Worked examples — Volume of revolution — disk method, washer method

This page is the drill floor for the disk & washer topic. The parent note built the two formulas. Here we throw every kind of situation at them, one at a time, and solve each fully. If you meet a rotation problem in an exam, it lives in one of the cells below.

Before we start, one reminder in plain words. A radius is just a distance: how far a point on the curve sits from the spinning line (the axis). Because a distance can never be negative, the radius is always the absolute value of the gap between the curve and the axis: $r=|f(x)-(\text{axis value})|.$ When the curve sits below the axis (so $f(x)$ is smaller than the axis value), this absolute value flips the sign for you — that is why in Example 4 you will see $2-f(x)$ instead of $f(x)-2$. Keep this in mind: radius is a length, never a signed number. Volume comes from adding up thin coins, $\pi r^{2}dx$. Everything else is bookkeeping about which distance and which variable.

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The scenario matrix

Every problem this topic can throw is one of these cells. The last column names the worked example that covers it.

Cell	What changes	Disk or Washer	Variable	Covered by
A. Region touches x-axis	radius $=f(x)$, no hole	Disk	$dx$	Ex 1
B. Region between two curves, above axis	outer/inner radii	Washer	$dx$	Ex 2
C. Rotate about y-axis	solve for $x=g(y)$	Disk	$dy$	Ex 3
D. Rotate about a horizontal line $y=k$	shift both radii by $k$	Washer	$dx$	Ex 4
E. Rotate about a vertical line $x=k$	shift radii, integrate in $y$	Washer	$dy$	Ex 5
F. Degenerate / zero case (curves meet, or region has no area)	volume collapses	—	—	Ex 6
G. Real-world word problem (a physical vase/tank)	translate words → curve	Disk	$dx$	Ex 7
H. Exam twist: outer/inner swap inside the interval (curves cross)	split the integral	Washer	$dx$	Ex 8
I. Axis passes through the region	no hole, use max distance	Disk	$dx$	Ex 9

The figure below is the map for the whole page. Read it like this: the gray solid lines are the x- and y-axes; each coloured dashed line is a different possible axis of rotation that the cells use — orange for the horizontal line $y=k$, red for $y=2$, green for the vertical line $x=3$. The blue arrow shows the single idea every cell shares: the radius is the distance from whichever axis you chose down to the curve. As you work through the examples, come back and find your cell's axis here.

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Example 1 — Cell A (region touches the x-axis, pure disk)

Forecast: the height grows fast (it's $x^2$), so squaring gives $x^4$ — expect a fairly big number, something like a few $\pi$. Guess before reading on.

The figure shows the region (green) and one representative red coin: because the region rests on the x-axis, the coin is solid — no hole.

1. Identify the radius. The region sits on the x-axis, so the radius of each coin is the curve's height, $r=f(x)=x^2$. Why this step? No gap between region and axis ⇒ solid coins ⇒ disk method.
2. Write the disk integral. $V=\pi {\int }_0^2[x^2{]}^{2}dx=\pi {\int }_0^2{x}^{4}dx.$ Why this step? Coin area $=\pi r^2=\pi (x^2{)}^2=\pi x^4$; sum with thickness $dx$.
3. Integrate. $V=\pi {\left[\frac{x^5}{5}\right]}_0^2=\pi \cdot \frac{32}{5}=\frac{32\pi }{5}.$ Why this step? Power rule: $\int x^{n}dx=x^{n+1}/(n+1)$, so $\int x^{4}dx=x^5/5$; then plug in $x=2$ (giving $32/5$) and $x=0$ (giving $0$) and subtract.

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Example 2 — Cell B (two curves above the axis, washer)

Forecast: $\sqrt{x}$ sits above $x$ on $(0,1)$, so it's the far curve. The gap is thin, so expect a small answer, maybe under $\pi$.

1. Decide outer vs inner. On $(0,1)$, $\sqrt{x}>x$ (test $x=\frac{1}{4}$: $\frac{1}{2}>\frac{1}{4}$). So $R=\sqrt{x}$, $r=x$. Why this step? Outer radius = distance to the far curve; the washer's hole comes from the near curve.
2. Square each radius separately, then subtract. $V=\pi {\int }_0^1([\sqrt{x}{]}^2-[x{]}^2)dx=\pi {\int }_0^1(x-x^2)dx.$ Why this step? Ring area $=\pi R^2-\pi r^2$, never $\pi (R-r{)}^2$.
3. Integrate. $V=\pi {\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi }{6}.$ Why this step? Integrate each power term separately ($\int xdx=x^2/2$, $\int x^{2}dx=x^3/3$); evaluate at $x=1$ (the lower limit $0$ contributes nothing) and combine over a common denominator $6$.

