Volumes by Slicing: Cross Sections and Examples

What Are Volumes By Slicing?

Volumes by slicing is the method of finding a solid's volume by cutting it into thin slices, writing the area of each slice as a function, and integrating that area along the solid. If a solid stretches from $x = a$ to $x = b$, and the cross-section cut by the plane perpendicular to the x-axis at $x$ has area $A(x)$, then

$$V = \int_{a}^{b} A(x),dx$$

Here is what each symbol means:

• $V$ is the volume of the solid, in cubic units.

• $A(x)$ is the area of the cross-section at position $x$, in square units.

• $dx$ is the thickness of one slice, so $A(x),dx$ is the volume of one thin slab.

• $a$ and $b$ are the positions of the first and last slices.

The only condition is that $A(x)$ must be integrable on $[a, b]$, which holds whenever it is continuous there. Nothing requires the solid to be round. That is the point of the method: the disk and washer methods are the special cases where every slice happens to be a circle or a ring.

How Is The Slicing Formula Derived?

The slicing formula is a definite integral built from a Riemann sum of slab volumes. Each slab is close to a prism, and a prism's volume is its face area times its thickness.

1. Split $[a, b]$ into $n$ pieces of width $\Delta x = \frac{b - a}{n}$, with points $x_0 = a, x_1, \ldots, x_n = b$.

2. Cut the solid at each point. The $k$-th slab has thickness $\Delta x$.

3. Pick a point $x_k^{}$in each piece. The slab's volume is about$A(x_k^{}),\Delta x$.

4. Add the slabs. In summation notation, $V \approx \sum_{k=1}^{n} A(x_k^{*}),\Delta x$.

5. Let $n \to \infty$. The sum becomes exact:

$$V = \lim_{n \to \infty} \sum_{k=1}^{n} A(x_k^{*}),\Delta x = \int_{a}^{b} A(x),dx$$

The formula also has a neat reading. Since the average value of a function $A$ on $[a, b]$ is $\frac{1}{b - a}\int_{a}^{b} A(x),dx$, the volume equals the solid's length times its average cross-section area. A square pyramid of side 6 and height 9 has volume 108, so its average slice area is $\frac{108}{9} = 12$.

How Do You Set Up A Volume By Slicing Problem?

Every slicing problem follows the same four steps, and the real work is step 2.

1. Draw the base and one slice. Sketch the base region, then draw one slice standing on it, perpendicular to the axis the problem names.

2. Find the slice's key length. Measure the side, base or diameter of the slice as a function of the slicing variable. For a slice perpendicular to the x-axis this is usually "top curve minus bottom curve".

3. Write the area $A(x)$. Put that length into the area formula for the slice's shape.

4. Integrate over the base. The limits are where the base region starts and stops along the slicing axis.

One adjustment matters. If the slices are perpendicular to the y-axis, the slice is described as a function of $y$, its length is "right curve minus left curve", and you integrate $A(y),dy$ between y-limits. My first instinct when writing these steps was to say "integrate in $x$". Actually, the variable is whichever axis the slices are perpendicular to; the base region only tells you the limits.

What Cross-Section Shapes Come Up Most Often?

Textbook problems use a short list of shapes, each with a fixed area formula in terms of one length $s$. Here $s$ is the length of the slice's edge that lies on the base.

The semicircle row trips up more students than any other, because the base length is the diameter and the radius is $\frac{s}{2}$. The semicircle formulas give $\frac{1}{2}\pi\left(\frac{s}{2}\right)^{2} = \frac{\pi}{8}s^{2}$, not $\frac{\pi}{2}s^{2}$.

How Are The Disk And Washer Methods Special Cases Of Slicing?

The disk and washer methods are volumes by slicing with circular slices. When a region under $y = f(x)$ spins about the x-axis, every slice is a disk of radius $f(x)$, so $A(x) = \pi\left(f(x)\right)^{2}$. When the region has a hole, every slice is a ring, so $A(x) = \pi\left(R(x)^{2} - r(x)^{2}\right)$.

That is all those methods are. The full technique, with axis choices and offset axes, lives on the pages for the disk method and the washer method, and the wider family is on volume of solids of revolution. Example 4 below puts the two side by side on the same base.

The reverse is not true. Most slicing problems have no axis of revolution at all: a solid with square slices standing on a parabola cannot be made by spinning anything. That is why the general formula comes first and the round cases come second.

What Is Cavalieri's Principle?

