9709 P3 §3.5 — Area & Volume of Revolution

💡 Lead In: From Areas to Volumes

The Scenario

In the last three lessons we mastered integration by parts (L1, L2) and substitution (L3). But integration is not just an algebraic exercise — it computes geometric quantities. Today we ask: what can integration tell us about shapes?

Consider the semicircle $y = \sqrt{4 - x^2}$ for $-2 \leqslant x \leqslant 2$. If you rotate this curve about the $x$-axis, it sweeps out a solid. What solid is it?

The semicircle $y = \sqrt{4 - x^2}$ makes a full turn about the $x$-axis and sweeps out a sphere of radius 2. Each cross-section perpendicular to the axis is a disk of radius $y$ and area $\pi y^2$; integrating these disks gives $V = \pi \int y^2 \, \text{d}x$.

The Question

The sphere above has radius $r = 2$, so by geometry its volume is $V = \dfrac{4}{3}\pi (2)^3 = \dfrac{32\pi}{3}$.

Can you write down an integral involving $y = \sqrt{4 - x^2}$ that gives this same volume? What is the integrand, and what are the limits?

Where does the $\pi$ come from? Where does the $y^2$ come from?

💡 The Bridge (click to reveal)

When $y = f(x)$ is rotated about the $x$-axis, each point on the curve traces out a circle of radius $f(x)$. The area of this circular cross-section is $\pi [f(x)]^2 = \pi y^2$.

Slicing the solid into thin disks of thickness $\text{d}x$, each disk has volume $\pi y^2 \, \text{d}x$. Integrating from $x = a$ to $x = b$:

$$\boxed{\;V = \pi \int_a^b y^2 \, \text{d}x\;}$$

For the sphere: $V = \pi \int_{-2}^{2} (4 - x^2) \, \text{d}x = \pi \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2} = \pi \left(\dfrac{32}{3}\right) = \dfrac{32\pi}{3}$. It matches!

📖 Key Concepts: Areas & Volumes by Integration

Definite integration computes signed areas. Today we extend this to two new geometric applications: the area between two curves, and the volume of a solid of revolution.

Key relation 1 — Area under a curve:

The area enclosed between the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$ (where $f(x) \geqslant 0$) is:

$$\boxed{\;A = \int_a^b f(x) \, \text{d}x\;}$$

If $f(x) \leqslant 0$ on $[a, b]$, the area is $\displaystyle\int_a^b |f(x)| \, \text{d}x = -\int_a^b f(x) \, \text{d}x$ (take the absolute value).

Key relation 2 — Area between two curves:

If $f(x) \geqslant g(x)$ on $[a, b]$, the area enclosed between the curves $y = f(x)$ (upper) and $y = g(x)$ (lower) is:

$$\boxed{\;A = \int_a^b \big[f(x) - g(x)\big] \, \text{d}x\;}$$

The key idea: always subtract the lower curve from the upper curve. If the curves cross, split the integral at the crossing points and check which is upper in each sub-interval.

Key relation 3 — Volume of revolution about the $x$-axis:

When the curve $y = f(x)$ is rotated about the $x$-axis from $x = a$ to $x = b$, the resulting solid has volume:

$$\boxed{\;V = \pi \int_a^b y^2 \, \text{d}x = \pi \int_a^b [f(x)]^2 \, \text{d}x\;}$$

Each cross-section perpendicular to the $x$-axis is a disk of radius $y$ and area $\pi y^2$. Integrating these disk areas gives the volume. About the $y$-axis, the formula is $V = \pi \displaystyle\int x^2 \, \text{d}y$.

Summary — the area & volume strategy:

Area under a curve: $A = \displaystyle\int_a^b f(x) \, \text{d}x$ (if $f \geqslant 0$).
Area between curves: $A = \displaystyle\int_a^b [f(x) - g(x)] \, \text{d}x$ (upper $-$ lower).
Volume about $x$-axis: $V = \pi \displaystyle\int_a^b [f(x)]^2 \, \text{d}x$.
Volume about $y$-axis: $V = \pi \displaystyle\int x^2 \, \text{d}y$ (express $x$ in terms of $y$).
Always check whether curves cross, and split the integral if needed.

Group Activity 1: Area & Volume Lab — Play, Probe, Predict

In groups of three, explore the lab together. Play with the controls first and watch the shaded area and the volume of revolution change. Then probe with the two questions as a team. Finally predict the two challenge values — you may set the controls to check yourselves, but agree on every answer before anyone clicks.

