Chapter 6 · Applications of Integration

6.2 Volumes by Slicing: Disks and Washers

Slice a solid like a loaf of bread, add up the slice areas: $V=\int_a^b A(x)\,dx$. For solids of revolution the slices are disks $\pi R^2$ or washers $\pi\left(R^2-r^2\right)$.

How much water does a vase hold? How much concrete goes into a tapered column, or how large is a tumour seen on a series of medical scans? For a box or a cylinder we multiply a base area by a height. Most real objects are not so obliging: their cross-sections change size and shape as we move along them.

The idea that rescues us is the one from Section 6.1, now in three dimensions. Cut the solid into thin slices perpendicular to an axis, like a loaf of bread. Each slice is almost a thin cylinder (a slab with constant cross-section), whose volume is area $\times$ thickness. Add up the slices and take the limit as the slices become infinitely thin: the sum turns into an integral of the cross-sectional area. When the solid is made by spinning a region about a line, every slice is a disk or a washer, and the area of each slice is simply $\pi\times(\text{radius})^2$.

Volume by Slicing

Let $S$ be a solid lying between the planes $x=a$ and $x=b$, and let $A(x)$ be the area of its cross-section by the plane perpendicular to the $x$-axis at $x$. Cut $[a,b]$ into $n$ pieces of width $\Delta x$ and pick $x_i^*$ in the $i$-th piece. The $i$-th slice is approximately a slab of base area $A(x_i^*)$ and thickness $\Delta x$, so $$V\approx\sum_{i=1}^{n}A(x_i^*)\,\Delta x .$$ This is a Riemann sum for $A$; refining the slices gives an integral.

📘 Definition 6.2 — Volume by slicing
Let $S$ be a solid lying between $x=a$ and $x=b$ whose cross-section perpendicular to the $x$-axis at $x$ has area $A(x)$, where $A$ is continuous on $[a,b]$. The volume of $S$ is $$V=\lim_{n\to\infty}\sum_{i=1}^{n}A(x_i^*)\,\Delta x=\int_a^bA(x)\,dx .$$ The same holds with $y$ in place of $x$ when the slices are perpendicular to the $y$-axis: $V=\int_c^dA(y)\,dy$.
🧭 Problem-solving strategy — volume by slicing
  1. Sketch the solid (or its base) and choose an axis. Draw one typical cross-section perpendicular to it.
  2. Express the size of that cross-section (a side, a radius, a height) in terms of the position $x$ — similar triangles and the equations of the boundary curves are the usual tools.
  3. Write $A(x)$ using the area formula for the shape: square $s^2$, equilateral triangle $\tfrac{\sqrt3}{4}s^2$, semicircle $\tfrac{\pi}{2}r^2$, disk $\pi r^2$, …
  4. Integrate $A(x)$ between the extreme positions $a$ and $b$, and check the units ($\text{length}^2\times\text{length}=\text{length}^3$).
✏️ Example 1 — The volume of a pyramid
A pyramid has a square base of side $4$ m and height $6$ m. Find its volume by slicing.
Axis Put the apex at the origin and let the $x$-axis run down the centre of the pyramid to the base at $x=6$. Each cross-section perpendicular to the axis is a square.
Similar triangles The side $s(x)$ grows linearly from $0$ at the apex to $4$ at the base: $\dfrac{s(x)}{x}=\dfrac46$, so $s(x)=\dfrac{2x}{3}$ and $A(x)=\dfrac{4x^2}{9}$.
Integrate $$V=\int_0^6\frac{4x^2}{9}\,dx=\frac49\cdot\frac{x^3}{3}\Big|_0^6=\frac49\cdot72=32\ \text{m}^3 .$$
Check The school formula $V=\tfrac13(\text{base area})(\text{height})=\tfrac13(16)(6)=32$. ✓ Repeating the argument with side $s$ and height $h$ gives $A(x)=\tfrac{s^2x^2}{h^2}$ and $V=\tfrac13s^2h$ — slicing explains where the $\tfrac13$ comes from.
✏️ Example 2 — Triangular cross-sections on a parabolic base
The base of a solid is the region between the parabola $y=4-x^2$ and the $x$-axis. Every cross-section perpendicular to the $x$-axis is an equilateral triangle with one side in the base. Find the volume.
Side The base runs from $x=-2$ to $x=2$. At position $x$ the side of the triangle is the height of the base region, $s(x)=4-x^2$.
Area An equilateral triangle of side $s$ has area $\tfrac{\sqrt3}{4}s^2$, so $A(x)=\dfrac{\sqrt3}{4}\left(4-x^2\right)^2$.
Integrate Using symmetry, $$\int_{-2}^{2}\left(4-x^2\right)^2dx=2\int_0^2\left(16-8x^2+x^4\right)dx=2\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{512}{15},$$ so $$V=\frac{\sqrt3}{4}\cdot\frac{512}{15}=\frac{128\sqrt3}{15}\approx14.78 .$$

