Why Slice Direction Matters
A base region in the plane can often be described using either vertical or horizontal slices. The direction affects how the boundary curves are expressed, the bounds of integration, and the formula for the length of each slice. A good choice usually makes those ingredients easier to state and check.
But convenience is not the only condition. The solid itself determines what its cross sections are. If a problem specifies square cross sections perpendicular to the \(x\)-axis, for instance, that specifies vertical slices. Replacing them with square cross sections perpendicular to the \(y\)-axis generally describes a different solid. You may choose between directions only when the problem permits either description or when you can correctly derive the cross-sectional areas for the same solid in the alternative direction.
Bounds, Slice Lengths, and Cross-Section Areas
A vertical slice is parallel to the \(y\)-axis. At a given \(x\), its length is the upper \(y\)-boundary minus the lower \(y\)-boundary. Its volume contribution is integrated with respect to \(x\), using \(dx\).
A horizontal slice is parallel to the \(x\)-axis. At a given \(y\), its length is the right \(x\)-boundary minus the left \(x\)-boundary. Its volume contribution is integrated with respect to \(y\), using \(dy\).
For square cross sections whose side is the slice length \(w\), the area is \(w^2\). Thus a vertical square-slice model has area \(A(x)=(\text{vertical slice length})^2\), while a horizontal square-slice model has area \(A(y)=(\text{horizontal slice length})^2\). Other cross-section shapes have their own area rules; the direction comparison still begins by finding the slice endpoints.
A Reliable Comparison Method
To compare the two directions, first describe the region with inequalities. For vertical slices, fix \(x\) and determine the lowest and highest \(y\)-values in the region. For horizontal slices, fix \(y\) and determine the leftmost and rightmost \(x\)-values. Solving the boundary equations is necessary, but also check which boundary is on which side and whether the interval changes across the region.
For vertical slices, fix \(x\); for horizontal slices, fix \(y\). Use the corresponding variable as the integration variable.
Read off upper and lower \(y\)-values for a vertical slice, or right and left \(x\)-values for a horizontal slice.
If the slice endpoints change formula or the region splits into separate intervals, divide the integral into pieces as needed.
Turn the slice length or other cross-sectional dimensions into an area, then integrate over the full range.
Prefer simpler bounds and area formulas when both choices describe the solid in the problem.
The comparison is not simply “which integral looks shorter?” A direction with fewer pieces may have more complicated functions, and a simple-looking integral is not valid if its bounds do not cover the region. The geometry and the solid’s cross-section rule are decisive.
Worked Example: A Region Between Two Curves
Worked Example: Comparing Square Slices Between Curves
Consider the base region bounded by \(y=x^2\) and \(y=2x\). Suppose we compare two candidate models: one with square cross sections perpendicular to the \(x\)-axis, and one with square cross sections perpendicular to the \(y\)-axis. Which direction gives the simpler area formula?
First find the intersections: \(x^2=2x\), so \(x(x-2)=0\), giving \(x=0\) and \(x=2\). For \(0\le x\le2\), \(2x\) is above \(x^2\), so a vertical slice has length \(2x-x^2\). The vertical square-slice model is therefore
Integrating term by term gives
Check the antiderivative: the derivative of \(\frac{4}{3}x^3-x^4+\frac{1}{5}x^5\) is \(4x^2-4x^3+x^4\), the integrand above.
For horizontal slices, the \(y\)-values range from \(0\) to \(4\). Solve each boundary equation for \(x\): \(y=x^2\) gives \(x=\sqrt y\), and \(y=2x\) gives \(x=y/2\). For \(0\le y\le4\), \(y/2\le\sqrt y\), so a horizontal slice runs from \(x=y/2\) to \(x=\sqrt y\). Its length is \(\sqrt y-y/2\), and the second candidate model gives
Thus
The derivative of the displayed antiderivative is \(y-y^{3/2}+y^2/4\), so the integration checks. The vertical width is polynomial, while the horizontal width involves a square root; for this square-section comparison, the vertical area formula is simpler. The two results differ because square slices in different directions define different solids, not because two computations of one fixed solid should disagree.
Worked Example: When Horizontal Bounds Simplify the Width
Worked Example: Choosing Horizontal Slices for a Curved Region
Let the base region be bounded by \(x=y^2\) and \(x=2y\), and compare square cross sections perpendicular to either axis. Find the bounds and see which direction gives the simpler width.
The curves meet where \(y^2=2y\), or \(y(y-2)=0\), so \(y=0\) and \(y=2\). For \(0\le y\le2\), the left boundary is \(x=y^2\) and the right boundary is \(x=2y\). A horizontal slice has length \(2y-y^2\). Its square-section volume is
Evaluate and check:
Differentiating the bracketed expression gives \(4y^2-4y^3+y^4\), as required.
For vertical slices, \(x\) ranges from \(0\) to \(4\). Solving for \(y\) gives \(y=\sqrt x\) from \(x=y^2\), and \(y=x/2\) from \(x=2y\). The slice runs from \(y=x/2\) to \(y=\sqrt x\), so its length is \(\sqrt x-x/2\). Consequently,
The derivative of the antiderivative is \(x-x^{3/2}+x^2/4=(\sqrt x-x/2)^2\), confirming the calculation. Here horizontal slices give a polynomial width directly, whereas vertical slices give a width involving a square root. If the solid’s description allows either square-slice model, the horizontal setup is algebraically simpler. As in the previous example, these different slice rules describe different solids.