Curve under study

Upper limit: $b$ = 2 (lower limit fixed at $a = 0$)

Probe 1 — What happens to the shaded area when the upper limit $b$ is moved to the right?

Probe 2 — Why does the volume use $y^{2}$ in the integrand?

Challenge 1 — For $y = 2x$ on $[0, 3]$, the shaded area is:

Challenge 2 — For $y = x^{2}$ on $[0, 3]$ rotated about the $x$-axis, $V = \pi\int_{0}^{3} x^{4}\,\text{d}x$ equals:

Group rule: nobody clicks until all three members agree on the answer. The amber dashed line marks the chosen upper limit $b$ — every strip to its left contributes to the area.

Individual Activity 2: Region Matcher — Dial In the Area and the Volume

Work on your own. Each mission shows a dotted target region. Predict first: say which curve and which upper limit you need. Then choose the function and slide the limit until your solid region sits exactly on the target, and press Check model. The read-out shows the area and the matching volume of revolution.

Function

Upper limit: $b$ = 1 (lower limit fixed at $a = 0$)

Dotted amber region = the target given by the mission. Solid emerald region = the region your controls select right now. Revolving that region about the $x$-axis gives $V = \pi\int_{a}^{b} y^{2}\,\text{d}x$.

Group · Guided Worked Example 1: Area Between $y = x + \dfrac{4}{x}$ and the $x$-axis Exercise 5A Q4

The graph of $y = x + \dfrac{4}{x}$ is shown below.

(i) Find the coordinates of the minimum point $P$ and the maximum point $Q$.

(ii) Find the area of each shaded region.

Exercise 5A Q4 diagram: the curve y = x + 4/x with shaded regions between the curve and the x-axis on -4 to -2 and on 2 to 5

👁 Reveal Solution

Part (i): Find $P$ and $Q$

$y = x + 4x^{-1}$, so $\dfrac{\text{d}y}{\text{d}x} = 1 - 4x^{-2} = 1 - \dfrac{4}{x^2}$.

Set $\dfrac{\text{d}y}{\text{d}x} = 0$: $\;1 = \dfrac{4}{x^2}$, so $x^2 = 4$, giving $x = 2$ or $x = -2$.

At $x = 2$: $\;y = 2 + \dfrac{4}{2} = 4$. At $x = -2$: $\;y = -2 + \dfrac{4}{-2} = -4$.

Second derivative: $\dfrac{\text{d}^2 y}{\text{d}x^2} = 8x^{-3} = \dfrac{8}{x^3}$.

At $x = 2$: $\dfrac{\text{d}^2 y}{\text{d}x^2} = \dfrac{8}{8} = 1 > 0$, so $P(2, 4)$ is a minimum.

At $x = -2$: $\dfrac{\text{d}^2 y}{\text{d}x^2} = \dfrac{8}{-8} = -1 < 0$, so $Q(-2, -4)$ is a maximum.

Part (ii): Area of the region in the first quadrant, $2 \leqslant x \leqslant 5$

On this interval the curve $y = x + \tfrac{4}{x}$ is above the $x$-axis, so the shaded area is the integral of the curve itself.

$\displaystyle A_1 = \int_2^5 \left(x + \frac{4}{x}\right) \text{d}x = \left[\frac{x^2}{2} + 4 \ln x\right]_2^5$

$\displaystyle = \left(\frac{25}{2} + 4 \ln 5\right) - \left(2 + 4 \ln 2\right) = \frac{21}{2} + 4 \ln \frac{5}{2}$.

Numerically: $A_1 = 10.5 + 4 \ln 2.5 \approx 14.17$.

Part (ii): Area of the region in the third quadrant, $-4 \leqslant x \leqslant -2$

For $x < 0$ the curve is below the $x$-axis, so its integral is negative; the shaded area is minus that integral.

$\displaystyle A_2 = -\int_{-4}^{-2} \left(x + \frac{4}{x}\right) \text{d}x = -\left[\frac{x^2}{2} + 4 \ln|x|\right]_{-4}^{-2}$

$\displaystyle = -\left[\left(2 + 4 \ln 2\right) - \left(8 + 8 \ln 2\right)\right] = 6 + 4 \ln 2$.

Numerically: $A_2 = 6 + 4 \ln 2 \approx 8.77$.

Final Answer

$P(2, 4)$ is the minimum, $Q(-2, -4)$ is the maximum.

