From a Base Region to a Solid
A planar region can serve as the base of a three-dimensional solid. Imagine cutting the solid with a succession of planes perpendicular to the \(x\)-axis. If every such cross section is a semicircle, the width of the base region at position \(x\) determines the diameter of that semicircle. The solid’s volume is found by adding the areas of all these slices across the interval.
The key modeling step is to distinguish the diameter from the radius. The distance across the base region is the diameter; the radius is half of that distance. This distinction determines the constant factor in the area formula and therefore in the final volume.
Deriving the Slice Area
The area of a full circle of radius \(r\) is \(\pi r^2\), so a semicircle has half that area. If its diameter is \(d\), then \(r=d/2\). Substituting the radius in terms of the diameter gives
At position \(x\), let the cross-sectional diameter be \(d(x)\). The area of that slice is therefore \(A(x)=\frac{\pi}{8}[d(x)]^2\). The slice principle for volume says that a solid whose cross-sectional area at \(x\) is \(A(x)\) has volume \(\int_a^b A(x)\,dx\), when the area function is integrable on the interval. Substituting the semicircle area gives the volume formula.
This follows by combining the geometry of each individual slice with the slice principle. It does not depend on whether the semicircle is built above or below the base plane: either placement has the same cross-sectional area. The direction of slicing does matter, however, because the diameter must be measured across the base in the plane of the chosen slice.
A Reliable Setup
For semicircular slices perpendicular to the \(x\)-axis, first identify the interval of \(x\)-values and the two boundaries of the base region. At each \(x\), subtract the lower boundary from the upper boundary to obtain the diameter. Then square that diameter, multiply by \(\pi/8\), and integrate across the interval. Keep the whole difference inside parentheses when squaring: in general, \([g(x)-f(x)]^2\) is not the same as \(g(x)^2-f(x)^2\).
Record the \(x\)-values over which the base region is sliced.
At each \(x\), calculate the full distance across the base region.
Use \(A(x)=\frac{\pi}{8}[d(x)]^2\), since \(d(x)\) is not the radius.
Integrate the slice area over the interval and give the volume in cubic units.
Worked Examples
Worked Example: A Triangular Base Region
The base region is bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=2\). Cross sections perpendicular to the \(x\)-axis are semicircles whose diameters lie across the base region. Find the volume.
For \(0\le x\le2\), the lower boundary is \(y=0\) and the upper boundary is \(y=x\). Thus the diameter is \(d(x)=x-0=x\). The cross-sectional area is
Integrate from \(x=0\) to \(x=2\):
Check the integral using the antiderivative: the derivative of \(\frac{\pi x^3}{24}\) is \(\frac{\pi}{8}x^2\), the area function being integrated. The diameter is nonnegative throughout the interval, and the volume is \(\frac{\pi}{3}\) cubic units.
Worked Example: A Curved Base Region
The base region lies between \(y=0\) and \(y=4-x^2\), from \(x=-2\) to \(x=2\). Semicircular cross sections perpendicular to the \(x\)-axis have their diameters across this base region. Find the volume.
The diameter is the upper boundary minus the lower boundary:
On \([-2,2]\), \(4-x^2\ge0\), so this gives a valid diameter throughout. The area and volume are
Expand the square and integrate each term:
For a check, an antiderivative of the expanded integrand \(\frac{\pi}{8}(16-8x^2+x^4)\) is \(\frac{\pi}{8}(16x-\frac{8x^3}{3}+\frac{x^5}{5})\); differentiating it recovers that integrand. The slice area is nonnegative, as required, and the volume is \(\frac{64\pi}{15}\) cubic units.
Worked Example: Two Curves Form the Diameter
The base region lies between \(y=x^2\) and \(y=x\) for \(0\le x\le1\). Cross sections perpendicular to the \(x\)-axis are semicircles. Find the volume.
Since \(x^2\le x\) on this interval, the lower curve is \(y=x^2\) and the upper curve is \(y=x\). The diameter is
Therefore the slice area and volume integral are
Expand the squared difference and evaluate:
As a check, differentiating \(\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\) gives \(x^2-2x^3+x^4=(x-x^2)^2\). The diameter is zero at both endpoints and positive between them, consistent with a nonnegative slice area. The volume is \(\frac{\pi}{240}\) cubic units.
Units and Geometric Checks
If the diameter is measured in length units, then \(d(x)^2\) has square-length units, so \(A(x)\) has area units. The integration variable \(dx\) contributes another length unit. Consequently, \(\int_a^b A(x)\,dx\) has cubic-length units, as a volume should. For instance, if the base coordinates are in centimetres, then \(A(x)\) is measured in square centimetres and \(V\) in cubic centimetres.
Before accepting a model, check that the diameter is a distance, not a signed difference that becomes negative on the interval. With boundary functions, use upper minus lower where the ordering is fixed. Also check that the resulting area is nonnegative, that the integral uses the full interval of slices, and that the final units are cubic.
Common Pitfall: Using the Diameter as the Radius
A common error is to substitute the diameter directly into the circle-area formula as though it were a radius. For a diameter \(d\), the incorrect calculation would be
This treats the radius as \(d\), even though the radius is \(d/2\). The correct calculation is
The incorrect area is four times too large because \(\frac{\frac12\pi d^2}{\frac18\pi d^2}=4\). For the triangular-base example, using the wrong formula would give \(V_{\text{wrong}}=\frac{\pi}{2}\int_0^2x^2\,dx=\frac{4\pi}{3}\), rather than the correct \(\frac{\pi}{3}\). Substituting the correct radius \(r=x/2\) into \(A=\frac12\pi r^2\) gives \(A(x)=\frac{\pi x^2}{8}\), confirming the correct volume setup.
Why the Diameter-Based Formula Is Useful
Once the cross sections are specified as semicircles, the varying geometry of the solid is captured by a single quantity: the diameter across the base region at each slice. The formula \(\frac{\pi}{8}[d(x)]^2\) turns that distance into an area, and the definite integral accumulates those areas along the slicing interval. This separates the problem into two clear parts: determine the diameter from the base, then integrate the corresponding slice areas.
The same procedure applies whether the diameter is a single curve’s height above an axis or the difference between two boundary curves. What matters is that it represents the full distance across the base at the chosen slice and that the cross section is a semicircle with that segment as its diameter.
Check Your Understanding
Use the diameter-first method to answer these questions. State any interval or nonnegativity condition needed for your setup.
- Derive the area of a semicircle whose diameter is \(d\), starting from the circle area formula.
- A base region lies between \(y=0\) and \(y=3x\) for \(0\le x\le1\). What is the diameter function for semicircular slices perpendicular to the \(x\)-axis?
- For a diameter function \(d(x)=2-x\) on \(0\le x\le2\), write the volume integral without evaluating it.
- Why is \(\frac{\pi}{8}[g(x)-f(x)]^2\), rather than \(\frac{\pi}{8}[g(x)^2-f(x)^2]\), the appropriate area when \(g\) is above \(f\)?
- If coordinates are measured in metres, what units should the cross-sectional area and the resulting volume have?
