Why Slice Direction Determines the Variable
When a solid is divided into thin slices, each slice has a position and a small thickness. The position tells us which coordinate labels the slice; that coordinate also determines the integration variable. Keeping these two ideas together is essential when setting up a volume integral.
In a base region drawn in the \(xy\)-plane, a vertical slice is parallel to the \(y\)-axis. Its position is recorded by a value of \(x\), so neighboring vertical slices are separated by a small change \(dx\). A horizontal slice is parallel to the \(x\)-axis; its position is recorded by a value of \(y\), and neighboring horizontal slices are separated by \(dy\).
Position, Direction, and Thickness
It helps to separate three questions that are easy to confuse:
- Which way does the slice extend in the base region? Vertical slices extend in the \(y\)-direction; horizontal slices extend in the \(x\)-direction.
- Which coordinate changes from one slice to the next? For vertical slices, \(x\) changes. For horizontal slices, \(y\) changes.
- What is the slice thickness? It is \(dx\) for vertical slices and \(dy\) for horizontal slices.
For a solid whose cross-sectional area at position \(x\) is \(A(x)\), the slicing principle gives a volume integral of the form \(\int A(x)\,dx\) when the slices are perpendicular to the \(x\)-axis. If the area at position \(y\) is \(B(y)\), slices perpendicular to the \(y\)-axis give \(\int B(y)\,dy\). The variable, area function, and bounds must all refer to the same coordinate.
Here \(A(x)\) and \(B(y)\) denote cross-sectional areas. They are functions of the coordinate that identifies the slice, not of the direction in which the slice extends within the base region.
Why the Variable Matches the Perpendicular Axis
The connection follows from how a definite integral adds thin slices. Suppose the solid is sliced by planes \(x=c\), where \(c\) varies across an interval \([a,b]\). Each plane has a cross-sectional area \(A(c)\). A thin slab near \(x=c\) has thickness approximately \(\Delta x\), so its volume is approximately \(A(c)\Delta x\). Adding these approximate slab volumes gives a Riemann sum.
As the slabs become thinner, the sum approaches the definite integral of the cross-sectional area over the range of \(x\). The thicknesses are changes in \(x\), which is why the integral uses \(dx\). For planes \(y=c\), the corresponding slabs have thickness \(\Delta y\), so the sum uses \(B(y_i^*)\Delta y\) and the integral uses \(dy\).
This reasoning assumes that the cross-sectional area is defined across the interval and is sufficiently well behaved for the definite integral to exist. The important point for choosing a variable is that the small thickness is measured along the axis perpendicular to the slices.
A Reliable Choice-of-Variable Procedure
Determine whether each slice is vertical or horizontal in the base region, or note the axis that the slicing planes are specified to be perpendicular to.
Vertical slices are positioned by \(x\); horizontal slices are positioned by \(y\).
Use an area function of the indexing coordinate and bounds describing that coordinate’s range.
The orientation usually determines the variable immediately. It does not, by itself, guarantee that one simple area formula works across the whole range. If the cross-sectional description changes, the volume may need to be split into multiple integrals. Choosing \(dx\) or \(dy\) is the first setup decision, not a substitute for checking the full region.
Worked Examples
Worked Example: Slices Perpendicular to the \(x\)-Axis
A solid has cross-sectional area \(A(x)=1+x\) at position \(x\), for \(0\le x\le 2\). The cross sections are perpendicular to the \(x\)-axis. Set up and evaluate the volume integral.
The planes \(x=c\) are perpendicular to the \(x\)-axis, so the slices are indexed by \(x\). Their thickness is \(dx\), and the stated range of positions is from \(x=0\) to \(x=2\). Therefore,
An antiderivative of \(1+x\) is \(x+\frac{x^2}{2}\). Evaluating at the bounds gives
The setup uses \(dx\) because the slices are perpendicular to the \(x\)-axis. As a check on the evaluation, differentiating \(x+\frac{x^2}{2}\) gives \(1+x\), the integrand. The volume is \(4\) cubic units if the lengths in the problem are measured in units.
Worked Example: Slices Perpendicular to the \(y\)-Axis
A solid has cross-sectional area \(B(y)=2+y\) at position \(y\), for \(0\le y\le 3\). The slices are perpendicular to the \(y\)-axis. Choose the variable and evaluate the volume.
The position of each slice is given by \(y\), so consecutive slices are separated by \(dy\). The matching integral is
An antiderivative is \(2y+\frac{y^2}{2}\). Thus,
The variable and area function both use \(y\), and the bounds \(0\) and \(3\) describe the stated \(y\)-range. Differentiating \(2y+\frac{y^2}{2}\) gives \(2+y\), confirming the evaluation. The volume is \(\frac{21}{2}\) cubic units when the length units are specified.
Worked Example: Reading the Bounds from the Slice Coordinate
The base region is a triangle with vertices \((0,0)\), \((2,0)\), and \((0,2)\). A solid over this base has cross-sectional area \(C(y)=3-y\) for horizontal slices. Choose the integration variable and bounds, then evaluate the resulting integral.
Horizontal slices are parallel to the \(x\)-axis and perpendicular to the \(y\)-axis. Therefore the integral uses \(dy\). The lowest and highest \(y\)-coordinates of the triangular base are \(0\) and \(2\), so those are the bounds. The supplied area function is already expressed in terms of \(y\):
An antiderivative of \(3-y\) is \(3y-\frac{y^2}{2}\). Therefore,
The bounds come from the range of \(y\), not from the horizontal extent of the triangle. Differentiating \(3y-\frac{y^2}{2}\) gives \(3-y\), which checks the evaluation. The volume is \(4\) cubic units if the base coordinates use units of length.
Common Mistake: Using the Slice’s Direction as the Variable
A frequent error is to use the coordinate in which the slice extends rather than the coordinate that identifies its position. For example, a horizontal slice extends in the \(x\)-direction, but its position changes as \(y\) changes. It is therefore indexed by \(y\), not \(x\).
In the preceding example, a wrong setup would be to treat the horizontal slices as though they were indexed by \(x\), and write \(\int_0^2(3-y)\,dx\). This is inconsistent: the integrand is given as a function of \(y\), while the differential is \(dx\). The stated horizontal-slice area and the \(y\)-range instead give
Do not decide the variable merely by asking which way the slice points across the page. Ask which coordinate changes from one slice to the next. That coordinate determines the differential and must also be used for the area function and bounds.
When the Choice Affects the Setup
A problem may allow more than one useful slicing direction, but the resulting setups need not be equally simple. One direction may produce a single area formula over one interval; another may require the region or area description to be split into pieces. The choice of variable is still determined by the chosen slice orientation:
| Slice orientation in the base | Perpendicular axis | Variable and thickness |
|---|---|---|
| Vertical | \(x\)-axis | \(x\) and \(dx\) |
| Horizontal | \(y\)-axis | \(y\) and \(dy\) |
Before integrating, check the setup as a whole: the slices have the stated orientation, the area function uses the coordinate that indexes them, and the bounds cover the range of that coordinate. This simple consistency check catches many errors before any integration is done.
Check Your Understanding
For each question, focus on the slice position and the coordinate that changes between neighboring slices.
- If cross-sectional planes are perpendicular to the \(x\)-axis, which differential should appear in the volume integral?
- Horizontal slices are parallel to which coordinate axis, and which coordinate indexes their positions?
- A cross-sectional area is given as \(D(y)\), and slices are horizontal for \(1\le y\le 4\). What is the matching integral form?
- Why is \(\int A(x)\,dx\) associated with slices perpendicular to the \(x\)-axis?
- What should be checked if the cross-sectional area changes its formula partway through the range?