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Example 3 — Cell C (rotate about the y-axis)

Forecast: the axis is now vertical, so we slice horizontally and integrate in $y$. Coins stack up the y-axis from $y=0$ to $y=4$.

The figure shows the flipped picture: the coins are now horizontal, and the radius runs sideways from the y-axis out to the curve $x=\sqrt{y}$.

1. Switch the slicing variable. Axis is vertical ⇒ thickness $dy$ ⇒ radius must be a horizontal distance $x$. Solve the curve for $x$: from $y=x^2$ we get $x=\sqrt{y}$. Why this step? A coin perpendicular to a vertical axis has thickness $dy$; its radius runs sideways, so we need $x$ as a function of $y$.
2. Radius = distance from y-axis to curve $=\sqrt{y}$. No hole (region touches the y-axis), so disk. $V=\pi {\int }_0^4[\sqrt{y}{]}^{2}dy=\pi {\int }_0^{4}ydy.$ Why this step? Limits are now $y$-values: the region spans $y=0$ to $y=4$.
3. Integrate. $V=\pi {\left[\frac{y^2}{2}\right]}_0^4=\pi \cdot \frac{16}{2}=8\pi .$ Why this step? $\int ydy=y^2/2$; substitute the upper limit $y=4$ (giving $16/2=8$) and subtract the lower limit's $0$.

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Example 4 — Cell D (rotate about a horizontal line $y=k$)

Forecast: the axis $y=2$ sits above both curves, so both curves are below it. The far curve from $y=2$ is the lower one ($x^2$), and the near curve is the upper one ($x$). Watch this reversal — it's the whole point of this cell.

1. Re-measure every radius from the new axis. Distance from $y=2$ down to a curve $y=f(x)$ is $|f(x)-2|=2-f(x)$ (both curves are under the line, so $f(x)-2$ is negative and the absolute value flips it to $2-f(x)$). Why this step? Radius = distance from axis to curve, and the axis moved to $y=2$; distance is always $\ge 0$, which is exactly what the absolute value guarantees.
2. Find outer vs inner. The point farthest from $y=2$ is the lowest curve. On $(0,1)$, $x^2<x$, so $x^2$ is lowest ⇒ farthest ⇒ $R=2-x^2$. The near curve is $x$ ⇒ $r=2-x$. Why this step? "Outer" means physically far from the axis, not "bigger $y$". Shifting the axis above the region flips which curve is outer.
3. Washer integral. $V=\pi {\int }_0^1[(2-x^2{)}^2-(2-x{)}^2]dx.$ Expand: $(2-x^2{)}^2=4-4x^2+x^4$, $(2-x{)}^2=4-4x+x^2$. Subtract: $(4-4x^2+x^4)-(4-4x+x^2)=x^4-5x^2+4x.$ Why this step? Squares first, subtract second; the constant $4$'s cancel, which is a good sign.
4. Integrate. $V=\pi {\int }_0^1(x^4-5x^2+4x)dx=\pi {\left[\frac{x^5}{5}-\frac{5x^3}{3}+2x^2\right]}_0^1=\pi \left(\frac{1}{5}-\frac{5}{3}+2\right).$ Combine: $\frac{1}{5}-\frac{5}{3}+2=\frac{3-25+30}{15}=\frac{8}{15}$. $V=\frac{8\pi }{15}.$ Why this step? Integrate term by term ($\int x^4=x^5/5$, $\int 5x^2=5x^3/3$, $\int 4x=2x^2$); at $x=1$ every power equals $1$, so the bracket is just $\frac{1}{5}-\frac{5}{3}+2$, and at $x=0$ it is $0$. Put over the common denominator $15$ to finish.

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Example 5 — Cell E (rotate about a vertical line $x=k$)

Forecast: vertical axis ⇒ slice in $y$. The axis $x=3$ is to the right of the whole region ($x$ runs $0$ to $2$), so the near edge is the curve $x=\sqrt{y}$ and the far edge is the y-axis line $x=0$.

The figure shows the region with the red vertical axis $x=3$ off to the right and a horizontal washer slice: its outer radius reaches all the way to $x=0$, its inner radius stops at the curve.

