Why Compare Slices?
Two solids can have very different shapes while containing the same amount of space. Directly measuring or describing each whole solid may be difficult, but it can be easier to compare their cross sections: the planar regions formed by cutting them at corresponding heights.
The key requirement is that the cross-sectional areas agree at every height in a common range. The shapes of the cross sections themselves need not agree. When the area functions match, integrating them over the height gives equal volumes.
Area as a Function of Height
Choose a vertical coordinate \(y\), and suppose each solid extends from height \(y=a\) to height \(y=b\). At height \(y\), cut the solid with a plane perpendicular to the \(y\)-axis. Write \(A(y)\) for the area of the resulting cross section. For a second solid, write \(B(y)\) for its cross-sectional area at that same height.
The interval and direction matter. A slice at height \(y\) in one solid must be compared with a slice at that same height in the other solid, using planes perpendicular to the same axis. If the solids extend over different intervals, one can sometimes choose a larger common interval and assign area zero where a solid has no slice, provided the resulting area functions are integrable.
The Equal-Slice Theorem
The theorem does not require the slices to have the same shape, orientation, or position within their planes. It requires their areas to match height by height. The condition that both solids use the same height interval ensures that the volume integrals cover corresponding slices throughout the solids.
Why the Theorem Works
The slice principle expresses each volume as the integral of its cross-sectional area. The conclusion then follows from the fact that equal integrable functions have equal integrals. The following partition argument makes the connection explicit.
Divide \([a,b]\) into subintervals with endpoints \(a=y_0<y_1<\cdots<y_n=b\). The thickness of the \(i\)th subinterval is \(\Delta y_i=y_i-y_{i-1}\).
Within a subinterval, a thin slab has approximate volume equal to its cross-sectional area times its thickness. Choosing a sample height \(y_i^*\) in that subinterval gives the sums \(\sum_{i=1}^n A(y_i^*)\Delta y_i\) and \(\sum_{i=1}^n B(y_i^*)\Delta y_i\).
Since \(A(y)=B(y)\) at every height, each corresponding pair of summands is equal: \(A(y_i^*)\Delta y_i=B(y_i^*)\Delta y_i\). Thus the two sums agree for every such partition and sample choice.
As the largest subinterval width tends to zero, the Riemann sums converge to the volume integrals. The equal sums therefore give equal limits.
In symbols, the argument is
The slab sums explain why the area comparison has geometric meaning; the integrals make the conclusion exact. This is often called the slice-area principle, or Cavalieri's principle in its equal-volume form.
Worked Example: Upright and Oblique Cylinders
Worked Example: Upright and Oblique Cylinders
Compare an upright circular cylinder of radius \(2\) and height \(5\) with an oblique cylinder made by translating a circular disk of radius \(2\) horizontally as it rises through height \(5\). Assume the translation does not change the disk's size.
Every horizontal cross section of the upright cylinder is a disk of radius \(2\). Its area is therefore
At each height, the oblique cylinder also has a disk of radius \(2\); its center may be shifted, but translation does not change area. Thus
The area functions agree throughout the common interval, so the equal-slice theorem applies. Integrating gives the volume of either cylinder:
As a check, the upright-cylinder formula gives \(\pi r^2h=\pi(2)^2(5)=20\pi\). For the oblique cylinder, its horizontal slices give the same integral, \(20\pi\). Both volumes are \(20\pi\) cubic units, approximately \(62.832\) cubic units to three decimal places.
This comparison illustrates why the theorem concerns areas rather than the apparent lean or overall outline of a solid. A horizontal shift of each slice changes its location, but not its area.
Worked Example: Square Slices and Rectangular Slices
Worked Example: Square Slices and Rectangular Slices
Consider two solids, each extending from \(y=0\) to \(y=2\). In the first, the horizontal cross section at height \(y\) is a square of side length \(y\). In the second, it is a rectangle with side lengths \(y/2\) and \(2y\). At \(y=0\), both slices degenerate to area zero.