$\boxed{A_1 = \dfrac{21}{2} + 4 \ln \dfrac{5}{2} \approx 14.17, \qquad A_2 = 6 + 4 \ln 2 \approx 8.77}$

The two areas are not equal: the first-quadrant region runs from $x = 2$ to $x = 5$ and is wider than the third-quadrant region, which runs from $x = -4$ to $x = -2$.

Individual · Guided Worked Example 2: Area Under $y = e^x$ and the Integral of $\ln x$ Exercise 5A Q5

The diagram shows the graph of $y = e^x$. The point $A$ has coordinates $(\ln 5, 0)$, $B$ has coordinates $(\ln 5, 5)$, and $C$ has coordinates $(0, 5)$. The curve meets the $y$-axis at $E$.

(i) Find the area of the region $OABE$ enclosed by the curve $y = e^x$, the $x$-axis, the $y$-axis, and the line $AB$. Hence find the area of the shaded region $EBC$.

(ii) The graph of $y = e^x$ is transformed into the graph of $y = \ln x$. Describe this transformation geometrically.

(iii) Using your answers to (i) and (ii), or otherwise, show that $\displaystyle\int_1^5 \ln x \, \text{d}x = 5 \ln 5 - 4$.

(iv) Deduce the values of (a) $\displaystyle\int_1^5 \ln(x^3) \, \text{d}x$ (b) $\displaystyle\int_1^5 \ln(3x) \, \text{d}x$.

Exercise 5A Q5 diagram: the curve y = e^x with the shaded region EBC inside the rectangle OABC

👁 Reveal Solution

Part (i): Area of $OABE$

The region $OABE$ is bounded by $y = e^x$, the $x$-axis ($y = 0$), the $y$-axis ($x = 0$), and the line $AB$ ($x = \ln 5$).

$\displaystyle A_{OABE} = \int_0^{\ln 5} e^x \, \text{d}x = \left[e^x\right]_0^{\ln 5} = e^{\ln 5} - e^0 = 5 - 1 = 4$.

Part (i): Area of $EBC$ (hence)

The rectangle $OABC$ has vertices $O(0,0)$, $A(\ln 5, 0)$, $B(\ln 5, 5)$, $C(0, 5)$.

$A_{OABC} = \ln 5 \times 5 = 5 \ln 5$.

The region $EBC$ is the part of the rectangle above the curve, so:

$A_{EBC} = A_{OABC} - A_{OABE} = 5 \ln 5 - 4$.

Part (ii): The transformation

$y = e^x$ and $y = \ln x$ are inverse functions. The graph of $y = \ln x$ is obtained from the graph of $y = e^x$ by reflecting in the line $y = x$.

Part (iii): Show $\int_1^5 \ln x \, \text{d}x = 5 \ln 5 - 4$

Under the reflection in $y = x$:

The curve $y = e^x$ (from $E(0, 1)$ to $B(\ln 5, 5)$) maps to $y = \ln x$ (from $(1, 0)$ to $(5, \ln 5)$).
The region $EBC$ maps to the region bounded by $y = \ln x$, the $x$-axis, $x = 1$, and $x = 5$.

Since reflection preserves area: $\displaystyle\int_1^5 \ln x \, \text{d}x = A_{EBC} = 5 \ln 5 - 4$. $\;\square$

Part (iv)(a): $\int_1^5 \ln(x^3) \, \text{d}x$

Using $\ln(x^3) = 3 \ln x$:

$\displaystyle\int_1^5 \ln(x^3) \, \text{d}x = 3 \int_1^5 \ln x \, \text{d}x = 3(5 \ln 5 - 4) = 15 \ln 5 - 12$.

Part (iv)(b): $\int_1^5 \ln(3x) \, \text{d}x$

Using $\ln(3x) = \ln 3 + \ln x$:

$\displaystyle\int_1^5 \ln(3x) \, \text{d}x = \int_1^5 \ln 3 \, \text{d}x + \int_1^5 \ln x \, \text{d}x = 4 \ln 3 + (5 \ln 5 - 4) = 5 \ln 5 + 4 \ln 3 - 4$.