1. Slice horizontally. Vertical axis ⇒ thickness $dy$, radius is a horizontal distance from $x=3$. Express the boundary as $x=\sqrt{y}$ (from $y=x^2$). Why this step? Coins perpendicular to a vertical axis stack in $y$.
2. Two radii from $x=3$. At height $y$, the region runs from $x=0$ (left edge) to $x=\sqrt{y}$ (curve). Distances from $x=3$ (each an absolute value, both edges left of the axis):
   • far edge $x=0$: $R=|0-3|=3$,
   • near edge $x=\sqrt{y}$: $r=|\sqrt{y}-3|=3-\sqrt{y}$. Why this step? Farther point ($x=0$) is farther from $x=3$, giving the outer radius; the absolute value keeps both radii positive.
3. Washer integral in $y$, from $y=0$ to $y=4$. $V=\pi {\int }_0^4[3^2-(3-\sqrt{y}{)}^2]dy.$ Expand $(3-\sqrt{y}{)}^2=9-6\sqrt{y}+y$. So $9-(9-6\sqrt{y}+y)=6\sqrt{y}-y$. $V=\pi {\int }_0^4(6\sqrt{y}-y)dy.$ Why this step? Squares first; the $9$'s cancel.
4. Integrate ($\int \sqrt{y}dy=\frac{2}{3}{y}^{3/2}$): $V=\pi {\left[6\cdot \frac{2}{3}{y}^{3/2}-\frac{y^2}{2}\right]}_0^4=\pi {\left[4y^{3/2}-\frac{y^2}{2}\right]}_0^4.$ At $y=4$: $y^{3/2}=8$, so $4\cdot 8-\frac{16}{2}=32-8=24$. $V=24\pi .$ Why this step? Write $\sqrt{y}=y^{1/2}$ so the power rule applies ($\int y^{1/2}=\frac{2}{3}{y}^{3/2}$); evaluate at $y=4$ using $4^{3/2}=(\sqrt{4}{)}^3=8$, and the lower limit $0$ again contributes nothing.

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Example 6 — Cell F (degenerate / zero volume)

Forecast: if two curves coincide, there is no region — zero area — so zero volume. Recognising degeneracy saves you from computing nonsense.

The figure makes the degeneracy visible: the two "curves" lie exactly on top of one another, so the shaded region has literally zero width.

1. First case: identical curves. $R=x$ and $r=x$, so $R^2-r^2=x^2-x^2=0$. $V=\pi {\int }_0^{1}0dx=0.$ Why this step? No gap between the curves ⇒ no washer thickness ⇒ no solid.
2. Second case: an interval of zero width. Integrating from $x=1$ to $x=1$: $V=\pi {\int }_1^1[(x^2+3{)}^2-(x^2{)}^2]dx=0.$ Why this step? ${\int }_a^a(\text{anything})dx=0$; a slice needs width $\Delta x>0$ to have volume.

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Example 7 — Cell G (real-world word problem: a bowl)

Forecast: it's Cell C in disguise (rotate about y-axis). Expect an answer with a $\pi$ and a clean fraction; units are cm³.

The figure shows the bowl's cross-section (the shaded parabola) and how spinning it about the y-axis carves out the paraboloid vessel.

1. Translate words to a curve and axis. "Rotate about the y-axis" ⇒ slice in $y$, radius $=x=\sqrt{y}$ (from $y=x^2$). The bowl runs from $y=0$ (bottom) to $y=9$ (since $x=3\Rightarrow y=9$). Why this step? The physical rim is at $x=3$, giving the top height $y=9$.
2. Disk method in $y$ (solid, touches axis): $V=\pi {\int }_0^9[\sqrt{y}{]}^{2}dy=\pi {\int }_0^{9}ydy.$ Why this step? The radius of each horizontal coin is $\sqrt{y}$, so its area is $\pi [\sqrt{y}{]}^2=\pi y$; stack the coins with thickness $dy$ from the bottom $y=0$ to the rim $y=9$.
3. Integrate. $V=\pi {\left[\frac{y^2}{2}\right]}_0^9=\pi \cdot \frac{81}{2}=\frac{81\pi }{2}{\text{cm}}^3.$ Why this step? $\int ydy=y^2/2$; plug in the rim height $y=9$ (giving $81/2$) and the bottom $y=0$ (giving $0$). The units stay cm³ because we integrated a cm² area over a cm height.

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Example 8 — Cell H (exam twist: curves cross, split the integral)

Forecast: the two curves cross at $x=1$. On $(0,1)$, $x>x^2$; on $(1,2)$, $x^2>x$. So the outer curve is different on each piece — you must split the integral at the crossing and pick the right $R$ on each part. Guess: two washer integrals, added.

The figure shows both curves over $[0,2]$, the crossing at $x=1$ marked, and shading in two colours to flag that the "top" curve changes.

