From a Region to a Solid
A plane region can form the base of a solid, but the base alone does not determine the solid’s volume. To specify the solid, we also need to know the shape of each cross section and how that shape changes as we move across the region.
For a region between two curves, a vertical slice at a particular \(x\)-value has a full length measured from the lower curve to the upper curve. That length becomes a dimension of the cross section. The cross-sectional shape then determines its area, and integrating those areas gives the volume.
Describing a Vertical Slice
Suppose a region extends from \(x=a\) to \(x=b\), with upper boundary \(y=f(x)\) and lower boundary \(y=g(x)\). Assume \(f(x)\ge g(x)\) throughout the interval. A vertical slice at \(x\) extends from \(g(x)\) to \(f(x)\), so its full length is
This is the entire distance across the region at that \(x\)-value. It is not automatically a radius, a diameter, or half of a side. What it represents depends on the cross-sectional shape specified in the problem.
For example, if the slice is a side of a square, the square’s side length is \(L(x)\). If the slice is the diameter of a semicircle, its radius is \(L(x)/2\). The area formula must use the dimension that the slice actually represents.
Why Integrating Cross-Sectional Areas Gives Volume
The integral adds the contributions of many thin slices. When a slice has thickness \(\Delta x\), its small volume is approximately its cross-sectional area multiplied by that thickness. Adding these approximate volumes across the interval leads to the definite integral.
Divide \([a,b]\) into thin subintervals, each with width \(\Delta x_i\).
Choose a sample value \(x_i^*\) in each subinterval. The piece’s volume is approximately \(A(x_i^*)\Delta x_i\).
The sum \(\sum_i A(x_i^*)\Delta x_i\) approaches \(\int_a^b A(x)\,dx\) as the widths shrink to zero.
This argument applies when the cross-sectional areas vary in an integrable way across the interval. It explains why the integrand is area, not slice length alone: each area is multiplied by a small thickness to represent a small volume.
Area Formulas in Terms of the Slice Length
Once \(L(x)\) is known, use the stated shape to form \(A(x)\). Three common cases illustrate how the same slice length leads to different integrands.
| Cross section | Meaning of \(L\) | Area \(A\) |
|---|---|---|
| Square | Side length | \(L^2\) |
| Equilateral triangle | Side length | \(\frac{\sqrt{3}}{4}L^2\) |
| Semicircle | Diameter | \(\frac{\pi}{8}L^2\) |
For the semicircle, the radius is \(L/2\). Its area is half the area of a circle with that radius, so \(A=\frac12\pi(L/2)^2=\frac{\pi}{8}L^2\). Using \(\pi L^2/4\) would instead give the area of a full circle with diameter \(L\).
Worked Examples
Worked Example: Square Cross Sections
The base is the region between \(y=4-x^2\) and \(y=0\) for \(-2\le x\le2\). Cross sections perpendicular to the \(x\)-axis are squares whose sides are the vertical slices. Find the volume.
Step 1: Find the full slice length. The upper curve is \(4-x^2\), and the lower curve is \(0\). Thus
On \([-2,2]\), this length is nonnegative, as required.
Step 2: Find the cross-sectional area. A square with side \(L(x)\) has area \(L(x)^2\), so
Step 3: Integrate.
Check: Differentiating the antiderivative gives \(16-8x^2+x^4=(4-x^2)^2\), the required area function. The side length is zero at \(x=\pm2\) and positive between them. The volume is \(\frac{512}{15}\) cubic units.
Worked Example: Equilateral-Triangle Cross Sections
The base is the region between \(y=2x\) and \(y=0\) for \(0\le x\le2\). Cross sections perpendicular to the \(x\)-axis are equilateral triangles with the vertical slice as a side. Find the volume.
Step 1: Find the full slice length.
Step 2: Use the equilateral-triangle area formula. A triangle with side length \(L\) has area \(\frac{\sqrt{3}}{4}L^2\). Therefore
Step 3: Integrate over the interval.
Check: The derivative of \(\sqrt{3}x^3/3\) is \(\sqrt{3}x^2\), the area function. At \(x=0\), the slice has zero length and area; the lengths are nonnegative through \(x=2\). The volume is \(\frac{8\sqrt{3}}{3}\) cubic units.
Worked Example: Semicircle Cross Sections
The base is the region between \(y=\sqrt{x}\) and \(y=x\) for \(0\le x\le1\). Cross sections perpendicular to the \(x\)-axis are semicircles whose diameters are the vertical slices. Find the volume.
Step 1: Identify the upper and lower curves. On \([0,1]\), \(\sqrt{x}\ge x\), so the full length is
Step 2: Use the diameter-based area formula. Since the slice is the diameter, the semicircle’s area is \(\frac{\pi}{8}L(x)^2\):
Step 3: Integrate.
Check: Differentiating the expression in brackets gives \(x-2x^{3/2}+x^2\), as required. The diameter is zero at both endpoints and nonnegative between them. The volume is \(\frac{\pi}{240}\) cubic units.
A Common Mistake: Using the Wrong Dimension
A frequent error is to treat the full slice length as a radius even when the problem says it is a diameter. For instance, if a slice of length \(L\) is the diameter of a semicircle, the radius is \(L/2\), not \(L\). The incorrect calculation would be
This is four times the correct semicircle area, because it uses a radius twice as large as the actual radius. The correction is to convert the specified slice dimension to the dimension required by the shape’s area formula:
Before integrating, state in words what the slice represents: a side, a diameter, a radius, or another specified dimension. Then write the area formula using that dimension.
A Reliable Setup Checklist
- State the interval of integration and identify the upper and lower curves on it.
- Write the full slice length as upper curve minus lower curve.
- Check that the length is nonnegative on the interval.
- Translate the slice length into the dimension required by the cross-sectional shape.
- Write the area function \(A(x)\) before writing the volume integral.
- Integrate across the interval, and report volume in cubic units.
If the upper and lower boundaries change on the interval, the full length may need a piecewise expression and the volume integral may need to be split at the change. The slice length must describe the region correctly at every point where the integral is evaluated.
Check Your Understanding
For each question, identify the slice length and the resulting area function before evaluating any integral.
- A region has upper curve \(y=3-x\) and lower curve \(y=0\) on \(0\le x\le3\). What is the side length of a square cross section perpendicular to the \(x\)-axis?
- If a vertical slice of length \(L(x)\) is the side of an equilateral triangle, what is the cross-sectional area?
- If a vertical slice of length \(L(x)\) is the diameter of a semicircle, what is its radius and area?
- Why does the volume integral use \(A(x)\,dx\), rather than just \(L(x)\,dx\)?
- For upper curve \(y=2\) and lower curve \(y=x\) on \(0\le x\le2\), write the volume integral for square cross sections perpendicular to the \(x\)-axis. Do not evaluate it.