The square's area is
The rectangle's area is the product of its side lengths:
Thus \(A(y)=B(y)\) for every \(y\in[0,2]\), although the cross-sectional shapes are different. The common volume is
To check the integration, differentiate the antiderivative: \(\frac{d}{dy}(y^3/3)=y^2\), the integrand. The first solid has volume \(8/3\) cubic units, and the second also has volume \(8/3\) cubic units by the equal-slice theorem. The result depends on equal areas, not on square and rectangle slices being interchangeable as shapes.
Worked Example: A Sphere and a Cylinder With a Double Cone Removed
Worked Example: A Sphere and a Cylinder With a Double Cone Removed
Compare a sphere of radius \(3\), centered at height \(y=0\), with a cylinder of radius \(3\) and height \(6\), extending from \(y=-3\) to \(y=3\), after removing a double cone from its interior. The double cone consists of two cones meeting at their common apex at \(y=0\), with their circular bases at \(y=-3\) and \(y=3\).
At height \(y\), the sphere's circular cross section has radius \(\sqrt{9-y^2}\), so its area is
The cylinder's cross-sectional disk has area \(9\pi\). In either cone, the radius of a horizontal circular slice at height \(y\) is \(|y|\): the radius grows linearly from \(0\) at the apex to \(3\) at the base. The removed slice therefore has area \(\pi|y|^2=\pi y^2\). The area remaining in the cylinder is
Hence \(A(y)=B(y)\) at every height in the shared interval. The sphere's volume from its slices is
For an independent check using the cylinder and cones, the cylinder has volume \(\pi(3)^2(6)=54\pi\). Each removed cone has volume \(\frac13\pi(3)^2(3)=9\pi\), so the two cones together have volume \(18\pi\). The remaining volume is \(54\pi-18\pi=36\pi\), as the slice comparison predicts. The sphere's volume is \(36\pi\) cubic units, approximately \(113.097\) cubic units to three decimal places.
What Must Match—and What Need Not
For a valid comparison, identify the direction of slicing, the common height interval, and the area of each slice as a function of height. Then compare the two area functions at corresponding heights. The cross sections do not have to be congruent, and the solids do not have to have matching boundaries or align in a visually similar way.
However, matching a few selected cross sections is not enough. The integral accumulates area throughout the interval, so a difference over even part of the height can change the total volume.
For a concrete illustration, suppose two integrable area functions on \([0,1]\) are
They agree at the endpoints, since \(A(0)=B(0)=1\) and \(A(1)=B(1)=1\). But their volumes differ:
Thus checking only the bottom and top slices would miss a difference in the interior. A reliable comparison establishes equality of the full area functions, not just equality at selected sample heights.
Why the Principle Is Useful
The slice-area principle allows a volume comparison without constructing a direct correspondence between points inside two solids. It is especially useful when one solid has a familiar description and another has awkward-looking geometry, but their slices are easy to compare. The sphere and cylinder-with-cones example is a strong instance: each slice calculation is simple, and matching the area functions proves the volume relation for the whole solids.
In practice, write down \(A(y)\) and \(B(y)\) before making a conclusion. Check that they refer to the same slicing direction and height, simplify both expressions, and verify equality throughout the interval. If they match and are integrable, the volume conclusion follows; if they do not, their integrals must be compared directly rather than assuming the solids have equal volume.
Check Your Understanding
Answer these questions using the area functions and conditions in the discussion.
- What must be shared by the two solids' slicing setup for the equal-slice theorem to apply?
- Must the cross sections at a given height have the same shape? Explain.
- Two solids have equal cross-sectional areas at the bottom and top but not at intermediate heights. Is that enough to conclude equal volumes? Why?
- A solid has cross-sectional area \(A(y)=3y\) for \(0\le y\le 4\). What volume does the slice integral give?
- In the sphere and cylinder-with-double-cone example, why does the removed slice have area \(\pi y^2\), including when \(y\) is negative?