The Disk Method

Many solids met in practice — bottles, bowls, nose cones, lenses, turned table legs — are made by spinning a flat shape about a line.

📘 Definition 6.3 — Solid of revolution
Let $\mathcal R$ be a region in the plane and $\ell$ a line in the same plane that does not pass through the interior of $\mathcal R$. The solid swept out when $\mathcal R$ is revolved once about $\ell$ is called a solid of revolution, and $\ell$ is its axis of revolution.

If the region lies between the curve $y=f(x)$ and the $x$-axis and we spin it about the $x$-axis, every cross-section perpendicular to the axis is a disk of radius $|f(x)|$. Its area is $A(x)=\pi[f(x)]^2$, and Definition 6.2 gives the volume at once.

📐 Theorem 6.3 — The disk method

Let $f$ be continuous on $[a,b]$, and let $S$ be the solid obtained by revolving the region between $y=f(x)$ and the $x$-axis, $a\le x\le b$, about the $x$-axis. Then $$V=\int_a^b\pi\left[f(x)\right]^2dx .$$

Likewise, if $g$ is continuous on $[c,d]$ and the region between $x=g(y)$ and the $y$-axis, $c\le y\le d$, is revolved about the $y$-axis, then $V=\displaystyle\int_c^d\pi\left[g(y)\right]^2dy$.

✏️ Example 3 — The volume of a sphere
Show that a sphere of radius $r$ has volume $\tfrac43\pi r^3$.
Model The sphere is obtained by revolving the upper half-disk under $y=\sqrt{r^2-x^2}$, $-r\le x\le r$, about the $x$-axis.
Disk area $A(x)=\pi\left(\sqrt{r^2-x^2}\right)^2=\pi\left(r^2-x^2\right)$ — the square root disappears, which is why disks are pleasant.
Integrate $$V=\int_{-r}^{r}\pi\left(r^2-x^2\right)dx=\pi\left[r^2x-\frac{x^3}{3}\right]_{-r}^{r}=\pi\left(\frac{2r^3}{3}+\frac{2r^3}{3}\right)=\frac43\pi r^3 .$$
Check For $r=2$ this gives $\tfrac{32\pi}{3}\approx33.51$; choose the sphere in Interactive 1 and watch the disk sums approach it. ✓
Interactive 1 — Building a Solid Disk by Disk

The curve $y=f(x)$ is revolved about the $x$-axis. Press Play to stack the disks one at a time — each has radius $f(x_i)$ (right end point) and thickness $\Delta x$, so volume $\pi[f(x_i)]^2\Delta x$. After each pass the number of disks doubles and the stack hugs the true solid more closely.

8
disks stacked—
running sum $\sum\pi r_i^2\,\Delta x$—
exact volume—
error (full stack)—
✏️ Example 4 — Revolving about the $y$-axis
The region bounded by $y=x^3$, the line $y=8$ and the $y$-axis is revolved about the $y$-axis. Find the volume.
Slice direction The axis is vertical, so the slices must be horizontal (perpendicular to the $y$-axis) and we integrate in $y$, from $y=0$ to $y=8$.
Radius A slice at height $y$ is a disk reaching from the $y$-axis to the curve. Solving $y=x^3$ for $x$: radius $=x=y^{1/3}$.
Integrate $$V=\int_0^8\pi\left(y^{1/3}\right)^2dy=\pi\int_0^8y^{2/3}\,dy=\pi\cdot\frac35y^{5/3}\Big|_0^8=\frac35\cdot32\pi=\frac{96\pi}{5}\approx60.32 .$$

The Washer Method

If the region being revolved does not touch the axis, each slice has a hole in the middle: it is a washer (an annulus), with an outer radius $R$ (distance from the axis to the far edge of the region) and an inner radius $r$ (distance to the near edge). Its area is the big disk minus the hole, $\pi R^2-\pi r^2$.