1. Find where the ordering swaps. Set $x=x^2\Rightarrow x(1-x)=0\Rightarrow x=0$ or $x=1$. So $x=1$ is the crossing inside $[0,2]$. Why this step? $R$ and $r$ are defined by which curve is farther from the axis; that can only change where the curves meet. Find those points first.
2. Assign $R,r$ on each piece.
   • On $[0,1]$: $x\ge x^2$, so $R=x$, $r=x^2$.
   • On $[1,2]$: $x^2\ge x$, so $R=x^2$, $r=x$. Why this step? Outer = farther from the x-axis = larger height (both curves are $\ge 0$ here). Test a point on each piece to be sure ($x=\frac{1}{2}$: $\frac{1}{2}>\frac{1}{4}$; $x=\frac{3}{2}$: $\frac{9}{4}>\frac{3}{2}$).
3. Two washer integrals, added. $V=\pi {\int }_0^1(x^2-x^4)dx+\pi {\int }_1^2(x^4-x^2)dx.$ Why this step? Each piece uses its own $R^2-r^2$ (always outer squared minus inner squared); a single integral with the wrong $R$ would give a nonsensical negative integrand on one piece.
4. Evaluate the first piece (this is Example 2's integrand): $\pi {\int }_0^1(x^2-x^4)dx=\pi {\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi }{15}.$
5. Evaluate the second piece. $\pi {\int }_1^2(x^4-x^2)dx=\pi {\left[\frac{x^5}{5}-\frac{x^3}{3}\right]}_1^2.$ At $x=2$: $\frac{32}{5}-\frac{8}{3}=\frac{96-40}{15}=\frac{56}{15}$. At $x=1$: $\frac{1}{5}-\frac{1}{3}=-\frac{2}{15}$. Subtract: $\pi \left(\frac{56}{15}-\left(-\frac{2}{15}\right)\right)=\pi \cdot \frac{58}{15}=\frac{58\pi }{15}.$ Why this step? Same antiderivative as piece 1 but with roles swapped, so the sign flips; evaluate between the new limits $1$ and $2$.
6. Add the pieces. $V=\frac{2\pi }{15}+\frac{58\pi }{15}=\frac{60\pi }{15}=4\pi .$ Why this step? Total volume is the sum of the two sub-solids; a common denominator makes the addition clean.

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Example 9 — Cell I (axis inside the region)

Forecast: because the axis runs through the region, the top half and bottom half sweep out the same cylinder — they overlap, they don't stack. The correct radius is the maximum distance, not top-minus-bottom. Expect a plain cylinder.

The figure shows the square crossing the axis and the single cylinder it sweeps — note there is no hole, because the region reaches the axis.

1. Spot the trap. The far edge is at distance $1$ above the axis; the near edge is the axis itself ($y=0$) because the region reaches it. So $R=1$, $r=0$ — a disk, not a washer. Why this step? When the axis lies inside the region, the solid is a full disk of radius = the largest distance to an edge; there is no hole.
2. Disk integral. $V=\pi {\int }_{-1}^1[1{]}^{2}dx=\pi {\int }_{-1}^{1}1dx.$ Why this step? Every coin has radius $1$ (constant), thickness $dx$, from $x=-1$ to $x=1$.
3. Integrate. $V=\pi [x{]}_{-1}^1=\pi (1-(-1))=2\pi .$ Why this step? $\int 1dx=x$; evaluating a constant integrand just measures the interval length, $1-(-1)=2$, so $V=2\pi$.

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Recall

Recall Which cell is which?

Rotate about x-axis, region on the axis (Cell A) ::: Disk in $dx$, radius $f(x)$. Rotate about y-axis (Cell C) ::: Disk/washer in $dy$, solve $x=g(y)$. Rotate about $y=k$ (Cell D) ::: Shift radii to $|f(x)-k|$; the farthest curve is outer. Rotate about $x=k$ (Cell E) ::: Slice in $dy$, radius is horizontal distance $|x-k|$. Two identical curves (Cell F) ::: Zero volume — no region. Curves cross inside the interval (Cell H) ::: Split the integral at every crossing; use outer²−inner² on each piece. Axis passes through the region (Cell I) ::: Disk of radius = max distance, not a washer.

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Connections

• Definite Integral as a Limit of Sums — every "sum the coins, let $\Delta x\to 0$" step is this.
• Area Between Curves — Cells B, D, E, H start from an area-between-curves picture.
• Shell Method — an alternative when solving for $x=g(y)$ is ugly (compare Cell C/E).
• Volume by Cross-Sections — disks and washers are the circular cross-section case.
• u-substitution — handy for expanding shifted-axis integrands (Cells D, E).