Choosing a Slice Direction

Why Slice Direction Matters

A lamina occupies a two-dimensional region, but its mass and first moments can often be calculated as one-dimensional iterated integrals. A vertical strip fixes \(x\) and varies \(y\); a horizontal strip fixes \(y\) and varies \(x\). Either direction can describe the same lamina, but the resulting bounds and integrands may differ substantially.

The useful question is not simply whether vertical or horizontal slices are possible. It is which direction gives the clearest description of the region and the most manageable integral. A convenient choice can avoid splitting the region into pieces, make boundary functions easier to write, or simplify how the density varies across each strip.

Definition: A region is vertically simple if it can be described as \(a\le x\le b\) and \(g_1(x)\le y\le g_2(x)\). It is horizontally simple if it can be described as \(c\le y\le d\) and \(h_1(y)\le x\le h_2(y)\). A region may have both descriptions, or may require splitting into pieces for one of them.

For a lamina with density \(\rho(x,y)\), the mass is the integral of density over the region. If the region is vertically simple, then

m=\int_a^b\int_{g_1(x)}^{g_2(x)}\rho(x,y)\,dy\,dx.

If it is horizontally simple, then

m=\int_c^d\int_{h_1(y)}^{h_2(y)}\rho(x,y)\,dx\,dy.

The inner integral accumulates the density across one strip; the outer integral adds the contributions of all strips. In examples below, coordinates and densities are understood in consistent normalized units, so the computed mass is in the corresponding mass units.

How to Compare the Two Directions

Begin with the boundary curves and ask what a typical strip encounters. A vertical slice should have a lower and upper \(y\)-boundary; a horizontal slice should have a left and right \(x\)-boundary. If a slice passes through separate parts of the lamina, the description needs multiple intervals and therefore multiple integrals.

Then examine the density. The density must be evaluated at the points of the strip, so the inner integral uses the variable that changes across that strip. A simple boundary description is helpful, but it is not the only consideration: a direction with more complicated limits may still lead to a simpler density expression or inner integration.

Count pieces: prefer a direction that represents the whole region with one pair of bounds, when the integrands are otherwise comparable.
Inspect the boundary formulas: choose bounds that are easy to solve for and integrate.
Inspect the density on a strip: determine whether substitution of the strip coordinate produces a manageable expression.
Consider the quantity being calculated: moments include extra factors of \(x\) or \(y\), so a direction that is convenient for mass need not be the most convenient for every moment.
Check the full setup: the limits must describe exactly the region, and the density must be integrated over the same region.

Takeaway: Slice direction is a modeling choice. Compare the complete iterated integrals—including bounds, density, and any moment factors—rather than choosing by the shape of the boundary alone.

Why the Two Setups Agree

When a region has both vertical and horizontal descriptions, either iterated integral gives the same mass, provided the density is continuous on the bounded region. This is the relevant form of the equality of iterated integrals: the two orders of integration compute the same double integral over the same region.

The reason can be understood from small area pieces. A tiny rectangle with side lengths \(\Delta x\) and \(\Delta y\) has area approximately \(\Delta x\Delta y\). If density is nearly constant on that rectangle, its mass is approximately \(\rho(x,y)\Delta x\Delta y\). Adding these contributions over finer and finer partitions gives the double integral. Grouping the pieces into vertical strips first or horizontal strips first changes the order of addition, not the mass being accumulated. Continuity on a bounded region ensures the iterated integrals are valid and agree.

Describe the region.
Write its boundary curves and identify the full range of each coordinate.

Test vertical strips.
Check whether each \(x\) gives one interval of \(y\)-values, or whether the range must be divided.

Test horizontal strips.
Check in the same way whether each \(y\) gives one interval of \(x\)-values.

Compare the integrals.
Include the density and any factors needed for moments; choose the clearer calculation.

Worked Examples

Worked Example: The Same Curved Region in Both Directions

Let \(R\) be the region between \(y=x^2\) and \(y=x\), for \(0\le x\le1\), and let its density be \(\rho(x,y)=2+x+y\). Set up and evaluate the mass using both directions.