📐 Theorem 6.4 — The washer method

Let $f$ and $g$ be continuous on $[a,b]$ with $f(x)\ge g(x)\ge0$. Revolving the region between $y=f(x)$ and $y=g(x)$, $a\le x\le b$, about the $x$-axis gives a solid of volume $$V=\int_a^b\pi\left(\left[f(x)\right]^2-\left[g(x)\right]^2\right)dx .$$

Other axes. If the region is revolved about a horizontal line $y=k$ that does not pass through its interior, let $R(x)$ and $r(x)$ be the distances from the line to the farther and nearer boundary curves. Then $V=\displaystyle\int_a^b\pi\left(R(x)^2-r(x)^2\right)dx$. For a vertical axis $x=h$ the slices are horizontal and $V=\displaystyle\int_c^d\pi\left(R(y)^2-r(y)^2\right)dy$, with the radii measured horizontally from $x=h$.

🧭 Problem-solving strategy — disks and washers
  1. Sketch the region, the axis of revolution, and one slice perpendicular to the axis. Horizontal axis $\Rightarrow$ vertical slices, integrate in $x$. Vertical axis $\Rightarrow$ horizontal slices, integrate in $y$.
  2. Find the limits of integration (usually from intersection points).
  3. Measure $R$ and $r$ from the axis: for the axis $y=k$ lying below the region, $R=\text{top}-k$ and $r=\text{bottom}-k$; for $y=k$ lying above it, $R=k-\text{bottom}$ and $r=k-\text{top}$. If the region touches the axis, $r=0$ (disks).
  4. Integrate $\pi\left(R^2-r^2\right)$.
✏️ Example 5 — A washer about the $x$-axis
The region between $y=x$ and $y=x^2$ is revolved about the $x$-axis. Find the volume.
Region $x=x^2$ at $x=0$ and $x=1$; on $[0,1]$ the line is above the parabola.
Radii A vertical slice at $x$ sweeps a washer with $R(x)=x$ (to the line) and $r(x)=x^2$ (to the parabola).
Integrate $$V=\int_0^1\pi\left(x^2-x^4\right)dx=\pi\left(\frac13-\frac15\right)=\frac{2\pi}{15}\approx0.4189 .$$
⚠️ Common mistakes with washers

1. $\pi(R-r)^2$ is not $\pi\left(R^2-r^2\right)$. In Example 5, $\pi\int_0^1\left(x-x^2\right)^2dx=\tfrac{\pi}{30}$ — only a quarter of the correct $\tfrac{2\pi}{15}$. Squaring the height of the strip gives the area of a disk whose radius is the strip, which is not the washer the strip sweeps out.

2. Radii are distances from the axis, not $y$-values. When the axis is $y=-1$, a curve at height $y$ is at distance $y+1$ from it.

3. Match the variable to the axis. Revolving about a vertical line needs horizontal slices and $dy$ — write the boundary curves as $x$ in terms of $y$ first.

✏️ Example 6 — Revolving about other lines
The region between $y=x$ and $y=x^2$ (as in Example 5) is revolved about (a) the line $y=-1$; (b) the line $x=-1$. Find each volume.
(a) Radii The axis $y=-1$ lies below the region. $R(x)=x-(-1)=x+1$ and $r(x)=x^2+1$.
(a) Integrate $$V=\pi\int_0^1\left[(x+1)^2-\left(x^2+1\right)^2\right]dx=\pi\int_0^1\left(2x-x^2-x^4\right)dx=\pi\left(1-\frac13-\frac15\right)=\frac{7\pi}{15}\approx1.4661 .$$
(b) Radii The axis $x=-1$ is vertical, so slice horizontally. At height $y$ ($0\le y\le1$) the region runs from the line $x=y$ to the parabola $x=\sqrt y$. Distances from $x=-1$: $R(y)=\sqrt y+1$, $r(y)=y+1$.
(b) Integrate $$V=\pi\int_0^1\left[\left(\sqrt y+1\right)^2-(y+1)^2\right]dy=\pi\int_0^1\left(2\sqrt y-y-y^2\right)dy=\pi\left(\frac43-\frac12-\frac13\right)=\frac{\pi}{2}\approx1.5708 .$$
Interactive 2 — Washer Cross-Section Viewer