Final Answers

$\boxed{A_{OABE} = 4, \quad A_{EBC} = 5 \ln 5 - 4}$

$\boxed{\displaystyle\int_1^5 \ln(x^3) \, \text{d}x = 15 \ln 5 - 12}$ and $\boxed{\displaystyle\int_1^5 \ln(3x) \, \text{d}x = 5 \ln 5 + 4 \ln 3 - 4}$

Verification: $5\ln5 - 4 \approx 4.047$; $15\ln5 - 12 \approx 12.142$; $5\ln5 + 4\ln3 - 4 \approx 8.442$. All confirmed by numerical integration.

✏️ Concept Quiz

Score: 0 / 4

1. The area enclosed between $y = f(x)$, the $x$-axis, $x = a$ and $x = b$ (with $f \geqslant 0$) is:

$f(b) - f(a)$

$[f(x)]_a^b$

$\\displaystyle\\int_a^b f(x) \\, \\text{d}x$

$\\displaystyle\\int_a^b f'(x) \\, \\text{d}x$

The definite integral $\int_a^b f(x) \, \text{d}x$ computes the signed area under $y = f(x)$ from $a$ to $b$. When $f \geqslant 0$, this equals the geometric area. The option $f(b) - f(a)$ is the result of the integral (the net change), not the area formula itself. The option $[f(x)]_a^b$ is notation for evaluation, not a statement of what the area is. The option $\int f'(x) \, \text{d}x$ integrates the derivative, giving $f(b) - f(a)$.

2. If $f(x) \geqslant g(x)$ on $[a, b]$, the area between the curves $y = f(x)$ and $y = g(x)$ is:

$\\displaystyle\\int_a^b [f(x) - g(x)] \\, \\text{d}x$

$\\displaystyle\\int_a^b [g(x) - f(x)] \\, \\text{d}x$

$\\displaystyle\\int_a^b f(x) \\cdot g(x) \\, \\text{d}x$

$\\displaystyle\\int_a^b |f(x) + g(x)| \\, \\text{d}x$

The area between two curves is $\int (\text{upper} - \text{lower}) \, \text{d}x$. Since $f \geqslant g$, the upper curve is $f$ and the lower is $g$, giving $\int [f - g] \, \text{d}x$. The trap $g - f$ reverses the order (gives a negative result). Multiplying ($f \cdot g$) or adding ($f + g$) does not compute the area between the curves.

3. The volume when $y = f(x)$ is rotated about the $x$-axis from $x = a$ to $x = b$ is:

$\\displaystyle\\int_a^b y^2 \\, \\text{d}x$

$2\\pi \\displaystyle\\int_a^b y \\, \\text{d}x$

$\\pi \\displaystyle\\int_a^b y \\, \\text{d}x$

$\\pi \\displaystyle\\int_a^b y^2 \\, \\text{d}x$

Each cross-section perpendicular to the $x$-axis is a disk of radius $y = f(x)$, with area $\pi y^2$. Integrating these disk areas gives $V = \pi \int_a^b y^2 \, \text{d}x$. The factor $\pi$ comes from the circular cross-section area. The trap without $\pi$ computes $\int y^2 \, \text{d}x$, which is not a volume. The trap $2\pi \int y \, \text{d}x$ uses the circumference $2\pi r$ instead of the area $\pi r^2$.

4. Rotating $y = \sqrt{4 - x^2}$ about the $x$-axis from $-2$ to $2$ gives a volume of:

$\\dfrac{16\\pi}{3}$

$\\dfrac{32\\pi}{3}$

$8\\pi$

$16\\pi$

$V = \pi \int_{-2}^2 (4 - x^2) \, \text{d}x = \pi \left[4x - \frac{x^3}{3}\right]_{-2}^2 = \pi \left[(8 - \frac{8}{3}) - (-8 + \frac{8}{3})\right] = \pi \cdot \frac{32}{3} = \frac{32\pi}{3}$. This is a sphere of radius 2, confirming $V = \frac{4}{3}\pi r^3 = \frac{32\pi}{3}$. The trap $\frac{16\pi}{3}$ is half the correct answer (integrating from 0 to 2 only). The trap $8\pi$ uses $\int y \, \text{d}x$ (forgetting the square).

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📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Q1 · Synthesis Standard

The region $R$ is bounded by the curve $y = x^2$, the $x$-axis, $x = 0$, and $x = 2$. Find the volume when $R$ is rotated about the $x$-axis.

💡 Hint

Use $V = \pi \int_0^2 y^2 \, \text{d}x$ with $y = x^2$, so $y^2 = x^4$.

👁 Reveal Solution

Step 1: Set up the integral

$y = x^2$, so $y^2 = (x^2)^2 = x^4$.