The curves meet where \(x^2=x\), so \(x=0\) or \(x=1\). For vertical slices, \(x^2\le y\le x\), giving

m=\int_0^1\int_{x^2}^{x}(2+x+y)\,dy\,dx.

Integrating with respect to \(y\) first gives

\begin{aligned} \int_{x^2}^{x}(2+x+y)\,dy &=(2+x)(x-x^2)+\frac{x^2-x^4}{2}\\ &=2x-\frac{x^2}{2}-x^3-\frac{x^4}{2}. \end{aligned}

Therefore

\begin{aligned} m &=\int_0^1\left(2x-\frac{x^2}{2}-x^3-\frac{x^4}{2}\right)dx\\ &=1-\frac16-\frac14-\frac1{10} =\frac{29}{60}. \end{aligned}

For horizontal slices, \(y\) ranges from \(0\) to \(1\). At a fixed \(y\), the inequalities \(x^2\le y\) and \(y\le x\), with \(x\ge0\), give \(y\le x\le\sqrt y\). Thus

m=\int_0^1\int_y^{\sqrt y}(2+x+y)\,dx\,dy.

The inner integral and its simplification are

\begin{aligned} \int_y^{\sqrt y}(2+x+y)\,dx &=(2+y)(\sqrt y-y)+\frac{y-y^2}{2}\\ &=2\sqrt y+y^{3/2}-\frac32y-\frac32y^2. \end{aligned}

Hence

\begin{aligned} m &=\int_0^1\left(2y^{1/2}+y^{3/2}-\frac32y-\frac32y^2\right)dy\\ &=\frac43+\frac25-\frac34-\frac12 =\frac{29}{60}. \end{aligned}

The two results agree. Here both descriptions use one integral, but the vertical bounds are polynomial while the horizontal bounds include \(\sqrt y\). The better choice depends on the rest of the calculation; the equality of the masses provides a useful check.

Worked Example: Choosing Horizontal Slices to Avoid a Split

Consider the region bounded by \(x=y^2\) and \(x=2-y\), with constant density \(3\). Find its mass, comparing both directions.

The intersections satisfy \(y^2=2-y\), so \((y+2)(y-1)=0\), and the intersection heights are \(y=-2\) and \(y=1\). Horizontal slices give one interval throughout: \(y^2\le x\le2-y\). Thus

\begin{aligned} m &=3\int_{-2}^{1}\int_{y^2}^{2-y}1\,dx\,dy\\ &=3\int_{-2}^{1}(2-y-y^2)\,dy\\ &=3\left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{-2}^{1}\\ &=3\left(\frac76-\left(-\frac{10}{3}\right)\right) =\frac{27}{2}. \end{aligned}

Vertical slices require two pieces. The parabola is \(y=\pm\sqrt x\). From \(x=0\) to \(x=1\), the slice runs from \(-\sqrt x\) to \(\sqrt x\). From \(x=1\) to \(x=4\), it runs from \(-\sqrt x\) to \(2-x\). The switch occurs at \((1,1)\), and the rightmost point is \((4,-2)\). Therefore

\begin{aligned} m &=3\left(\int_0^1 2\sqrt x\,dx+ \int_1^4(2-x+\sqrt x)\,dx\right)\\ &=3\left(\frac43+ \left[2x-\frac{x^2}{2}+\frac23x^{3/2}\right]_1^4\right)\\ &=3\left(\frac43+\frac{19}{6}\right) =\frac{27}{2}. \end{aligned}

Both calculations give \(27/2\), but horizontal slicing uses a single integral while vertical slicing requires two. Horizontal slices are the more efficient choice here.

Worked Example: Comparing Slices for Mass and First Moments

Let \(R\) be the triangle \(0\le x\le1\), \(0\le y\le1-x\), with density \(\rho(x,y)=1+x+2y\). Find its mass and centroid using vertical slices, then check using horizontal slices.