Cavalieri's principle says that two solids between the same pair of parallel planes have equal volumes if every plane parallel to those planes cuts them in cross-sections of equal area. In calculus terms, the volume depends only on the area function $A(x)$, not on the shape or position of each slice.

$$A_{1}(x) = A_{2}(x) \text{ for every } x \text{ in } [a, b] \quad \Longrightarrow \quad V_{1} = V_{2}$$

A stack of coins shows it. Push the stack so it leans, and every coin keeps its area, so the volume does not change. The same reason makes an oblique prism or cylinder have the same volume as a straight one with the same base and height.

• The sphere, by comparison. Put a hemisphere of radius $r$ beside a cylinder of radius $r$ and height $r$ with a cone removed from it. At height $y$, the hemisphere slice has area $\pi(r^{2} - y^{2})$, and the ring left in the cylinder has area $\pi r^{2} - \pi y^{2}$. The slices match at every height.

• So the volumes match. The hemisphere's volume is the cylinder minus the cone, $\pi r^{3} - \frac{1}{3}\pi r^{3} = \frac{2}{3}\pi r^{3}$, and a whole sphere has volume $\frac{4}{3}\pi r^{3}$.

• The napkin ring. Drill a hole through the centre of a ball, leaving a band of height $h$. Every slice of the band has area $\pi\left(\frac{h^{2}}{4} - y^{2}\right)$, with no trace of the ball's radius, so the band's volume is $\frac{\pi h^{3}}{6}$ for a marble or for a planet.

What Cavalieri meant by an "indivisible" is harder to pin down than the principle itself, and this is one place where the history is more slippery than the mathematics. The integral version above is the clean modern form, and in multivariable calculus it becomes the idea behind iterated integrals.

Examples Of Volumes By Slicing

Example 1

A pyramid has a square base of side 6 and height 9. Find its volume by slicing.

Slice parallel to the base. At height $y$ above the tip, similar triangles give a square of side $\frac{6y}{9} = \frac{2y}{3}$:

$$A(y) = \left(\frac{2y}{3}\right)^{2} = \frac{4y^{2}}{9}$$

$$V = \int_{0}^{9} \frac{4y^{2}}{9},dy = \frac{4}{9} \cdot \frac{9^{3}}{3}$$

$$V = 108$$

This matches the volume formula for a pyramid, $\frac{1}{3}(\text{base area})(\text{height}) = \frac{1}{3}(36)(9) = 108$, which slicing has just proved.

Final answer: $108$ cubic units.

Example 2

The base is the region between $y = x$ and $y = x^{2}$ for $0 \leq x \leq 1$. Cross-sections perpendicular to the x-axis are semicircles. Find the volume.

The slice stands on the segment from $y = x^{2}$ up to $y = x$, of length $s = x - x^{2}$. The tempting move is to treat that length as the radius:

$$V_{\text{wrong}} = \int_{0}^{1} \frac{\pi}{2}\left(x - x^{2}\right)^{2},dx = \frac{\pi}{60} \approx 0.0524$$

Test it against a simple bound. The widest slice has length $\frac{1}{4}$ at $x = \frac{1}{2}$, so every slice fits inside a semicircle of diameter $\frac{1}{4}$, with area $\frac{\pi}{128}$. The solid is 1 unit long, so its volume is at most $\frac{\pi}{128} \approx 0.0245$. The answer $0.0524$ is more than double that ceiling, so it cannot be right.

The segment is the diameter, and the radius is half of it:

$$A(x) = \frac{\pi}{8}\left(x - x^{2}\right)^{2} = \frac{\pi}{8}\left(x^{2} - 2x^{3} + x^{4}\right)$$

$$V = \frac{\pi}{8}\left[\frac{x^{3}}{3} - \frac{x^{4}}{2} + \frac{x^{5}}{5}\right]_{0}^{1} = \frac{\pi}{8} \cdot \frac{1}{30}$$

$$V = \frac{\pi}{240}$$

Final answer: $\frac{\pi}{240} \approx 0.0131$ cubic units, a quarter of the wrong value.

Example 3

The base is the disk $x^{2} + y^{2} \leq 4$. Find the volume when cross-sections perpendicular to the x-axis are (a) squares, (b) equilateral triangles, (c) semicircles.

This answers the Brainly question "Find the volume of the solid whose base is bounded by the circle x^2 + y^2 = 4 with the indicated cross sections". At position $x$, the slice stands on the full chord from $y = -\sqrt{4 - x^{2}}$ to $y = \sqrt{4 - x^{2}}$:

$$s = 2\sqrt{4 - x^{2}}, \qquad s^{2} = 4\left(4 - x^{2}\right)$$

$$\int_{-2}^{2} 4\left(4 - x^{2}\right),dx = 4\left(16 - \frac{16}{3}\right) = \frac{128}{3}$$

(a) Squares: $V = \frac{128}{3} \approx 42.6667$.