Left: the region, the axis of revolution (dashed) and the slice at $x$, with the outer radius $R$ and inner radius $r$ measured from the axis. Right: the washer that this slice sweeps out, with area $\pi\left(R^2-r^2\right)$. Change the axis and watch $R$ and $r$ change; the accumulated volume is $\int\pi\left(R^2-r^2\right)dx$ from the left end up to the slice.

region, axis and slice
the washer swept by the slice
slice at $x$—
outer radius $R$—
inner radius $r$—
$A(x)=\pi(R^2-r^2)$—
volume so far / total—

Exercises

I
Volumes by slicing (known cross-sections).
  • 1 The base is the triangle with vertices $(0,0)$, $(4,0)$ and $(0,4)$. Cross-sections perpendicular to the $x$-axis are squares.
  • 2 Same base as part 1; cross-sections perpendicular to the $x$-axis are semicircles with their diameter in the base.
  • 3 The base is the disk $x^2+y^2\le9$; cross-sections perpendicular to the $x$-axis are squares.
  • 4 The base is the disk $x^2+y^2\le9$; cross-sections perpendicular to the $x$-axis are equilateral triangles.
  • 5 The base is the region between $y=x^2$ and $y=1$; cross-sections perpendicular to the $y$-axis are squares.
  • 6 Engineering. A concrete footing is $1.5$ m tall. Its horizontal cross-sections are squares whose side shrinks linearly from $2$ m at the bottom to $1$ m at the top. How much concrete is needed?
1 The hypotenuse is $y=4-x$, so the side is $s=4-x$: $$V=\int_0^4(4-x)^2dx=\left[-\frac{(4-x)^3}{3}\right]_0^4=\frac{64}{3}\approx21.33 .$$
2 Diameter $4-x$, radius $\tfrac{4-x}{2}$, $A(x)=\tfrac{\pi}{2}\left(\tfrac{4-x}{2}\right)^2=\tfrac{\pi}{8}(4-x)^2$: $$V=\frac\pi8\cdot\frac{64}{3}=\frac{8\pi}{3}\approx8.378 .$$
3 The chord at $x$ has length $s=2\sqrt{9-x^2}$, so $A(x)=4\left(9-x^2\right)$: $$V=\int_{-3}^{3}4\left(9-x^2\right)dx=4(54-18)=144 .$$
4 $A(x)=\tfrac{\sqrt3}{4}s^2=\tfrac{\sqrt3}{4}\cdot4\left(9-x^2\right)=\sqrt3\left(9-x^2\right)$: $$V=\sqrt3\int_{-3}^{3}\left(9-x^2\right)dx=36\sqrt3\approx62.35 .$$
5 At height $y$ ($0\le y\le1$) the base runs from $x=-\sqrt y$ to $x=\sqrt y$, so $s=2\sqrt y$ and $A(y)=4y$: $$V=\int_0^14y\,dy=2 .$$
6 At height $h$ m the side is $s(h)=2-\tfrac{2h}{3}$ m. $$V=\int_0^{1.5}\left(2-\frac{2h}{3}\right)^2dh=\left[-\frac12\left(2-\frac{2h}{3}\right)^3\right]_0^{1.5}=-\frac12(1-8)=3.5\ \text{m}^3 .$$ (Check: the frustum formula $\tfrac h3\left(A_1+A_2+\sqrt{A_1A_2}\right)=0.5(4+1+2)=3.5$. ✓)
II
The disk method. Find the volume of the solid obtained by revolving the region about the given axis.
  • 1 $y=x^2$, $y=0$, $0\le x\le2$; about the $x$-axis
  • 2 $y=e^{-x}$, $y=0$, $0\le x\le1$; about the $x$-axis
  • 3 $y=\sin x$, $y=0$, $0\le x\le\pi$; about the $x$-axis
  • 4 $y=\dfrac1x$, $y=0$, $1\le x\le4$; about the $x$-axis
  • 5 $y=\sec x$, $y=0$, $0\le x\le\tfrac\pi4$; about the $x$-axis
  • 6 $y=x^2$, $y=9$, $x=0$ ($x\ge0$); about the $y$-axis
  • 7 $x=\dfrac1y$, $x=0$, $1\le y\le3$; about the $y$-axis
  • 8 Revolve $y=\dfrac{r}{h}x$, $0\le x\le h$, about the $x$-axis to show that a cone has volume $\tfrac13\pi r^2h$. Then find the volume of a cone of radius $3$ cm and height $5$ cm.
  • 9 Sport. Model a rugby ball as the solid formed by revolving the upper half of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ about the $x$-axis. Show $V=\tfrac43\pi ab^2$ and evaluate it for $a=14$ cm, $b=9$ cm.