For vertical slices, the mass in a strip is

\int_0^{1-x}(1+x+2y)\,dy=2-2x.

Thus \(m=\int_0^1(2-2x)\,dx=1\). The first moment about the \(y\)-axis is \(M_y=\iint_R x\rho\,dA\), and the first moment about the \(x\)-axis is \(M_x=\iint_R y\rho\,dA\). Direct integration gives

\begin{aligned} M_y&=\int_0^1x(2-2x)\,dx =1-\frac23=\frac13,\\ M_x&=\int_0^1\int_0^{1-x}y(1+x+2y)\,dy\,dx\\ &=\int_0^1\left(\frac{(1+x)(1-x)^2}{2} +\frac{2(1-x)^3}{3}\right)dx\\ &=\int_0^1\left(\frac76-\frac52x+\frac32x^2-\frac16x^3\right)dx\\ &=\frac76-\frac54+\frac12-\frac1{24} =\frac38. \end{aligned}

The centroid is \((\bar x,\bar y)=(M_y/m,M_x/m)=(1/3,3/8)\).

For a horizontal check, \(0\le y\le1\) and \(0\le x\le1-y\). The mass in a horizontal strip is

\int_0^{1-y}(1+x+2y)\,dx=\frac32-\frac32y^2.

It follows that \(m=\int_0^1(3/2-3y^2/2)\,dy=1\). The moments are

\begin{aligned} M_y &=\int_0^1\int_0^{1-y}x(1+x+2y)\,dx\,dy\\ &=\int_0^1\left(\frac56-y-\frac12y^2+\frac23y^3\right)dy =\frac13,\\ M_x &=\int_0^1y\left(\frac32-\frac32y^2\right)dy =\frac34-\frac38=\frac38. \end{aligned}

Both directions reproduce the mass and moments, and therefore the same centroid. This example also shows why the whole calculation matters: the density and moment factors are part of the comparison, not just the geometric bounds.

A Common Pitfall

A frequent error is to reverse a boundary but keep the old coordinate range. For example, in the region between \(y=x^2\) and \(y=x\), one might incorrectly write the horizontal bounds as \(x^2\le y\le x\). Those are vertical-slice bounds: they describe \(y\) in terms of a fixed \(x\), not \(x\) in terms of a fixed \(y\).

For horizontal slices, solve the inequalities for \(x\). From \(x^2\le y\), together with \(x\ge0\), we obtain \(x\le\sqrt y\); from \(y\le x\), we obtain \(x\ge y\). Also, the height range is \(0\le y\le1\). The corrected bounds are therefore \(y\le x\le\sqrt y\), exactly as used in the first example.

Check before integrating: Choose a value strictly inside the outer range and test whether the proposed inner bounds run from the left edge to the right edge, or from the bottom edge to the top edge, of the actual region.

Choosing Efficiently in Practice

There is no rule that one direction is always superior. A useful comparison is to write down both candidate descriptions before committing to either. Note where each boundary changes, count the resulting pieces, and inspect the density after substituting the strip coordinate. Then choose the setup with the simpler overall calculation.

If both choices look equally simple, either is valid; use the one whose limits are easiest to verify. If one direction needs several pieces, that does not make it wrong, but it raises the possibility that the other direction is more efficient. When calculating mass or moments, comparing the final values from the alternative setup can also reveal a missed region, incorrect boundary, or algebra error.

Check Your Understanding

Use the boundary descriptions and integrands to decide which setup is appropriate before calculating.

What is the difference between a vertically simple and a horizontally simple region?
For the region between \(y=x^2\) and \(y=x\), why does a horizontal slice have \(x\)-limits \(y\) and \(\sqrt y\)?
What feature of the region in the second worked example makes horizontal slices more efficient?
When comparing directions for a first moment, which additional factors must be included in each integrand?
What hypotheses allow the two orders of integration to give the same mass for a lamina?

What You'll Learn