Volumes of Revolution
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.
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.
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 $\pi$.
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.
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.
- a) Calculate the volume of the solid generated using the integral $V = \pi \int_{0}^{5} [f(x)]^2 \, dx$.
- b) What geometric solid has been generated? Check the result using the formula for the volume of that solid.
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.
- a) What geometric solid is generated and between what limits must you integrate?
- b) Calculate its volume using the corresponding integral.
- c) Check the result using the formula $V = \frac{4}{3} \pi r^3$.
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.
- a) Calculate its volume using the integral $V = \pi \int_{0}^{3} x^2 \, dx$.
- b) Identify the radius and height of the cone and check the result using the formula $V = \frac{1}{3} \pi r^2 h$.
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.
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:
«$V = \pi \int_{-1}^{1} (x^2 + 1) \, dx = \pi \left[ \frac{x^3}{3} + x \right]_{-1}^{1} = \pi \left( \frac{1}{3} + 1 + \frac{1}{3} + 1 \right) = \frac{8\pi}{3}\ \text{u}^3$»
- a) Identify the error in his approach and explain what it consists of.
- b) Calculate the volume correctly.
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 $g(x)$.
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.
- a) Find the intersection points of both curves.
- b) Set up the volume as the difference of two integrals (washer method) and calculate it.
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 $x = 9$.
- a) Calculate the interior volume of the cup in $\text{cm}^3$, rounding to the nearest integer.
- 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.
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).
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).
- 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.
- b) Calculate the volume of the funnel.
- 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.
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.
- a) Calculate the volume of material needed to manufacture the part, assuming it is solid between the axis and the profile.
- 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.
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 $x = 2$.
- a) Calculate the volume of the solid generated by rotating around the $OX$ axis.
- 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.)
- c) Although the rotating region is the same, the two volumes do not coincide. Explain why with reasons.
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:
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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 $y^2$.
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If the limit of integration is infinite, first calculate the volume up to $b$ and then find the limit as $b\to+\infty$.
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The volume can be finite even if the area of the region that rotates is infinite: this is the paradox of Torricelli's trumpet.
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.
- a) Rearrange $y^2$ in the equation of the ellipse and set up the integral giving the volume of the ellipsoid.
- b) Prove that the volume of an ellipsoid of revolution is $V = \frac{4}{3} \pi a b^2$.
- 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.
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.
- a) Calculate the volume $V(b)$ of the section of the trumpet between $x = 1$ and $x = b$.
- b) Find $\lim_{b \to +\infty} V(b)$ and interpret the result: is the volume of the full trumpet finite?
- 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.
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.