Horizontal Base Segments and Volume
A solid can be described by its two-dimensional base region and the shape of each cross section. In this tutorial, each cross section is built on a horizontal segment in the base region: the segment lies at a fixed \(y\)-value, and its endpoints vary in the \(x\)-direction.
This setup calls for \(y\)-bounds. At each height \(y\), we find the left and right endpoints of the horizontal segment, use their difference as its length, and calculate the area of the specified cross section. Integrating those areas over the \(y\)-interval gives the volume.
Describing a Horizontal Segment
Suppose the base region extends from \(y=a\) to \(y=b\). For a particular value of \(y\), draw a horizontal segment across the region. If its left endpoint has \(x\)-coordinate \(L(y)\), and its right endpoint has \(x\)-coordinate \(R(y)\), then its length is
The inequality matters: it confirms that the designated left endpoint is not to the right of the designated right endpoint. If the region’s boundaries exchange roles as \(y\) changes, the correct left and right functions may differ on different parts of the interval. In that case, split the integral where the roles switch.
The specified cross-sectional shape determines an area from the segment length \(s(y)\). For example, if the cross section is a square, its area is \(s(y)^2\). If it is a semicircle with the horizontal segment as its diameter, its radius is \(s(y)/2\), so its area is \(\frac{\pi}{8}s(y)^2\). The segment is a diameter in that example, not a radius.
Why Integrating Cross-Sectional Areas Gives Volume
Each cross section at height \(y\) has area \(A(y)\). Imagine dividing the interval from \(a\) to \(b\) into thin strips of height \(\Delta y\). Over a sufficiently thin strip, the solid’s volume is approximated by a slab whose base area is \(A(y_i^*)\), for some sample height \(y_i^*\), and whose thickness is \(\Delta y\). Its approximate volume is \(A(y_i^*)\Delta y\).
Adding these slab estimates and taking the limit as the largest strip width approaches zero gives the definite integral. This is the same area-accumulation principle used to build volume integrals; here, the variable and bounds are specifically chosen to match the horizontal segments.
When a cross section’s area depends only on the length of its horizontal base segment, first write \(s(y)=R(y)-L(y)\), then substitute \(s(y)\) into the appropriate shape formula. For instance, square cross sections give \(A(y)=[R(y)-L(y)]^2\). The integral’s differential is \(dy\), because the slices are stacked through changes in \(y\).
Worked Example: Square Cross Sections
Worked Example: Square Cross Sections
The base region is bounded by \(y=0\), \(y=2\), \(x=2\), and \(x=\frac{y^2}{4}\). Cross sections perpendicular to the \(y\)-axis are squares. Find the volume.
At a fixed \(y\) from \(0\) to \(2\), the left endpoint is \(L(y)=\frac{y^2}{4}\), and the right endpoint is \(R(y)=2\). The horizontal base length is
For \(0\le y\le 2\), this length is nonnegative: it decreases from \(2\) to \(2-1=1\). The area of a square with side length \(s(y)\) is \(s(y)^2\). Therefore,
Integrate from \(y=0\) to \(y=2\):
To check the integration, differentiate the antiderivative: \(\frac{d}{dy}\left(4y-\frac{y^3}{3}+\frac{y^5}{80}\right)=4-y^2+\frac{y^4}{16}\), the integrand above. The endpoint evaluation gives \(86/15\) cubic units.
Boundary Switches Require Piecewise Lengths
Sometimes two boundary curves exchange which one is on the left and which one is on the right. Their intersection heights identify where the order can change. On each resulting interval, determine the order of the endpoints before writing the length. The length must remain right minus left on every piece.
Worked Example: Semicircular Cross Sections with Switching Boundaries
The base region lies between the lines \(x=y\) and \(x=2-y\), for \(0\le y\le 2\). Cross sections perpendicular to the \(y\)-axis are semicircles whose diameters are the horizontal segments in the base. Find the volume.