Worked Example: Avoiding a Split Integral
Worked Example: A Region with a V-Shaped Boundary
Consider the triangular region \(-1\le x\le1\), \(|x|\le y\le1\). Compare square cross sections perpendicular to each axis. The absolute value makes the vertical lower boundary change formula at \(x=0\), while the horizontal endpoints have one formula.
For vertical slices, \(x\) runs from \(-1\) to \(1\), and the slice length is \(1-|x|\). The width has different expressions on either side of zero: it is \(1+x\) for \(-1\le x\le0\) and \(1-x\) for \(0\le x\le1\). Thus the model requires two pieces:
For the first piece, an antiderivative is \((1+x)^3/3\), which evaluates from \(-1\) to \(0\) to \(1/3\). For the second, an antiderivative is \(-(1-x)^3/3\), which evaluates from \(0\) to \(1\) to \(1/3\). Differentiating these antiderivatives gives the corresponding squared widths.
For horizontal slices, \(y\) ranges from \(0\) to \(1\). The inequality \(|x|\le y\) is equivalent to \(-y\le x\le y\), so each horizontal slice has length \(2y\). There is just one integral:
The derivative of \(4y^3/3\) is \(4y^2=(2y)^2\), confirming the integral. The horizontal bounds are simpler because one formula covers the full range. Again, the volume differs because the cross-sectional rule has been applied in a different direction.
A Pitfall: Solving for the Wrong Boundary
When converting a vertical description to horizontal bounds, it is not enough to reuse the same curve formulas with \(x\) and \(y\) exchanged. The equations must be solved for the coordinate that varies along the slice, and the resulting endpoints must match the geometry.
Common Mistake: Reusing the Original Formulas as Horizontal Bounds
In the region between \(y=x^2\) and \(y=2x\), a mistaken horizontal setup might claim that the \(x\)-endpoints are \(y^2\) and \(2y\). At \(y=1\), this gives \(x=1\) and \(x=2\). But the correct boundary equations give \(x=\sqrt y=1\) and \(x=y/2=1/2\), so the actual horizontal slice runs from \(1/2\) to \(1\).
The error is that \(y=x^2\) and \(y=2x\) express \(y\) in terms of \(x\); they are not already expressions for the horizontal slice's \(x\)-endpoints. Solve them: \(y=x^2\) gives \(x=\sqrt y\) on this region, and \(y=2x\) gives \(x=y/2\). Then check which is left and which is right before forming the width \(\sqrt y-y/2\).
What Makes a Direction the Better Choice?
A direction is often preferable when it produces a single interval of integration, simple boundary functions, and a slice length that leads to a manageable area expression. It may also be preferable if the other direction requires splitting the region into pieces. These are useful comparisons, not permission to alter a specified solid.
| Question | What to check |
|---|---|
| Are the bounds simple? | Can one pair of boundary formulas cover the entire integration interval? |
| Is the slice length manageable? | Does subtracting the endpoints give a clear expression? |
| Does the region split? | Would the chosen direction require multiple integrals or piecewise bounds? |
| Does this describe the required solid? | Does the cross-sectional rule match the direction and dimensions specified? |
| Can the setup be checked? | Do sample slices have the predicted endpoints, and does the area remain nonnegative? |
A brief derivation explains why the width checks matter. If a vertical slice at \(x\) extends from \(y=g(x)\) to \(y=f(x)\), with \(g(x)\le f(x)\), its length is \(f(x)-g(x)\). If its cross section is a square, its area must therefore be \((f(x)-g(x))^2\). The volume contribution over a small interval of width \(\Delta x\) is approximately that area times \(\Delta x\). Adding these contributions and taking the limit gives the slice integral \(\int (f(x)-g(x))^2\,dx\). For horizontal slices, the same reasoning uses right endpoint minus left endpoint and thickness \(\Delta y\). The integration variable and bounds follow from the direction of the slices.
This also explains two useful checks: a slice length must be nonnegative throughout its stated bounds, and an area must have the correct units of square length before integration gives cubic length. If either check fails, revisit the endpoint order, bounds, or area formula.
Summary
To choose a slice direction, describe the region using vertical and horizontal slices, solve for the corresponding endpoints, and compare the resulting bounds and widths. Then use the cross-sectional area rule to form the volume integral. Favor a simpler valid setup, but do not change the direction of a specified cross section merely to make the algebra easier: that can change the solid being modeled.
Check Your Understanding
Use the region and slice rules carefully; distinguish a simpler calculation from a different solid.
- For a vertical slice, which coordinate is fixed, and what is the order for subtracting the endpoints?
- What must be solved when converting curve boundaries into horizontal slice endpoints?
- Why might horizontal slices require one integral when vertical slices require two?
- If a problem specifies square cross sections perpendicular to the \(x\)-axis, may you replace them with square cross sections perpendicular to the \(y\)-axis just because the resulting integral looks simpler? Explain.
- What two checks can help reveal an incorrect slice setup before evaluating its integral?