Check: see the solid and compare a numerical volume with your answer
radius function—
numerical $\pi\int(\text{radius})^2$—
exact answer—
1 $$V=\pi\int_0^2x^4\,dx=\pi\frac{x^5}{5}\Big|_0^2=\frac{32\pi}{5}\approx20.11 .$$
2 $$V=\pi\int_0^1e^{-2x}dx=\pi\left[-\frac12e^{-2x}\right]_0^1=\frac\pi2\left(1-e^{-2}\right)\approx1.358 .$$
3 Use $\sin^2x=\tfrac12(1-\cos2x)$: $$V=\pi\int_0^{\pi}\sin^2x\,dx=\frac\pi2\left[x-\frac{\sin2x}{2}\right]_0^{\pi}=\frac{\pi^2}{2}\approx4.935 .$$
4 $$V=\pi\int_1^4x^{-2}dx=\pi\left[-\frac1x\right]_1^4=\pi\left(1-\frac14\right)=\frac{3\pi}{4}\approx2.356 .$$
5 $$V=\pi\int_0^{\pi/4}\sec^2x\,dx=\pi\Big[\tan x\Big]_0^{\pi/4}=\pi\approx3.142 .$$
6 Horizontal slices, radius $x=\sqrt y$, $0\le y\le9$: $$V=\pi\int_0^9y\,dy=\frac{81\pi}{2}\approx127.2 .$$
7 Radius $\tfrac1y$: $$V=\pi\int_1^3y^{-2}dy=\pi\left(1-\frac13\right)=\frac{2\pi}{3}\approx2.094 .$$
8 $$V=\pi\int_0^h\frac{r^2}{h^2}x^2dx=\frac{\pi r^2}{h^2}\cdot\frac{h^3}{3}=\frac13\pi r^2h .$$ With $r=3$ cm, $h=5$ cm: $V=\tfrac13\pi(9)(5)=15\pi\approx47.12$ cm³.
9 $y^2=b^2\left(1-\tfrac{x^2}{a^2}\right)$, so $$V=\pi\int_{-a}^{a}b^2\left(1-\frac{x^2}{a^2}\right)dx=\pi b^2\left(2a-\frac{2a}{3}\right)=\frac43\pi ab^2 .$$ For $a=14$, $b=9$: $V=\tfrac43\pi(14)(81)=1512\pi\approx4750$ cm³ $\approx4.75$ litres.
III
Washers and other axes. Sketch the region, the axis and a typical slice before you integrate.
  • 1 $y=\sqrt x$ and $y=x$; about the $x$-axis
  • 2 $y=2x$ and $y=x^2$; about the $x$-axis
  • 3 $y=2x$ and $y=x^2$; about the $y$-axis
  • 4 $y=x^2$ and $y=4$; about the line $y=4$
  • 5 $y=x^2$ and $y=4$; about the line $y=-1$
  • 6 $y=x$ and $y=x^2$; about the line $x=1$
  • 7 $y=\sqrt x$, $y=0$ and $x=4$; about the line $x=4$
  • 8 $y=\sin x$ and $y=0$, $0\le x\le\pi$; about the line $y=-1$
  • 9 Design. The inside of a bowl is formed by revolving $y=\tfrac{x^2}{4}$ about the $y$-axis ($x$, $y$ in cm). How much does it hold when filled to a depth of $8$ cm?
  • 10 For the bowl of part 9, find the volume $V(h)$ when the water is $h$ cm deep, and the depth at which the bowl holds exactly half of the amount in part 9.
  • 11 Challenge. A cylindrical hole of radius $1$ cm is drilled along a diameter through a solid ball of radius $2$ cm. Find the volume of the bead that remains.
1 On $[0,1]$, $\sqrt x\ge x$: $R=\sqrt x$, $r=x$. $$V=\pi\int_0^1\left(x-x^2\right)dx=\pi\left(\frac12-\frac13\right)=\frac\pi6\approx0.5236 .$$
2 Meet at $x=0,2$; $R=2x$, $r=x^2$: $$V=\pi\int_0^2\left(4x^2-x^4\right)dx=\pi\left(\frac{32}{3}-\frac{32}{5}\right)=\frac{64\pi}{15}\approx13.40 .$$
3 Horizontal slices, $0\le y\le4$: the region runs from the line $x=\tfrac y2$ to the parabola $x=\sqrt y$, so $R=\sqrt y$, $r=\tfrac y2$: $$V=\pi\int_0^4\left(y-\frac{y^2}{4}\right)dy=\pi\left(8-\frac{16}{3}\right)=\frac{8\pi}{3}\approx8.378 .$$
4 The region touches the axis $y=4$, so the slices are disks of radius $4-x^2$, $-2\le x\le2$: $$V=\pi\int_{-2}^{2}\left(4-x^2\right)^2dx=\frac{512\pi}{15}\approx107.2 .$$
5 Axis below: $R=4-(-1)=5$, $r=x^2+1$: $$V=\pi\int_{-2}^{2}\left[25-\left(x^2+1\right)^2\right]dx=\pi\int_{-2}^{2}\left(24-2x^2-x^4\right)dx=\pi\left(96-\frac{32}{3}-\frac{64}{5}\right)=\frac{1088\pi}{15}\approx227.9 .$$
