Why Slice Horizontally?
A region’s area can be built by adding the areas of many thin rectangles. When those rectangles are horizontal, each one has a small height and a horizontal length. If the region’s left and right boundaries can be described in terms of the vertical coordinate \(y\), then each slice’s length is found from those boundary functions.
The important point is that the rectangle’s length is measured along the \(x\)-direction, while its small thickness is measured along the \(y\)-direction. As a result, the length is the right \(x\)-coordinate minus the left \(x\)-coordinate, and the thickness is \(dy\). This leads to an integral with respect to \(y\).
Describing a Horizontal Slice
Suppose a bounded region extends vertically from \(y=c\) to \(y=d\), where \(c<d\). Assume that for every \(y\) in this interval, a horizontal line through the region meets its left boundary at \(x=L(y)\) and its right boundary at \(x=R(y)\). The functions \(L\) and \(R\) give the \(x\)-coordinates of the two ends of the slice.
“Right minus left” is not an arbitrary subtraction rule. Both boundaries give \(x\)-coordinates at the same height. The distance between two points on a horizontal line is the larger \(x\)-coordinate minus the smaller \(x\)-coordinate. Since \(R(y)\) is the right endpoint and \(L(y)\) is the left endpoint, that distance is \(R(y)-L(y)\).
The formula also depends on the boundaries being consistently ordered: \(L(y)\le R(y)\). If the order changes within the interval, a single expression \(R(y)-L(y)\) may not describe the slice length everywhere. The formula developed here applies directly when the same left and right boundary functions describe the region across the full vertical interval.
Deriving the Horizontal-Slice Formula
Divide the vertical interval \([c,d]\) into \(n\) smaller intervals. Write their heights as \(\Delta y_i\), and choose a sample height \(y_i^*\) in each interval. At that sample height, the horizontal slice has length \(R(y_i^*)-L(y_i^*)\). A thin rectangle with this length and height \(\Delta y_i\) has area approximately \([R(y_i^*)-L(y_i^*)]\Delta y_i\).
Adding these rectangle areas gives an approximation to the region’s area. As the largest slice height approaches zero, the rectangles fill the region with increasingly small gaps and overhangs. For continuous boundary functions, the sums converge to a definite integral. The slice thicknesses are changes in \(y\), so the resulting integral is with respect to \(y\).
The integral adds slice lengths over the vertical range. A slice at one exact value of \(y\) has zero height and thus contributes no area by itself; the area comes from accumulating the lengths across the interval. This is why the bounds are \(c\) and \(d\), the lowest and highest \(y\)-values of the region.
Worked Examples
Worked Example: A Region Between a Curve and the \(y\)-Axis
Consider the region bounded by \(x=0\), \(x=y^2\), \(y=0\), and \(y=2\). Use horizontal slices to find its area.
For \(0\le y\le2\), the left boundary is \(L(y)=0\) and the right boundary is \(R(y)=y^2\). The slice length is therefore \(y^2-0=y^2\). The horizontal-slice formula gives
An antiderivative of \(y^2\) is \(\frac{y^3}{3}\), so
Check: differentiating \(\frac{y^3}{3}\) gives \(y^2\), the integrand. Also, on the interval \(0\le y\le2\), \(y^2\ge0\), so the right boundary is indeed no farther left than \(x=0\). The area is \(\frac{8}{3}\) square units.
Worked Example: A Region Between Two Lines
Consider the region bounded by \(x=y+1\), \(x=3\), \(y=0\), and \(y=2\). Find its area using horizontal slices.
At height \(y\), the left boundary is \(L(y)=y+1\) and the right boundary is \(R(y)=3\). The slice length is \(3-(y+1)=2-y\). Thus
An antiderivative of \(2-y\) is \(2y-\frac{y^2}{2}\). Evaluating at both bounds gives
Check: differentiating \(2y-\frac{y^2}{2}\) gives \(2-y\). The slice length is nonnegative on the stated interval: \(2-y\ge0\) for \(0\le y\le2\), and it decreases to zero at \(y=2\). The area is \(2\) square units.
Worked Example: A Curved Region with Zero-Width End Slices
Consider the region between \(x=y^2\) and \(x=2y\), from \(y=0\) to \(y=2\). Find its area with horizontal slices.
The proposed left and right boundaries are \(L(y)=y^2\) and \(R(y)=2y\). Their difference is
For \(0\le y\le2\), both factors \(y\) and \(2-y\) are nonnegative. Thus \(2y-y^2\ge0\), confirming that \(x=2y\) is the right boundary and \(x=y^2\) is the left boundary throughout this interval. The area is
An antiderivative is \(y^2-\frac{y^3}{3}\). Therefore,
Check: differentiating \(y^2-\frac{y^3}{3}\) gives \(2y-y^2\). The width is zero at \(y=0\) and \(y=2\), and nonnegative between them, as required. The area is \(\frac{4}{3}\) square units.
Why the Differential Is \(dy\)
In each example, the small rectangle was horizontal. Its long side measured a distance along the \(x\)-axis, but its thin side measured a small change in height. If two neighboring horizontal levels are \(y\) and \(y+\Delta y\), the rectangle’s thickness is \(\Delta y\). In the limiting sum, that thickness is represented by \(dy\).
Consequently, \(R(y)-L(y)\) supplies the horizontal length and \(dy\) supplies the vertical thickness. Their product has units of area: if the coordinates are measured in centimeters, the length is in centimeters and \(dy\) is in centimeters, so the product is in square centimeters. The variable of integration records which coordinate is being accumulated; here it is \(y\).
A Common Mistake: Reversing the Boundaries
A frequent error is to subtract left from right in the wrong order. For the region in the second example, a mistaken calculation might be
A negative result cannot be the geometric area of this region. The mistake is that \((y+1)-3\) is left minus right, giving a negative slice length. Correcting the order gives
Before integrating, identify which boundary is left and which is right at a typical height. Then check that the proposed slice length \(R(y)-L(y)\) is nonnegative over the interval.
When This Formula Applies
The formula \(A=\int_c^d[R(y)-L(y)]\,dy\) is especially useful when a horizontal line passes through the region in one continuous segment and the left and right boundaries can be represented consistently as functions of \(y\). The vertical bounds \(c\) and \(d\) must cover the region, and the slice length must be nonnegative throughout that range.
If a horizontal slice has several separated pieces, or if the left and right boundaries switch within the vertical range, one unchanged width expression may not describe all slices. The central reasoning remains the same—measure each horizontal piece from left to right and accumulate over \(y\)—but the setup must account for any change in the region’s description.
Check Your Understanding
Use the slice description and boundary order to answer each question.
- A region has left boundary \(x=1-y\), right boundary \(x=4\), and vertical extent \(0\le y\le2\). What expression gives the length of a horizontal slice?
- Why is the thickness of a horizontal slice represented by \(dy\), rather than \(dx\)?
- For boundaries \(x=y^2\) and \(x=3y\) on \(0\le y\le2\), write the horizontal-slice area integral, without evaluating it.
- What does a negative value of \(R(y)-L(y)\) indicate about the proposed boundary order?
- In the formula \(A=\int_c^d[R(y)-L(y)]\,dy\), what do \(c\) and \(d\) represent?