The boundary lines meet where \(y=2-y\), so \(2y=2\), and therefore \(y=1\). For \(0\le y\le 1\), \(y\le 2-y\); thus the left endpoint is \(y\), and the right endpoint is \(2-y\). The diameter is
For \(1\le y\le 2\), \(2-y\le y\); now the left endpoint is \(2-y\) and the right endpoint is \(y\). The diameter is
A semicircle with diameter \(d\) has radius \(d/2\), so its area is \(\frac12\pi(d/2)^2=\frac{\pi}{8}d^2\). Use the two diameter formulas on their respective intervals:
For a check, the two antiderivatives differentiate to \(4-8y+4y^2=(2-2y)^2\) and \(4y^2-8y+4=(2y-2)^2\), respectively. The evaluated contributions are each \(\pi/6\), totaling \(\pi/3\) cubic units. The split is essential: using one endpoint order over the entire interval would make the diameter negative on one side, even though a physical length cannot be negative.
Worked Example: Equilateral Triangular Cross Sections
Worked Example: Equilateral Triangular Cross Sections
The base region is bounded by \(x=0\), \(y=x^2\), and \(y=4\). Cross sections perpendicular to the \(y\)-axis are equilateral triangles, with each horizontal segment as a side. Find the volume.
For a fixed \(y\) between \(0\) and \(4\), the curve \(y=x^2\) gives \(x=\sqrt{y}\) on this region, while the other boundary is \(x=0\). Therefore, \(L(y)=0\), \(R(y)=\sqrt{y}\), and the side length is
An equilateral triangle of side length \(s\) has area \(\frac{\sqrt{3}}{4}s^2\). Thus the cross-sectional area is
Integrating gives
As a check, differentiating the antiderivative \(\frac{\sqrt{3}}{8}y^2\) gives \(\frac{\sqrt{3}}{4}y\), exactly the cross-sectional area. The volume is \(2\sqrt{3}\) cubic units.
A Common Mistake: Reversing the Endpoint Order
Horizontal segment length is not left endpoint minus right endpoint. For example, suppose a segment runs from \(x=0\) on the left to \(x=2\) on the right. The incorrect calculation \(0-2=-2\) produces a negative value, which cannot be its length. The correct calculation is
In a volume problem, using the wrong order can produce negative cross-sectional areas if the length is squared only after other operations, or can obscure which boundary is which. First identify the left and right endpoints on the interval in question; then compute \(R(y)-L(y)\). If their roles switch, split the interval and keep the same right-minus-left rule on each piece.
A Reliable Setup Process
Read the region’s lowest and highest \(y\)-values to determine the integration interval.
At a fixed \(y\), write the left endpoint as \(x=L(y)\) and the right endpoint as \(x=R(y)\).
Find heights where boundary curves intersect or where the region’s endpoint formulas change, and split the interval there.
On each interval, calculate \(R(y)-L(y)\), then apply the given cross-sectional shape formula.
Integrate area with respect to \(y\), evaluate every interval, and differentiate the antiderivative to verify the integrand.
This method matters because the cross-sectional shape and the orientation of its base determine the correct area function and integration variable. Writing \(A(y)\) explicitly before integrating makes those choices visible and helps prevent sign errors. A final reasonableness check is also useful: the area should be nonnegative throughout the interval, and the volume should be nonnegative.
Check Your Understanding
For each question, identify the endpoint order before forming a volume integral.
- A horizontal segment runs from \(x=1-y\) on the left to \(x=3\) on the right. What is its length?
- A square cross section has horizontal base length \(R(y)-L(y)\). What is its area?
- A semicircular cross section uses its horizontal base segment as a diameter of length \(d(y)\). What is its area?
- The boundaries are \(x=y\) and \(x=4-y\) for \(0\le y\le 4\). At what height do their left-right roles switch?
- Why must a volume integral be split if the identity of the left and right boundary functions changes?
