From a Base Region to Rectangular Slices
A solid with a prescribed base and a known cross-sectional shape can be studied one slice at a time. For rectangular cross sections, the main geometric step is to find the length of the rectangle’s base from the base region, then use the stated relationship between that base and the rectangle’s height.
For example, if each cross section has height three times its base length, then a base of length \(s\) produces a rectangle with area \(3s^2\). The cross-sectional area changes as the slice moves through the base region. Integrating that area across the region gives the solid’s volume.
Set Up the Cross-Sectional Area
Suppose a slice is perpendicular to the \(x\)-axis. At a particular value of \(x\), the slice cuts the base region in a segment whose length is \(s(x)\). This segment is the base of the rectangle. If the rectangle’s height is \(k\) times that length, its height is \(ks(x)\), so its area is
If the base region is bounded above by \(y=f(x)\) and below by \(y=g(x)\), and \(f(x)\ge g(x)\) throughout the interval, then the vertical segment has length \(s(x)=f(x)-g(x)\). Thus
For slices perpendicular to the \(y\)-axis, the same geometric reasoning applies with a horizontal segment. If the segment runs from \(x=u(y)\) to \(x=v(y)\), with \(v(y)\ge u(y)\), its length is \(s(y)=v(y)-u(y)\), and the cross-sectional area is \(A(y)=k[v(y)-u(y)]^2\).
The differential in the integral must measure the distance between neighboring slices: use \(dx\) for slices perpendicular to the \(x\)-axis and \(dy\) for slices perpendicular to the \(y\)-axis. The bounds must cover the corresponding interval of slices through the base.
Why Integrating the Areas Gives Volume
Over a short interval of width \(\Delta x\), a thin slice of the solid has approximately the same cross-sectional area as the rectangle at a representative point \(x_i\). Its volume is therefore approximately \(A(x_i)\Delta x\). Adding such estimates across the interval gives a Riemann sum. As the widths of the slices approach zero, the sum approaches the definite integral of the cross-sectional area.
At each coordinate, determine the length of the segment cut from the base region.
Multiply the slice length by the stated height-to-base ratio \(k\).
Multiply the rectangle’s base by its height to get \(A=k s^2\).
Use the differential and bounds that match the direction and extent of the slices.
This reasoning assumes that the cross sections are perpendicular to the chosen axis and that the base length and area vary continuously over the interval, as they do in the examples below. The definite integral adds the volumes of the slices; it does not require the solid itself to have a simple outer shape.
Worked Examples
Worked Example: Height Three Times the Base
The base of a solid is the region between \(y=2\) and \(y=x^2\), for \(-\sqrt{2}\le x\le\sqrt{2}\). Cross sections perpendicular to the \(x\)-axis are rectangles whose height is three times their base. Find the volume.
Step 1: Find the base length. The upper curve is \(y=2\) and the lower curve is \(y=x^2\) on this interval, so the vertical segment has length
This length is nonnegative because \(x^2\le 2\) throughout the stated interval.
Step 2: Find the rectangular area. The height is \(3s(x)=3(2-x^2)\), so
Step 3: Integrate across the interval.
Let \(a=\sqrt{2}\). Since \(a^3=2a\) and \(a^5=4a\), the antiderivative at \(a\) is
The antiderivative is odd, so its value at \(-a\) is \(-32a/15\). Therefore
Check: An antiderivative of \(3(4-4x^2+x^4)\) is \(12x-4x^3+\frac{3x^5}{5}\); differentiating gives \(12-12x^2+3x^4=3(2-x^2)^2\). Substituting the bounds gives \(64\sqrt{2}/5\), as calculated. If lengths are measured in units, the volume is \(64\sqrt{2}/5\) cubic units.
Worked Example: A Height Equal to Half the Base
The base region is bounded by \(y=x\), \(y=0\), \(x=0\), and \(x=2\). Cross sections perpendicular to the \(x\)-axis are rectangles whose height is half their base. Find the volume.
Step 1: Find the base length. At a fixed \(x\) between 0 and 2, the vertical segment extends from \(y=0\) to \(y=x\), so
Step 2: Find the area. The height is \(x/2\), giving
Step 3: Integrate.
Check: Differentiating \(x^3/6\) gives \(x^2/2\), the cross-sectional area. Evaluating the antiderivative at \(x=2\) and \(x=0\) gives \(4/3\). The volume is \(4/3\) cubic units when the base measurements are in units.
Worked Example: Horizontal Slices and a Noninteger Ratio
The base is the region between \(x=y^2\) and \(x=4\), for \(-2\le y\le2\). Cross sections perpendicular to the \(y\)-axis are rectangles whose height is \(3/2\) times their base. Find the volume.
Step 1: Find the horizontal base length. Each horizontal segment begins at \(x=y^2\) and ends at \(x=4\), so
Because \(y^2\le4\) on the interval, this length is nonnegative.
Step 2: Find the area. The rectangle’s height is \(\frac32(4-y^2)\), so
Step 3: Integrate with respect to \(y\).
At \(y=2\), the expression in brackets is
At \(y=-2\), it is \(-256/15\). Thus
Check: Differentiating \(16y-\frac{8y^3}{3}+\frac{y^5}{5}\) gives \(16-8y^2+y^4=(4-y^2)^2\). Including the factor \(3/2\) recovers the stated area function, and evaluating at the bounds gives \(256/5\) cubic units.
Common Pitfall: Using the Height as the Area
A frequent error is to calculate the rectangle’s height but stop before multiplying by its base. Suppose the base length is 3 and the height is twice the base. The height is 6, but the area is not 6. It is
More generally, if \(h=2s\), then the area is \(A=sh=s(2s)=2s^2\), not \(2s\). The correct formula is quadratic in the base length because both rectangle dimensions depend on that length. A useful check is to write both dimensions before multiplying: base \(s\), height \(ks\), area \(ks^2\).
Also check that the segment length is nonnegative throughout the interval. For vertical slices, subtract lower \(y\)-value from upper \(y\)-value; for horizontal slices, subtract left \(x\)-value from right \(x\)-value. Reversing either subtraction can produce a negative length, which cannot represent a rectangle’s dimension.
Why the Height Ratio Matters
The ratio \(k\) scales every cross-sectional area directly. If the base region and slice direction remain fixed, doubling the height-to-base ratio doubles \(A\) at every slice, and therefore doubles the integral for volume. The base length itself is squared in the area formula, so a slice twice as long produces four times the area when the same ratio \(k\) is used.
These relationships provide useful checks on a setup. The area must have square units: \(s^2\) has square units and \(k\) is a dimensionless ratio. Integrating area over a length adds one more unit of length, producing cubic units for volume. The integrand must also reflect the actual rectangle dimensions, not just the height ratio or just the base segment.
Check Your Understanding
For each question, identify the slice length and the correct area function before evaluating an integral.
- A vertical slice of a base region has length \(5-x\). If each rectangular cross section has height four times its base, what is \(A(x)\)?
- The base segment at \(x\) extends from \(y=1\) to \(y=x+1\), for \(0\le x\le2\). Rectangles have height half their base. Write the volume integral.
- A horizontal slice has length \(y+2\), and the rectangle’s height is three times its base. What is the cross-sectional area?
- If the base length is \(s\) and the rectangle’s height is \(ks\), explain why the area is \(ks^2\), not \(ks\).
- What units result when cross-sectional area is integrated with respect to a coordinate measured in units of length?