6 Horizontal slices, $0\le y\le1$: region from $x=y$ to $x=\sqrt y$; the axis $x=1$ is to the right, so $R=1-y$, $r=1-\sqrt y$: $$V=\pi\int_0^1\left[(1-y)^2-\left(1-\sqrt y\right)^2\right]dy=\pi\int_0^1\left(y^2-3y+2\sqrt y\right)dy=\pi\left(\frac13-\frac32+\frac43\right)=\frac\pi6\approx0.5236 .$$
7 Horizontal slices, $0\le y\le2$: the region runs from $x=y^2$ to $x=4$ and touches the axis, so the slices are disks of radius $4-y^2$: $$V=\pi\int_0^2\left(4-y^2\right)^2dy=\pi\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{256\pi}{15}\approx53.62 .$$
8 $R=\sin x+1$, $r=1$: $$V=\pi\int_0^{\pi}\left(2\sin x+\sin^2x\right)dx=\pi\left(4+\frac\pi2\right)=4\pi+\frac{\pi^2}{2}\approx17.50 .$$
9 Horizontal slices: $x^2=4y$, so a slice at height $y$ is a disk of area $\pi x^2=4\pi y$. $$V=\int_0^84\pi y\,dy=2\pi y^2\Big|_0^8=128\pi\approx402.1\ \text{cm}^3 .$$
10 $V(h)=\int_0^h4\pi y\,dy=2\pi h^2$ cm³. Half of $128\pi$ is $64\pi$: $2\pi h^2=64\pi$, $h=\sqrt{32}=4\sqrt2\approx5.66$ cm — well above half the depth, because the bowl is wider at the top.
11 Put the ball as $x^2+y^2\le4$ revolved about the $x$-axis, with the hole along the $x$-axis. A slice at $x$ is a washer with $R=\sqrt{4-x^2}$, $r=1$, which exists while $\sqrt{4-x^2}\ge1$, i.e. $|x|\le\sqrt3$. $$V=\pi\int_{-\sqrt3}^{\sqrt3}\left(3-x^2\right)dx=\pi\left(6\sqrt3-2\sqrt3\right)=4\sqrt3\,\pi\approx21.77\ \text{cm}^3 .$$
IV
Thinking about volume.
  • 1 For the region between $y=x$ and $y=x^2$ revolved about the $x$-axis, a student writes $V=\pi\displaystyle\int_0^1\left(x-x^2\right)^2dx$. Evaluate the student's integral, compare it with Example 5, and explain the error.
  • 2 Cavalieri's principle. Explain, using $V=\int_a^bA(x)\,dx$, why a straight stack of coins and the same stack pushed sideways into a slanted pile have the same volume.
  • 3 The region under $y=\sqrt x$, $0\le x\le4$, is revolved (a) about the $x$-axis, (b) about the $y$-axis. Which solid is larger, and by how much?
1 $\pi\int_0^1\left(x^2-2x^3+x^4\right)dx=\pi\left(\tfrac13-\tfrac12+\tfrac15\right)=\tfrac{\pi}{30}\approx0.1047$, a quarter of the correct $\tfrac{2\pi}{15}$. The student used the area of a disk whose radius is the strip length $R-r$; the slice really sweeps a washer of area $\pi R^2-\pi r^2=\pi(R-r)(R+r)$, which is larger because the strip sits away from the axis.
2 At every height the two piles have the same cross-section (one coin, area $\pi r^2$), so $A(y)$ is the same function for both and $\int A(y)\,dy$ gives the same volume. Volume depends only on cross-sectional areas, not on how the slices are lined up.
3 (a) Disks: $V=\pi\int_0^4x\,dx=8\pi\approx25.13$. (b) Horizontal slices, $0\le y\le2$: the region runs from $x=y^2$ to $x=4$, so washers with $R=4$, $r=y^2$: $V=\pi\int_0^2\left(16-y^4\right)dy=\pi\left(32-\tfrac{32}{5}\right)=\tfrac{128\pi}{5}\approx80.42$. The second solid is larger by $\tfrac{88\pi}{5}\approx55.29$ — the region is far from the $y$-axis, so it sweeps out large circles.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How do radiologists measure the volume of a tumour or an organ from a stack of medical images, and how accurate is the quick "ellipsoid" formula they often use?