(b) Equilateral triangles: $V = \frac{\sqrt{3}}{4} \cdot \frac{128}{3} = \frac{32\sqrt{3}}{3} \approx 18.4752$.

(c) Semicircles: $V = \frac{\pi}{8} \cdot \frac{128}{3} = \frac{16\pi}{3} \approx 16.7552$, which is exactly half of a ball of radius 2.

The step students skip here is doubling the half-chord. Writing $s = \sqrt{4 - x^{2}}$, the distance from the x-axis to the circle, quarters every answer, because $s$ gets squared.

Final answer: (a) $\frac{128}{3}$, (b) $\frac{32\sqrt{3}}{3}$, (c) $\frac{16\pi}{3}$ cubic units.

Example 4

The base is the region under $y = \sqrt{x}$ for $0 \leq x \leq 4$. Find the volume when cross-sections perpendicular to the x-axis are squares, and compare it with spinning the same region about the x-axis.

The square stands on the segment from $0$ to $\sqrt{x}$, so $A(x) = \left(\sqrt{x}\right)^{2} = x$:

$$V_{\text{squares}} = \int_{0}^{4} x,dx = 8$$

Spinning the region instead makes every slice a disk of radius $\sqrt{x}$, so $A(x) = \pi x$:

$$V_{\text{disks}} = \int_{0}^{4} \pi x,dx = 8\pi$$

Same base, same limits, same integral up to the factor in front. The only thing that changed is the area formula for one slice.

Final answer: $8$ cubic units with squares, $8\pi \approx 25.1327$ with disks.

Example 5

The base is the region under $y = e^{-x^{2}}$ for $0 \leq x \leq 1$. Cross-sections perpendicular to the x-axis are squares. Find the volume.

$$A(x) = \left(e^{-x^{2}}\right)^{2} = e^{-2x^{2}}$$

$$V = \int_{0}^{1} e^{-2x^{2}},dx$$

This integral is non-elementary: $e^{-2x^{2}}$ has no antiderivative made of polynomials, roots, exponentials, logs and trig functions. Its exact value uses the error function, $\frac{\sqrt{2\pi}}{4}\operatorname{erf}\left(\sqrt{2}\right)$, which is itself defined by an integral. The practical answer comes from numerical integration:

$$\text{Simpson's rule, } n = 4: \quad V \approx 0.5981$$

$$\text{Simpson's rule, } n = 8: \quad V \approx 0.5981$$

The two estimates agree to four decimal places ($0.59808$ and $0.59814$), and high-precision quadrature gives $0.598144$.

Final answer: $V \approx 0.5981$ cubic units (no elementary closed form).

Example 6

A wedge is cut from a cylinder of radius 3 m by two planes: one perpendicular to the cylinder's axis, the other crossing it at 45° along a diameter. Find the wedge's volume.

This is a Physics Forums thread, "Volume by Slicing: A cylindrical wedge, help needed please", where the poster could not find the width of the slice. Put the diameter on the x-axis, with the half-disk base above it, and slice perpendicular to the diameter. Each slice is a right triangle with one leg along the base, of length $\sqrt{9 - x^{2}}$, and the other leg its height, $\sqrt{9 - x^{2}}\tan 45^{\circ} = \sqrt{9 - x^{2}}$:

$$A(x) = \frac{1}{2}\left(9 - x^{2}\right)$$

$$V = \int_{-3}^{3} \frac{1}{2}\left(9 - x^{2}\right),dx = \frac{1}{2}\left(54 - 18\right)$$

$$V = 18$$

Slicing the other way, parallel to the diameter, gives rectangles of width $2\sqrt{9 - y^{2}}$ and height $y$, and $\int_{0}^{3} 2y\sqrt{9 - y^{2}},dy = 18$ as well. Two directions, one volume.

Final answer: $18,\text{m}^{3}$.

Why Do Volumes By Slicing Matter?

"Most solids are not spun on a lathe" is the problem this method solves. Before slicing, volume formulas existed shape by shape: one for the cube, one for the cylinder, one for the cone, each proved separately.

• Archimedes compared the sphere with a cylinder. Archimedes proved that a sphere's volume is two-thirds of its circumscribing cylinder, and his Method says some results first became clear to him "by a mechanical method" before he proved them geometrically.

• Cavalieri turned slices into a rule. Comparing solids slice by slice gave volumes for shapes with no formula at all.

• Calculus made the rule a computation. $V = \int A(x),dx$ turns "compare the slices" into "integrate the slice area", for any solid whose slices you can describe.

The same thinking runs from the area between two curves, where you add thin strips, to volume, where you add thin slabs.

Where Are Volumes By Slicing Used In The Real World?

