another for \(1\le x\le2\), how should their contributions be represented in a shell calculation?
- For a fixed \(x\), explain why the shell contributions of disjoint vertical intervals can be added.
