From a Base Slice to a Semicircular Cross Section
A solid can be described by the two-dimensional shapes that appear when it is sliced. In this setting, each slice is a semicircle, and the segment cut from the base region is the semicircle’s diameter. The important step is therefore to express the length of that segment as a function of the slicing variable.
Once that length is known, the area of each cross section follows from a fixed formula. Integrating those areas across the region’s bounds gives the volume. This uses the slicing principle: the volume is the integral of cross-sectional area with respect to distance through the solid.
Deriving the Area Formula
Let \(d\) be the diameter of a semicircle, with \(d\ge 0\). Its radius is half its diameter, so \(r=d/2\). A semicircle has half the area of a full circle. Since a circle of radius \(r\) has area \(\pi r^2\), the semicircle’s area is:
This derivation explains why the coefficient is \(\pi/8\), rather than \(\pi/4\): the radius is \(d/2\), and the shape is only half of a circle. Because \(d\) is squared, the cross-sectional area is proportional to the square of the base-slice length.
Choosing the Slice Length and Bounds
For vertical slices, the diameter is the vertical distance across the base region: upper boundary minus lower boundary. For horizontal slices, it is the horizontal distance: right boundary minus left boundary. The length must be nonnegative throughout the chosen interval. If the region changes boundaries or the order of its edges, the area expression may need to change as well.
When the cross sections are perpendicular to the \(x\)-axis, integrate with respect to \(x\); when they are perpendicular to the \(y\)-axis, integrate with respect to \(y\). The corresponding volume integral is the area of each cross section accumulated over the interval:
Here the bounds describe the full interval of slices through the base region. The integration variable measures distance, so if lengths are measured in units, \(A\) has units squared and the resulting volume has units cubed.
Use vertical slices for cross sections perpendicular to the \(x\)-axis, and horizontal slices for cross sections perpendicular to the \(y\)-axis.
Subtract the lower boundary from the upper boundary, or the left boundary from the right boundary, as appropriate.
Substitute the diameter into \(A=\frac{\pi}{8}d^2\).
Integrate the area expression over the full interval of slices to obtain the volume.
Worked Examples
Worked Example: Vertical Slices Under a Parabola
A base region is bounded by \(y=0\), \(y=4-x^2\), \(x=0\), and \(x=2\). Cross sections perpendicular to the \(x\)-axis are semicircles whose diameters are the vertical segments in the base. Find the volume.
At each \(x\) from \(0\) to \(2\), the upper boundary is \(4-x^2\) and the lower boundary is \(0\). Thus the diameter is:
The cross-sectional area is therefore:
Integrate this area over the interval \(0\le x\le 2\). First expand the square:
An antiderivative of the polynomial is \(16x-\frac{8}{3}x^3+\frac{1}{5}x^5\). Evaluating at both bounds gives:
Check: Differentiating the antiderivative gives \(16-8x^2+x^4\), the expanded integrand. At \(x=0\), the diameter is \(4\); at \(x=2\), it is \(0\), consistent with the base region tapering to a point. The volume is \(\frac{32\pi}{15}\) cubic units.
Worked Example: Horizontal Slices
A base region is bounded by \(x=0\), \(x=1-y\), \(y=0\), and \(y=1\). Cross sections perpendicular to the \(y\)-axis are semicircles whose diameters are the horizontal segments in the base. Find the volume.
For \(0\le y\le 1\), the right boundary is \(x=1-y\) and the left boundary is \(x=0\). The horizontal diameter is:
Thus each cross-sectional area is:
Integrate with respect to \(y\) over the stated bounds. Expanding the square gives \((1-y)^2=1-2y+y^2\), so:
Check: The derivative of \(y-y^2+\frac{y^3}{3}\) is \(1-2y+y^2\), as required. The diameter is \(1\) at \(y=0\) and \(0\) at \(y=1\), matching the region’s horizontal widths. The volume is \(\frac{\pi}{24}\) cubic units.
Worked Example: A Diameter Between Two Curves
The base region is bounded by \(y=x\) and \(y=x^2\) for \(0\le x\le 1\). Cross sections perpendicular to the \(x\)-axis are semicircles whose diameters are the vertical segments in the region. Find the volume.
On \(0\le x\le 1\), \(x\ge x^2\), so \(y=x\) is the upper boundary and \(y=x^2\) is the lower boundary. The diameter is:
Substitute this length into the semicircle area formula and square the entire diameter:
Integrating from \(0\) to \(1\) gives:
Check: Differentiating \(\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\) gives \(x^2-2x^3+x^4\), which is the expanded area integrand after factoring out \(\frac{\pi}{8}\). The diameter is \(0\) at both endpoints and positive between them, as the two boundaries meet at \(x=0\) and \(x=1\). The resulting volume is \(\frac{\pi}{240}\) cubic units.
Common Pitfall: Using the Diameter as the Radius
The diameter is not the radius. If a slice has diameter \(d\), its radius is \(d/2\). A common error is to use the full-circle area formula with \(d\) in place of the radius, or to forget that a semicircle is half of a circle.
For example, in the first worked example the diameter is \(4-x^2\). Incorrectly treating it as the radius would give:
which is twice the correct semicircle area. Integrating that incorrect expression would give \(\frac{64\pi}{15}\), not \(\frac{32\pi}{15}\). The correction is to use radius \(\frac{4-x^2}{2}\) and take half the circle’s area:
A second useful check is to test a simple slice. If its diameter is \(2\), its radius is \(1\), and its semicircle area must be \(\pi/2\). The formula gives \(\frac{\pi}{8}(2)^2=\frac{\pi}{2}\), confirming the coefficient.
Why the Setup Matters
The shape of the cross section alone does not determine its area; its diameter changes from slice to slice. Finding the correct slice length connects the two-dimensional base region to the three-dimensional solid. A reversed boundary subtraction can produce a negative length, while using the wrong slice direction can give a distance that is not the diameter described in the problem.
Before integrating, check that the diameter is a nonnegative distance on the entire interval and that it matches the boundaries of the base. Then square that whole distance in the area formula. The final integral accumulates the cross-sectional areas across the region’s bounds, so its result has units of volume.
Check Your Understanding
For each question, identify the diameter before using the semicircle area formula.
- A vertical slice runs from \(y=0\) to \(y=3-x\). Write its semicircular cross-sectional area as a function of \(x\).
- A horizontal slice runs from \(x=y^2\) to \(x=2\). Write the diameter as a function of \(y\), and state the condition needed for it to be nonnegative.
- Why does a semicircle with diameter \(d\) have area \(\frac{\pi}{8}d^2\), rather than \(\frac{\pi}{4}d^2\)?
- If \(d(x)=2x\) for \(0\le x\le 1\), set up, but do not evaluate, the volume integral for semicircular cross sections perpendicular to the \(x\)-axis.
- For the region between \(y=x\) and \(y=x^2\) on \(0\le x\le 1\), explain why the vertical diameter is \(x-x^2\), not \(x^2-x\).
