From a Base Chord to a Slice
A solid can be described by its base region and the shapes of its cross-sections. In this tutorial, each cross-section is either a semicircle or a full circle, and its diameter is the chord cut from the base region by the slicing line. The key step is to translate that chord length into an area before integrating.
The chord is not the radius. If its length is \(d\), the radius of the circular slice is \(d/2\). Confusing these quantities changes the area, and therefore the volume, by a constant factor.
We will use vertical slices first. A vertical slice at \(x\) is perpendicular to the \(x\)-axis, and its chord is measured vertically through the base region. Horizontal slices are perpendicular to the \(y\)-axis, and their chords are measured horizontally. In either case, the volume is the integral of the cross-sectional area over the interval that indexes the slices.
Deriving the Slice-Area Formulas
The area of a full circle with radius \(r\) is \(\pi r^2\). Since the chord is the diameter, \(r=d/2\). Substitution gives
A semicircle is one half of a circle with the same radius. Its area is therefore half the full-circle area:
These formulas explain why the diameter must be identified before integration: the chord length is squared in the area expression. If the chord varies from slice to slice, the area varies too.
Why Integrating the Areas Gives Volume
Suppose the slices perpendicular to the \(x\)-axis fill the interval \([a,b]\), and the cross-sectional area at \(x\) is \(A(x)\). A thin slab of thickness \(\Delta x\) near \(x\) has approximate volume \(A(x)\Delta x\). Adding the volumes of many such slabs approximates the whole solid’s volume. As the slab widths shrink, this sum approaches the definite integral of the cross-sectional area.
For circular or semicircular slices, substitute the corresponding area formula into this volume formula. Thus the chord determines \(A(x)\), and integration accumulates the areas across the solid. For horizontal slices, the same reasoning uses slab thickness \(\Delta y\) and gives \(V=\int_c^e A(y)\,dy\).
This argument requires the slices to cover the solid across the stated interval and the area function to be integrable there. In the examples below, the chord lengths are continuous and nonnegative on their intervals, so the area functions are continuous and the integrals are well-defined.
Vertical Slices
Suppose the base region is bounded above by \(y=u(x)\) and below by \(y=\ell(x)\), for \(a\le x\le b\), with \(u(x)\ge \ell(x)\). The vertical chord length is the top value minus the bottom value:
Once \(d(x)\) is known, use the semicircle or full-circle formula for the cross-sectional area. The limits of integration are the leftmost and rightmost \(x\)-values of the base region.
Worked Example: Semicircular Slices Above a Parabolic Base
The base region is bounded by \(y=0\) and \(y=4-x^2\). Vertical slices form semicircles whose diameters are the base-region chords. Find the volume.
The curves meet where \(4-x^2=0\). Thus \(x^2=4\), so the bounds are \(x=-2\) and \(x=2\). At a given \(x\), the upper boundary is \(4-x^2\) and the lower boundary is \(0\). Therefore, the chord length is
The cross-sections are semicircles, so their area is
Integrate across the interval:
At \(x=2\), the bracketed expression is \(32-\frac{64}{3}+\frac{32}{5}=\frac{256}{15}\). At \(x=-2\), it is \(-32+\frac{64}{3}-\frac{32}{5}=-\frac{256}{15}\). Their difference is \(\frac{512}{15}\), so
Check: the antiderivative differentiates to \(16-8x^2+x^4=(4-x^2)^2\), and the area factor \(\pi/8\) has been retained. The chord is nonnegative on \([-2,2]\), so every cross-sectional area is nonnegative; the volume is positive as expected.
Horizontal Slices
For horizontal slices, the chord runs from the left boundary to the right boundary at a fixed height \(y\). If these boundaries are \(x=L(y)\) and \(x=R(y)\), with \(R(y)\ge L(y)\), then
The resulting area is integrated with respect to \(y\), over the bottom and top heights of the base region. Notice that horizontal chords are measured in the \(x\)-direction even though the integration variable is \(y\).
Worked Example: Circular Slices From a Triangular Base
Let the base region be bounded by \(x=0\), \(x=y\), \(y=0\), and \(y=2\). Horizontal slices form full circles whose diameters are the base chords. Find the volume.
At each height \(y\) from \(0\) to \(2\), the left boundary is \(x=0\) and the right boundary is \(x=y\). The chord length is therefore
Since the slices are full circles, their cross-sectional area is
Integrating over the height of the region gives
Check: \(\frac{d}{dy}(y^3/3)=y^2\), and \(A(y)\ge0\) for \(0\le y\le2\). The result has the expected volume units because a cross-sectional area is integrated over a length.
Worked Example: Semicircular Slices With a Curved Horizontal Chord
Consider the base region \(0\le y\le1\), \(0\le x\le y^2\). Horizontal slices form semicircles with the base chords as diameters. Find the volume.
At height \(y\), the left endpoint of the chord is \(x=0\), and the right endpoint is \(x=y^2\). Hence
The cross-section is a semicircle, so its area is
Integrate from the bottom height \(0\) to the top height \(1\):
Check: differentiating \(y^5/5\) gives \(y^4\), and the chord and cross-sectional area are nonnegative throughout the interval. At \(y=0\), the chord and area are zero, consistent with the boundary geometry.
A Common Diameter-and-Radius Mistake
A frequent error is to treat the chord length as the radius. For example, if a semicircular slice has chord length \(d\), an incorrect area calculation is \(\frac{\pi d^2}{2}\). That expression uses \(d\) as the radius and also halves the circle, so it overstates the correct area by a factor of four.
The correction is to identify the radius first: \(r=d/2\). Then substitute that radius into the circle-area formula, and halve only if the cross-section is a semicircle. A second check is geometric: a semicircle cannot have more area than the full circle with the same diameter.
Choosing and Checking the Integral
Before integrating, verify that the chord length corresponds to the direction of the slices. For vertical slices, subtract bottom from top and integrate in \(x\). For horizontal slices, subtract left from right and integrate in \(y\). In both cases, square the chord length only after writing it as a nonnegative length, and use \(\pi/8\) for a semicircle or \(\pi/4\) for a full circle.
Decide whether the slices are vertical or horizontal; this determines the integration variable.
Subtract bottom from top for a vertical chord, or left from right for a horizontal chord.
Use \(\pi d^2/8\) for semicircles and \(\pi d^2/4\) for full circles, because \(d\) is the diameter.
Use the endpoint values of the coordinate indexing the slices, then check the antiderivative and the sign of the result.
Units provide another useful check. If the base coordinates are measured in units of length, then \(d^2\) has units of area. Integrating an area with respect to a length gives units of volume. The formulas describe geometric cross-sections, not slices created by rotating the base region; the shape of each slice is specified directly as a semicircle or full circle.
Check Your Understanding
For each question, identify the chord length before deciding on the area formula.
- A vertical slice has a base chord of length \(3\). What are the areas of the semicircular and full circular slices with that diameter?
- A base region has top boundary \(y=5-x\) and bottom boundary \(y=1\) on \(0\le x\le4\). Write the chord length for vertical slices.
- A horizontal chord extends from \(x=1\) to \(x=2y+1\). Write its length as a function of \(y\), and give the area of a full circular slice.
- For semicircular slices with horizontal chord length \(d(y)=y\) on \(0\le y\le3\), write the volume integral. You do not need to evaluate it.
- Why is \(\pi d^2/2\) not the correct area of a semicircle whose diameter is \(d\)?
