How much wine can a barrel hold? What is the capacity of a vase shaped like a parabola? How do engineers design satellite dishes? These objects are all solids of revolution β three-dimensional shapes created by rotating a curve around an axis. Just as integration finds areas by summing infinitely many thin rectangles, it finds volumes by summing infinitely many thin disks. The disk method and washer method are two of the most elegant applications of definite integrals, turning complex 3D volume problems into straightforward single-variable calculus.
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If you rotate the curve $y = \sqrt{x}$ from $x = 0$ to $x = 4$ around the x-axis, what 3D shape do you get? How might you find its volume using what you know about areas of circles?
Core Content
When a curve $y = f(x)$ from $x = a$ to $x = b$ is rotated around the x-axis, each point traces out a circle. At position $x$, the radius of the circle is $f(x)$, so the cross-sectional area is $\pi [f(x)]^2$.
Summing these circular areas gives the volume:
V = Ο β«_a^b [f(x)]^2 dx
Example: Find the volume when $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated around the x-axis.
V = Ο β«_0^4 (βx)^2 dx = Ο β«_0^4 x dx = Ο [x^2/2]_0^4 = Ο Β· 8 = 8Ο
Example around y-axis: Find the volume when $x = y^2$ from $y = 0$ to $y = 2$ is rotated around the y-axis.
V = Ο β«_0^2 (y^2)^2 dy = Ο β«_0^2 y^4 dy = Ο [y^5/5]_0^2 = Ο Β· 32/5 = 32Ο/5
When the region being rotated does not touch the axis of revolution, the solid has a hole through the middle. Each cross-section is a washer (a disk with a hole):
V = Ο β«_a^b ([R(x)]^2 - [r(x)]^2) dx
where $R(x)$ is the outer radius and $r(x)$ is the inner radius.
Example: Find the volume when the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ is rotated around the x-axis.
R(x) = x (outer), r(x) = x^2 (inner)
V = Ο β«_0^1 (x^2 - x^4) dx = Ο [x^3/3 - x^5/5]_0^1 = Ο(1/3 - 1/5) = 2Ο/15
Common error: Some students try $\pi \int_0^1 (x - x^2)^2 \, dx$. This is wrong! The washer method subtracts the squares, not the square of the difference. $(x - x^2)^2 \neq x^2 - x^4$.
The region bounded by $y = x^2$ and $y = 4$ is rotated around the y-axis. Find the volume.
The volume of the solid.
Method: Disk method around y-axis.
Express x in terms of y: x = βy
At y, radius = βy
V = Ο β«_0^4 (βy)^2 dy = Ο β«_0^4 y dy
= Ο [y^2/2]_0^4 = Ο Β· 8 = 8Ο
$8\pi$ cubic units.
Find the volume when $y = x^3$ from $x = 0$ to $x = 1$ is rotated around the x-axis.
Answer:
$V = \pi \int_0^1 (x^3)^2 \, dx = \pi \int_0^1 x^6 \, dx = \pi [\frac{x^7}{7}]_0^1 = \frac{\pi}{7}$.
$V = \pi \int_a^b [f(x)]^2 \, dx$
$V = \pi \int_c^d [g(y)]^2 \, dy$
$V = \pi \int_a^b (R^2 - r^2) \, dx$
Square first, then subtract β not $(R-r)^2$
Rotating $y = \sqrt{x}$ from $x = 0$ to $x = 4$ around the x-axis produces a paraboloid β a shape like a satellite dish or a wine glass. At each $x$, the cross-section is a circle with radius $\sqrt{x}$, so the area is $\pi (\sqrt{x})^2 = \pi x$. Summing these circular areas gives $V = \pi \int_0^4 x \, dx = 8\pi$. The key insight is that every volume-of-revolution problem reduces to: (1) find the radius as a function of position, (2) write the cross-sectional area $\pi r^2$, (3) integrate.
5 random questions from a replayable lesson bank β feedback shown immediately
Short Answer
8. Find the volume when $y = x^2$ from $x = 0$ to $x = 3$ is rotated around the x-axis. Show all working. 3 MARKS
9. Find the volume when the region bounded by $y = x$ and $y = x^2$ is rotated around the x-axis. Show all working. 4 MARKS
10. A wine barrel has profile $y = 10\sqrt{1 - (\frac{x}{30})^2}$ for $-30 \leq x \leq 30$ (in cm), rotated around the x-axis. (a) Identify the shape. (b) Set up the integral for the volume. (c) Evaluate it and convert to litres (1000 cmΒ³ = 1 L). 4 MARKS
1. $V = \pi \int_0^2 x^4 \, dx = \pi [\frac{x^5}{5}]_0^2 = \frac{32\pi}{5}$.
2. $V = \pi \int_0^9 x \, dx = \pi [\frac{x^2}{2}]_0^9 = \frac{81\pi}{2}$.
3. Intersections at $x = 0, 1$. $V = \pi \int_0^1 (x^2 - x^4) \, dx = \pi [\frac{x^3}{3} - \frac{x^5}{5}]_0^1 = \frac{2\pi}{15}$.
1. The cross-sectional area of a washer is $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$. The square of the difference $(R-r)^2$ would give the wrong area.
2. $V = \pi \int_0^4 (2\sqrt{x})^2 \, dx = \pi \int_0^4 4x \, dx = \pi [2x^2]_0^4 = 32\pi \approx 100.5$ litres.
3. Because the radius is measured horizontally from the y-axis, so we need $x$ as a function of $y$ to describe how far from the axis each point is.
Q8 (3 marks): $V = \pi \int_0^3 (x^2)^2 \, dx = \pi \int_0^3 x^4 \, dx$ [1] $= \pi [\frac{x^5}{5}]_0^3$ [1] $= \pi \cdot \frac{243}{5} = \frac{243\pi}{5}$ [1].
Q9 (4 marks): Intersections: $x = x^2 \Rightarrow x(1-x) = 0$, so $x = 0, 1$ [1]. On $[0, 1]$, $x \geq x^2$, so $R(x) = x$, $r(x) = x^2$ [1]. $V = \pi \int_0^1 (x^2 - x^4) \, dx$ [1] $= \pi [\frac{x^3}{3} - \frac{x^5}{5}]_0^1 = \pi(\frac{1}{3} - \frac{1}{5}) = \frac{2\pi}{15}$ [1].
Q10 (4 marks): (a) An ellipsoid (prolate spheroid) [1]. (b) $V = \pi \int_{-30}^{30} 100(1 - \frac{x^2}{900}) \, dx$ [1] $= 100\pi [x - \frac{x^3}{2700}]_{-30}^{30}$ [1] $= 100\pi [(30 - 10) - (-30 + 10)] = 100\pi \cdot 40 = 4000\pi \approx 12566$ cmΒ³ β 12.6 L [1].
Climb platforms by calculating volumes of revolution using disk and washer methods. Pool: lesson 7.
Tick when you've finished all activities and checked your answers.