Any time a solid is known through its cross-sections, slicing is how its volume gets computed.

• Earthworks: road and railway builders survey cross-sections of the ground at intervals along the route and add slab volumes to estimate the earth to cut and fill.

• Medical imaging: CT and MRI scanners produce a stack of image slices, and an organ's volume is estimated by adding each slice's area times the slice spacing.

• 3D printing: a printer builds a part layer by layer, and the software computes material from the area of each layer.

• Architecture and design: domes, roofs and sculptures with square or irregular plans have no rotation axis, so their volumes come from slice areas.

Where Solutions Go Off The Rails With Volumes By Slicing

Mistake 1: Using The Base Width As The Semicircle's Radius

Where it slips in: Semicircle cross-sections, where the slice stands on a segment of the base.

Don't do this: Write $A = \frac{\pi}{2}s^{2}$ with $s$ the segment length; Example 2 shows it makes the volume four times too big.

The correct way: The segment is the diameter, so the radius is $\frac{s}{2}$ and $A = \frac{\pi}{8}s^{2}$. The memorizer who learned "semicircle area is half of $\pi r^{2}$" applies it correctly and still gets this wrong, because nobody asked which length is $r$.

Mistake 2: Slicing In The Wrong Direction

Where it slips in: Problems that say "cross-sections perpendicular to the y-axis".

Don't do this: Write the slice length as top curve minus bottom curve and integrate in $x$ out of habit.

The correct way: Slices perpendicular to the y-axis are horizontal, so their length is right curve minus left curve, written in $y$, with y-limits. The rusher who sets up every problem in $x$ hits this wall first; a ten-second sketch of one slice prevents it.

Mistake 3: Using The Half-Chord On A Circular Base

Where it slips in: Bases bounded by a circle or ellipse, as in Example 3.

Don't do this: Take $s = \sqrt{r^{2} - x^{2}}$, the distance from the axis to the edge.

The correct way: The slice spans the whole chord, $s = 2\sqrt{r^{2} - x^{2}}$. The first instinct is to read the upper half of the circle as the slice, and because $s$ is squared, that single missing factor of 2 shrinks the volume by 4.

Mistake 4: Mixing Units Between The Base And The Height

Where it slips in: Word problems where the base is measured in centimetres and the length in metres.

Don't do this: Multiply an area in square centimetres by a thickness in metres and call the result cubic metres.

The correct way: Convert every length to one unit before writing $A(x)$. Volume carries three length factors, so one wrong unit can throw the answer off by a factor of 100 or more.

Practice Problems On Volumes By Slicing

1. The base is the region between $y = x^{2}$ and $y = 1$. Cross-sections perpendicular to the x-axis are squares. Find the volume.
   Answer to Problem 1: $\frac{16}{15}$ cubic units.

2. The base is the triangle with vertices $(0, 0)$, $(2, 0)$ and $(0, 2)$. Cross-sections perpendicular to the x-axis are semicircles. Find the volume.
   Answer to Problem 2: $\frac{\pi}{3}$ cubic units.

3. The base is a semicircular region of radius 5. Cross-sections perpendicular to the base and parallel to the diameter are squares. Find the volume.
   Answer to Problem 3: $\frac{1000}{3}$ cubic units.

4. The base is the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} \leq 1$. Cross-sections perpendicular to the x-axis are squares. Find the volume.
   Answer to Problem 4: $96$ cubic units.

5. The base is the region between $y = x$ and $y = x^{2}$. Cross-sections perpendicular to the x-axis are isosceles right triangles with one leg on the base. Find the volume.
   Answer to Problem 5: $\frac{1}{60}$ cubic units.

6. A wedge is cut from a cylinder of radius 2 by a plane through a diameter of the base at 30° to the base. Find its volume.
   Answer to Problem 6: $\frac{16\sqrt{3}}{9}$ cubic units.

Key Takeaways

• Volumes by slicing finds any solid's volume by integrating its cross-section area: $V = \int_{a}^{b} A(x),dx$.

• The hard step is the slice: find its length from the base, then turn that length into an area with the right shape formula.

• Semicircles stand on their diameter, and circular bases need the full chord, not the half-chord.

• The disk and washer methods are slicing with circular slices, and Cavalieri's principle says equal slices mean equal volumes.

• When the slice area has no elementary antiderivative, a checked numeric value is the right answer.

Practice These Before Moving On

Work through Practice Problems 1 to 4 first, then 5 and 6; if the slice length trips you up, return to "How Do You Set Up A Volume By Slicing Problem?", and if a semicircle answer looks too big, reread Example 2. Want a live Bhanzu trainer to walk through more slicing problems with you? Book a free demo class:

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