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Secondary 6 · Mathematics
Integrals · Definite integral

Volumes of Revolution

Before you start, make sure you know: Volume of a geometric solid (Secondary Year 4 (second cycle, academic track))

A woodturner in La Vega makes a wooden piece by spinning a block on a lathe: the shape that comes out is a curve "going around" an axis. A rectangle that spins gives a cylinder, a right triangle gives a cone, and a semicircle gives a sphere. How can you calculate the volume of a piece whose shape has no known formula? The definite integral lets you do it.

The Disk Method

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

Approximation of the volume of a solid of revolution around the x-axis using stacked circular disks of radius f(x) and thickness dx each disk: radius f(x), thickness dx x-axis
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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 with the Formulas You Already Know

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

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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
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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
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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 the Square

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

Comparison between the incorrect formula without the square and the correct formula with the square of the disk radius π∫f(x)dx ✗ π∫[f(x)]²dx ✓ the disk radius is f(x); its area is π·radius² = π[f(x)]²
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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 $f(x)\ge g(x)$ (and not between a curve and the axis), each cross-section is no longer a solid disk 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
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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
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The profile of a ceramic cup is modeled by the curve $y = \sqrt{x}$, with $x$ and $y$ measured in centimeters, rotating around the $OX$ axis between $x = 0$ and $x = 9$.…

The profile of a ceramic cup is modeled by the curve $y = \sqrt{x}$, with $x$ and $y$ measured in centimeters, rotating around the $OX$ axis between $x = 0$ and $x = 9$.

  1. a) Calculate the interior volume of the cup in $\text{cm}^3$, rounding to the nearest integer.
  2. b) Knowing that $1\ \text{cm}^3 = 1\ \text{ml}$, what is the capacity of the cup in milliliters? Does this seem like a reasonable capacity for a cup? Comment on the result.
Solve

Rotating Around the y-Axis

If the rotation is around the $y$-axis, the logic is the same but with the roles of $x$ and $y$ swapped: solve for $x$ as a function of $y$ and the disks stack vertically.

Washer Around the y-Axis

Sometimes the region does not reach the $y$-axis: there is a gap between the axis and the region, and the horizontal cross-sections are washers. Then you integrate with respect to $y$, using the outer radius $R(y)$ and the inner radius $r(y)$ (horizontal distances to the $y$-axis).

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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 decimeters).…

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 decimeters).

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
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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 centimeters), 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
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Consider the plane region bounded by the parabola $y = x^2$, the $OX$ axis and the line $x = 2$. a) Calculate the volume of the solid generated by rotating around the…

Consider the plane region bounded by the parabola $y = x^2$, the $OX$ axis and the line $x = 2$.

  1. a) Calculate the volume of the solid generated by rotating around the $OX$ axis.
  2. b) Calculate the volume of the solid generated by rotating around the $OY$ axis. (Hint: subtract from the cylinder of radius $2$ and height $4$ the solid generated by the region between the curve and the $OY$ axis.)
  3. c) Although the rotating region is the same, the two volumes do not coincide. Explain why with reasons.
Solve

Famous Solids and Infinite Volumes

The disk method is used to prove well-known formulas and to study solids that extend to infinity. The ideas are these:

  • For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, solve for $y^2=b^2\left(1-\frac{x^2}{a^2}\right)$ and integrate from $-a$ to $a$. You don't need to take a root: the integrand is already $y^2$.

  • If the limit of integration is infinite, first calculate the volume up to $b$ and then find the limit as $b\to+\infty$.

  • The volume can be finite even if the area of the region that rotates is infinite: this is the paradox of Torricelli's trumpet.

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A rugby ball is shaped like an ellipsoid of revolution: its cross-section is the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which rotates around the $OX$ axis, as…

A rugby ball is shaped like an ellipsoid of revolution: its cross-section is the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which rotates around the $OX$ axis, as shown in the figure.

Y X a −a b −b O rotation around the OX axis y = f(x) ≥ 0
  1. a) Rearrange $y^2$ in the equation of the ellipse and set up the integral giving the volume of the ellipsoid.
  2. b) Prove that the volume of an ellipsoid of revolution is $V = \frac{4}{3} \pi a b^2$.
  3. c) A regulation ball has a length of $30\ \text{cm}$ ($a = 15\ \text{cm}$) and a maximum diameter of $18\ \text{cm}$ ($b = 9\ \text{cm}$). Calculate its volume and express it in liters, rounding to one decimal place.
Solve
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The infinite region bounded by the hyperbola $y = \frac{1}{x}$, the $OX$ axis and the line $x = 1$ rotates around the $OX$ axis, generating an infinite solid known as…

The infinite region bounded by the hyperbola $y = \frac{1}{x}$, the $OX$ axis and the line $x = 1$ rotates around the $OX$ axis, generating an infinite solid known as Torricelli's trumpet, as shown in the figure.

y x 1 0 1 2 3 4 (1, 1) y = 1/x to +∞ rotates around the OX axis
  1. a) Calculate the volume $V(b)$ of the section of the trumpet between $x = 1$ and $x = b$.
  2. b) Find $\lim_{b \to +\infty} V(b)$ and interpret the result: is the volume of the full trumpet finite?
  3. c) Calculate the area of the region that has rotated, that is, $\lim_{b \to +\infty} \int_{1}^{b} \frac{1}{x} dx$. Compare with the previous part and comment on the painter's paradox: you can fill the trumpet with paint, but you cannot paint its surface.
Solve

Go back to the beginning: the piece that comes off the lathe is a solid of revolution, and its volume is obtained by integrating the area $\pi[f(x)]^2$ of its disks.

Summary

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