$\displaystyle V = \pi \int_0^2 x^4 \, \text{d}x$.

Step 2: Integrate and evaluate

$\displaystyle V = \pi \left[\frac{x^5}{5}\right]_0^2 = \pi \left(\frac{32}{5} - 0\right) = \frac{32\pi}{5}$.

Final Answer

$\boxed{\dfrac{32\pi}{5}}$

Numerically: $\frac{32\pi}{5} \approx 20.11$.

Q2 · Synthesis Medium

Find the area enclosed between the curves $y = x^2$ and $y = x$.

💡 Hint

First find where the curves intersect by solving $x^2 = x$.

On this interval, determine which curve is the upper one by testing a point.

👁 Reveal Solution

Step 1: Find the intersection points

$x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0$, so $x = 0$ or $x = 1$.

Step 2: Identify upper and lower curves

Test $x = \frac{1}{2}$: $y = x = \frac{1}{2}$ and $y = x^2 = \frac{1}{4}$. So $x > x^2$ on $(0, 1)$, meaning $y = x$ is the upper curve.

Step 3: Compute the area

$\displaystyle A = \int_0^1 (x - x^2) \, \text{d}x = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.

Final Answer

$\boxed{\dfrac{1}{6}}$

Q3 · Synthesis Medium

The region $R$ is bounded by the curve $y = \sqrt{x}$, the $x$-axis, $x = 0$, and $x = 4$. Find the volume when $R$ is rotated about the $x$-axis.

💡 Hint

Use $V = \pi \int_0^4 y^2 \, \text{d}x$ with $y = \sqrt{x}$, so $y^2 = x$.

The integral simplifies to $\pi \int_0^4 x \, \text{d}x$.

👁 Reveal Solution

Step 1: Set up the integral

$y = \sqrt{x}$, so $y^2 = (\sqrt{x})^2 = x$.

$\displaystyle V = \pi \int_0^4 x \, \text{d}x$.

Step 2: Integrate and evaluate

$\displaystyle V = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \left(\frac{16}{2} - 0\right) = 8\pi$.

Final Answer

$\boxed{8\pi}$

Numerically: $8\pi \approx 25.13$.

Q4 · Synthesis Challenging

The region $R$ is bounded by the curve $y = x(2 - x)$ and the $x$-axis. Find the volume when $R$ is rotated about the $x$-axis.

(Hint: first expand $y = 2x - x^2$, then square it carefully before integrating.)

💡 Hint 1

Find where $y = x(2 - x)$ meets the $x$-axis by solving $x(2 - x) = 0$.

These are your limits of integration.

💡 Hint 2

Expand $y = 2x - x^2$, then compute $y^2 = (2x - x^2)^2 = 4x^2 - 4x^3 + x^4$.

Now integrate term by term: $V = \pi \int (4x^2 - 4x^3 + x^4) \, \text{d}x$.

👁 Reveal Solution

Step 1: Find the limits

$y = x(2 - x) = 0$ when $x = 0$ or $x = 2$. So the limits are $0$ and $2$.

Step 2: Expand and square $y$

$y = 2x - x^2$.

$y^2 = (2x - x^2)^2 = 4x^2 - 4x^3 + x^4$.

Step 3: Set up and evaluate the integral

$\displaystyle V = \pi \int_0^2 (4x^2 - 4x^3 + x^4) \, \text{d}x = \pi \left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]_0^2$.

$= \pi \left[\left(\frac{32}{3} - 16 + \frac{32}{5}\right) - 0\right] = \pi \left[\frac{160 - 240 + 96}{15}\right] = \pi \cdot \frac{16}{15} = \frac{16\pi}{15}$.

Final Answer

$\boxed{\dfrac{16\pi}{15}}$

Numerically: $\frac{16\pi}{15} \approx 3.35$.

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Exit Ticket

Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.

I can find the area between two curves using $A = \displaystyle\int_a^b [f(x) - g(x)] \, \text{d}x$, identifying which curve is upper and which is lower. I can find the volume of revolution about the $x$-axis using $V = \pi \displaystyle\int_a^b y^2 \, \text{d}x$, including expanding and squaring the integrand. I can use the inverse-function relationship between $e^x$ and $\ln x$ (reflection in $y = x$) to deduce integrals involving $\ln x$. I can set up and evaluate volume integrals for curves rotated about the $x$-axis, including polynomials and square-root functions.