Deriving Area From Horizontal Slices

Why Slice Horizontally?

A region’s area can be estimated by covering it with thin rectangles. If the rectangles run horizontally, each one has a small vertical thickness and a horizontal width determined by the region’s left and right boundaries. Adding these rectangle areas suggests a formula that integrates horizontal widths as the height changes.

This viewpoint is useful whenever a region is naturally described from left to right at each fixed height. The essential question is: for a given \(y\), where does the horizontal slice begin, and where does it end?

Describing the Region

Suppose a bounded plane region extends vertically from \(y=c\) to \(y=d\), where \(c<d\). At a height \(y\) in this interval, assume the horizontal cross-section of the region is the interval from \(x=L(y)\) to \(x=R(y)\), with \(L(y)\le R(y)\). Here \(L(y)\) is the left boundary and \(R(y)\) is the right boundary.

The expression \(R(y)-L(y)\) is a length: it measures how far the slice extends horizontally. The differential \(dy\) represents a small vertical thickness. Thus, width multiplied by thickness gives the approximate area of a thin horizontal strip.

Continuity of both \(L\) and \(R\) on \([c,d]\) is a convenient sufficient condition: it ensures that the width is continuous, and hence Riemann integrable. Continuity is not necessary. More generally, the width \(R-L\) may be discontinuous and still be Riemann integrable; in that case the integral gives the area when the region is described appropriately by its horizontal slices.

Deriving the Formula from Thin Rectangles

First consider the common case in which \(L\) and \(R\) are continuous on \([c,d]\). Divide the vertical interval into \(n\) subintervals. Write their endpoints as \(c=y_0<y_1<\cdots<y_n=d\), and let \(\Delta y_i=y_i-y_{i-1}\) be the thickness of the \(i\)-th horizontal strip.

Choose a sample height \(y_i^*\) in each subinterval. At that height, the region’s width is \(R(y_i^*)-L(y_i^*)\). A rectangle with that width and thickness \(\Delta y_i\) has area \(\bigl(R(y_i^*)-L(y_i^*)\bigr)\Delta y_i\). Adding these estimates gives a Riemann sum.

Why does this sum approach the region’s area? Within a thin strip, the left and right boundaries may move as \(y\) changes. But continuity on the closed interval means that both boundary functions are uniformly continuous. As the largest strip thickness becomes small, the movement of each boundary within every strip becomes small too. The rectangle taken at the sample height therefore differs from the actual portion of the region in that strip only by narrow slivers along its left and right edges.

More precisely, in the \(i\)-th strip, let \(\omega_{L,i}\) and \(\omega_{R,i}\) be the maximum changes in \(L\) and \(R\) across that subinterval. The total area of the boundary slivers is at most \((\omega_{L,i}+\omega_{R,i})\Delta y_i\). Consequently, the difference between the region’s area and the sum of the sample-rectangle areas is bounded by

Uniform continuity makes the maximum boundary changes tend to zero as the largest subinterval length tends to zero. The displayed error bound therefore tends to zero. The sums converge to the Riemann integral of the width, which establishes the formula.

Worked Examples

Conditions and Interpretation

The formula depends on describing the region correctly. At each height in the stated vertical extent, the slice must run from the chosen left boundary to the chosen right boundary. If a region has more than one separate interval in a horizontal slice, one interval from a single \(L(y)\) to a single \(R(y)\) may include points outside the region. In that situation, describe the slice using its separate intervals and add their widths before integrating.

The condition \(L(y)\le R(y)\) ensures that each width is nonnegative. If the boundaries are accidentally reversed, the resulting integral may be negative. A negative value signals a signed difference in the wrong order, not a negative geometric area.

Also keep the integration variable and thickness consistent. Horizontal slices have vertical thickness \(dy\), so the integral is with respect to \(y\). The integrand is a horizontal length, \(R(y)-L(y)\). The product of these quantities has units of area.

A Common Mistake: Reversing the Boundaries

Suppose the boundaries in the curved example are \(x=y^2\) and \(x=y\) on \([0,1]\). A common error is to subtract the right boundary from the left and write

This is wrong for geometric area because \(y^2\) is the left boundary and \(y\) is the right boundary on this interval. The correction is to use right minus left:

Before integrating, check the ordering of the boundaries at heights in the interval. A quick test at an interior value, such as \(y=\frac12\), gives \(y^2=\frac14<\frac12=y\), confirming which boundary is on the left and which is on the right.

Why the Slice Formula Matters

The formula is more than a way to calculate area. It expresses a geometric region as a continuous accumulation of one-dimensional widths. The sum of rectangle areas becomes an integral because the widths are sampled across the region’s vertical extent and the approximation error tends to zero.

This also explains the structure of the formula: right boundary minus left boundary gives the width, and integrating that width with respect to \(y\) adds the contributions from all heights. Keeping those geometric roles clear helps prevent sign, variable, and boundary-order errors.