Why it matters. Whether a tumour is shrinking under treatment is judged by its volume. CT and MRI scanners produce cross-sectional images at fixed spacing; adding up slice areas times spacing is exactly the slicing method $V=\int A(z)\,dz$ (often called the Cavalieri method in medicine).

What to do.

  1. Sample case: a lesion appears on seven scan slices $0.5$ cm apart, with cross-sectional areas $0,\ 2.1,\ 3.6,\ 4.2,\ 3.8,\ 2.4,\ 0$ cm² at $z=0,0.5,\dots,3.0$ cm.
  2. Estimate the volume with the trapezoidal rule and with Simpson's rule (Section 5.4).
  3. Clinicians often use the ellipsoid formula $V\approx\tfrac{\pi}{6}LWH$. With length $L=3.0$ cm, width $W=2.6$ cm and height $H=2.0$ cm, compute this estimate and compare.
  4. Explain where the formula $\tfrac{\pi}{6}LWH$ comes from: it is $\tfrac43\pi abc$ with semi-axes $a=\tfrac L2$, $b=\tfrac W2$, $c=\tfrac H2$ — prove the case $b=c$ with the disk method (Exercise II.9).

Tools: Calculator or spreadsheet

Hand in: A half-page report with the three volume estimates, the percentage differences between them, and one reason why the slice method is preferred when the shape is irregular.

Going further. The same method estimates the volume of lakes from depth-contour maps, of trees from trunk girths measured at intervals, and of ship hulls from cross-section drawings. Pick one and find real data to try it on.

Instructor note: trapezoidal estimate $8.05$ cm³, Simpson estimate $\approx8.27$ cm³, ellipsoid estimate $\tfrac\pi6(3)(2.6)(2)\approx8.17$ cm³ — all within about $3\%$. Students should notice that the ellipsoid formula assumes a smooth, symmetric shape.

🧪 Self-check quiz

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Q1. Revolving the region under $y=f(x)$, $a\le x\le b$, about the $x$-axis gives volume

Q2. A washer with outer radius $R$ and inner radius $r$ has area

Q3. Revolving the region under $y=x$, $0\le x\le1$, about the $x$-axis gives a cone of volume

Q4. A region lying above the line $y=-2$ is revolved about $y=-2$. If its top boundary is $y=f(x)$, the outer radius is

Q5. A solid has as its base the disk $x^2+y^2\le1$, and its cross-sections perpendicular to the $x$-axis are squares. Its volume is

← 6.1 Areas between Curves6.3 Volumes by Cylindrical Shells →