Back to homeMathematics · 2nd Bach. Science · Change subject
Start for free
BAC2 tech · Mathematics
Integrals · Definite integral

Volumes of revolution

When a plane region rotates about an axis, it generates a solid in space: a rotating rectangle gives a cylinder, a right-angled triangle gives a cone, and a semicircle gives a sphere. The definite integral lets you calculate the volume of any solid of this type, not just those with a known formula.

Before you start, make sure you know: Special products (2nd ESO)
Before you start, make sure you know: Volume of a solid (3rd ESO)

The disc method

Just as you approximated an area with rectangles, now you approximate a volume with stacked discs (very thin cylinders). Each disc, at height $x$, has radius $f(x)$ and thickness $dx$: its volume is $\pi [f(x)]^2\,dx$. Adding up (integrating) all the discs between $a$ and $b$ gives the total volume.

Approximation of the volume of a solid of revolution about the OX axis using stacked circular discs of radius f(x) and thickness dx each disc: radius f(x), thickness dx OX axis
Practice★☆☆Not started
The region under the curve $y = \sqrt{x}$ between $x = 0$ and $x = 4$ rotates around the $OX$ axis, as shown in the figure. Set up the corresponding integral…

The region under the curve $y = \sqrt{x}$ between $x = 0$ and $x = 4$ rotates around the $OX$ axis, as shown in the figure.

y x 0 1 2 3 4 1 2 (4, 2) y = √x y = √x rotates around the OX axis

Set up the corresponding integral $V = \pi \int_{a}^{b} [f(x)]^2 \, dx$ and calculate the volume of the solid of revolution generated, expressing the result in terms of $\pi$.

Solve

Checking against formulae you already know

Before applying it to new curves, check it with the usual shapes: if the "curve" that rotates is a horizontal line, you get a cylinder; if it is a sloping line through the origin, a cone; if it is a semicircle, a sphere. The result of the integral must match the geometric formula.

Practice★☆☆Not started
The constant function $f(x) = 2$ between $x = 0$ and $x = 5$ rotates around the $OX$ axis. a) Calculate the volume of the solid generated using the integral…

The constant function $f(x) = 2$ between $x = 0$ and $x = 5$ rotates around the $OX$ axis.

  1. a) Calculate the volume of the solid generated using the integral $V = \pi \int_{0}^{5} [f(x)]^2 \, dx$.
  2. b) What geometric solid has been generated? Check the result using the formula for the volume of that solid.
Solve
Practice★☆☆Not started
The semicircle $y = \sqrt{9 - x^2}$ rotates around the $OX$ axis. a) What geometric solid is generated and between what limits must you integrate? b) Calculate its…

The semicircle $y = \sqrt{9 - x^2}$ rotates around the $OX$ axis.

  1. a) What geometric solid is generated and between what limits must you integrate?
  2. b) Calculate its volume using the corresponding integral.
  3. c) Check the result using the formula $V = \frac{4}{3} \pi r^3$.
Solve
Practice★☆☆Not started
The segment of the line $y = x$ between $x = 0$ and $x = 3$ rotates around the $OX$ axis generating a cone. a) Calculate its volume using the integral…

The segment of the line $y = x$ between $x = 0$ and $x = 3$ rotates around the $OX$ axis generating a cone.

  1. a) Calculate its volume using the integral $V = \pi \int_{0}^{3} x^2 \, dx$.
  2. b) Identify the radius and height of the cone and check the result using the formula $V = \frac{1}{3} \pi r^2 h$.
Solve

The most common mistake: forgetting to square

The typical exam mistake is to write $V=\pi\int f(x)\,dx$ instead of $V=\pi\int [f(x)]^2\,dx$. It seems a small detail, but it completely changes the result: without the square, what you calculate is $\pi$ times an area, not a volume.

Geometric comparison between the plane region under a curve and the three-dimensional volume generated by rotating it about the OX axis π∫f(x)dx ✗ π∫[f(x)]²dx ✓ the radius of the disc is f(x); its area is π·radius² = π[f(x)]²
Practice★★★Not started
Marcos calculated the volume of the solid generated by the region under the curve $y = x^2 + 1$ between $x = -1$ and $x = 1$ when rotating around the $OX$ axis as…

Marcos calculated the volume of the solid generated by the region under the curve $y = x^2 + 1$ between $x = -1$ and $x = 1$ when rotating around the $OX$ axis as follows:

«$V = \pi \int_{-1}^{1} (x^2 + 1) \, dx = \pi \left[ \frac{x^3}{3} + x \right]_{-1}^{1} = \pi \left( \frac{1}{3} + 1 + \frac{1}{3} + 1 \right) = \frac{8\pi}{3}\ \text{u}^3$»

  1. a) Identify the error in his approach and explain what it consists of.
  2. b) Calculate the volume correctly.
Solve

The washer method: volume between two curves

If the region that rotates lies between two curves $f(x)\ge g(x)$ (rather than between a curve and the axis), each cross-section is no longer a solid disc but a ring (a "washer"): a large circle of radius $f(x)$ with a hole of radius $g(x)$.

Circular cross-section with a concentric hole illustrating a washer with outer radius f(x) and inner radius g(x) washer: large radius f(x), small radius g(x) hole
Practice★★☆Not started
The region enclosed between the parabola $y = x^2$ and the line $y = x$, shaded in the figure, rotates around the $OX$ axis generating a hollow solid. a) Find the…

The region enclosed between the parabola $y = x^2$ and the line $y = x$, shaded in the figure, rotates around the $OX$ axis generating a hollow solid.

x y 0,5 1 0,5 1 (0, 0) (1, 1) y = x y = x²
  1. a) Find the intersection points of both curves.
  2. b) Set up the volume as the difference of two integrals (washer method) and calculate it.
Solve

Rotating about the OY axis

If the rotation is about the $OY$ axis instead of $OX$, the logic is identical but with the roles of $x$ and $y$ swapped: solve for $x$ in terms of $y$, and the discs are now stacked "vertically".

Practice★★☆Not started
A conical laboratory funnel is obtained by revolving the line $y = 2x$, with $0 \le x \le 2$, about the $OY$-axis, as shown in the figure (measurements in decimetres).…

A conical laboratory funnel is obtained by revolving the line $y = 2x$, with $0 \le x \le 2$, about the $OY$-axis, as shown in the figure (measurements in decimetres).

X Y rotates around the OY axis 1 2 2 4 0 (2, 4) y = 2x x = 2
  1. a) Express $x$ in terms of $y$ and set up the integral $V = \pi \int_{c}^{d} [g(y)]^2 \, dy$ that gives the volume, stating the limits of integration.
  2. b) Calculate the volume of the funnel.
  3. c) Check the result using the formula for the volume of a cone, $V = \frac{1}{3} \pi r^2 h$, identifying the radius and the height of the funnel.
Solve
Practice★★☆Not started
The nozzle of a model aircraft engine is a part whose outer wall follows the profile of the hyperbola $y = \frac{2}{x}$ between $x = 1$ and $x = 4$ (measurements in…

The nozzle of a model aircraft engine is a part whose outer wall follows the profile of the hyperbola $y = \frac{2}{x}$ between $x = 1$ and $x = 4$ (measurements in centimetres), when rotating around the $OX$ axis.

  1. a) Calculate the volume of material needed to manufacture the part, assuming it is solid between the axis and the profile.
  2. b) If the material has a density of $2,5\ \text{g/cm}^3$, what is the mass of the part? Round to one decimal place of a gram.
Solve

Summary

See all the exercises in this section