When a plane region rotates around a certain line, it forms a three-dimensional solid. To find the volumes of solids of revolution, we use the definite integral, which connects the areas of cross-sections with the total volume of the solid. First, we will look at the main formulas and then apply them to examples of different levels of difficulty.
From a Plane Region to a Three-Dimensional Solid: How a Solid of Revolution Is Formed
Let us start with the basic idea. Suppose we have a region in the plane bounded by the graph of a function, a coordinate axis, and two lines. If this region is rotated around a certain axis, each of its points moves along a circle.
As a result, the plane region forms a three-dimensional solid. Its shape depends on the region itself and on the line chosen as the axis of rotation.
For example, suppose the region lies between the graph of the function \( y=f(x) \) and the \( x \)-axis, and the region is rotated around this axis. In this case, the distance from the graph to the axis determines the radius of the circular cross-sections of the resulting solid.
Why does the definite integral appear here? We can imagine the solid as a collection of very thin layers. For each layer, we can approximately calculate its volume and then add all these values together. As we pass to the limit, this sum leads to a definite integral.
Therefore, to find the volume, we need to determine the area of a cross-section of the solid and integrate it over the corresponding interval.
Rotation Around the X-Axis: Formula Using Circular Cross-Sections
Consider the function
\[ y=f(x), \]
which is continuous on the interval \( [a,b] \). Suppose the region bounded by its graph, the \( x \)-axis, and the lines \( x=a \) and \( x=b \) rotates around the \( x \)-axis.
Fix some value of \( x \). The cross-section of the solid by a plane perpendicular to the \( x \)-axis will be a circle. Its radius is equal to the distance from the graph of the function to the axis of rotation:
\[ R(x)=|f(x)|. \]
Let us denote the area of the cross-section by \( A(x) \). Then
\[ A(x)=\pi\cdot(R(x))^2. \]
Since
\[ (|f(x)|)^2=(f(x))^2, \]
we obtain
\[ A(x)=\pi\cdot(f(x))^2. \]
The volume of the solid is
\[ V=\int_a^b A(x)\,dx. \]
Therefore,
\[ V=\pi\int_a^b (f(x))^2\,dx. \]
Why is the function squared? The reason is directly related to the geometry of a circle. The value \( f(x) \) determines the distance to the axis of rotation, while the area of a circle depends on the square of its radius.
Rotation Around the Y-Axis: Changing the Variable of Integration
What changes if the region rotates around the \( y \)-axis? In this case, it is convenient to consider horizontal cross-sections and describe the boundary of the region as a function of \( y \).
Suppose
\[ x=g(y), \]
and the variable \( y \) takes values from \( c \) to \( d \).
The radius of the circular cross-section is
\[ R(y)=|g(y)|. \]
Therefore, its area is
\[ A(y)=\pi\cdot(g(y))^2. \]
Integrating with respect to \( y \), we obtain
\[ V=\int_c^d A(y)\,dy, \]
or
\[ V=\pi\int_c^d (g(y))^2\,dy. \]
Thus, the main difference lies in the variable of integration. For cross-sections perpendicular to the \( y \)-axis, the radius is written as a function of \( y \), and the integration is performed with respect to this variable.
Annular Cross-Sections: When the Solid Has an Inner Cavity
Not every cross-section of a solid of revolution is a solid circle. If a hollow space remains inside the solid after rotation, the cross-section has the shape of an annulus.
In this case, we need to determine two radii. The outer radius \( R(x) \) is the larger distance from the axis of rotation to the boundary of the region, while the inner radius \( r(x) \) is the smaller distance.
The area of the annular cross-section is equal to the difference between the areas of two circles:
\[ A(x)=\pi\cdot(R(x))^2-\pi\cdot(r(x))^2. \]
Or
\[ A(x)=\pi\cdot\left((R(x))^2-(r(x))^2\right). \]
Therefore, the volume is
\[ V=\pi\int_a^b \left((R(x))^2-(r(x))^2\right)\,dx. \]
Why do we need to subtract the areas? The outer circle includes the entire cross-section together with the inner cavity. Since this cavity is not part of the solid, its area must be subtracted.
At the same time, the outer and inner radii should not be determined simply by which graph lies above or below the other. What matters is the actual distance from each boundary of the region to the axis of rotation.
Volumes of Solids of Revolution: Practical Calculations Using the Integral
After reviewing the main formulas, let us move on to calculations. In each problem, we will first determine the geometry of the region and the required radii, and then set up the corresponding definite integral.
Example 1. Find the volume of the solid formed by rotating the region bounded by the graph
\[
y=x,
\]
the \( x \)-axis, and the lines
\[
x=0,\qquad x=2
\]
around the \( x \)-axis.
On the interval \( [0,2] \), the function is nonnegative, so the radius of the circular cross-section is
\[ R(x)=x. \]
Its area is
\[ A(x)=\pi\cdot x^2. \]
Then
\[ V=\int_0^2 A(x)\,dx = \pi\int_0^2 x^2\,dx. \]
Let us find the antiderivative:
\[ \int x^2\,dx=\frac{x^3}{3}. \]
Therefore,
\[ V=\pi\left.\frac{x^3}{3}\right|_0^2. \]
Substitute the limits:
\[ V = \pi\left( \frac{2^3}{3} – \frac{0^3}{3} \right) = \frac{8\pi}{3}. \]
Thus,
\[ V=\frac{8\pi}{3}. \]
Example 2. Find the volume of the solid formed by rotating the region bounded by the graph
\[
y=x^2,
\]
the \( x \)-axis, and the lines
\[
x=0,\qquad x=2
\]
around the \( x \)-axis.
On the given interval,
\[ R(x)=x^2. \]
Therefore, the area of the cross-section is
\[ A(x)=\pi\cdot(x^2)^2 = \pi\cdot x^4. \]
Hence,
\[ V=\pi\int_0^2 x^4\,dx. \]
Let us find the antiderivative:
\[ \int x^4\,dx=\frac{x^5}{5}. \]
Then
\[ V=\pi\left.\frac{x^5}{5}\right|_0^2. \]
Substitute the limits:
\[ V = \pi\left( \frac{2^5}{5} – \frac{0^5}{5} \right) = \frac{32\pi}{5}. \]
Thus,
\[ V=\frac{32\pi}{5}. \]
Note. The radius is equal to \( x^2 \), but in the formula for the area of a circle, it is squared once again. That is why \( x^4 \) appears under the integral.
Example 3. Find the volume of the solid formed by rotating the region bounded by the curve
\[
x=\sqrt{y},
\]
the \( y \)-axis, and the lines
\[
y=0,\qquad y=4
\]
around the \( y \)-axis.
Since the rotation is around the \( y \)-axis, we integrate with respect to \( y \).
The radius is
\[ R(y)=\sqrt{y}. \]
Therefore,
\[ A(y)=\pi\cdot(\sqrt{y})^2 = \pi\cdot y. \]
Hence,
\[ V=\pi\int_0^4 y\,dy. \]
Let us find the antiderivative:
\[ \int y\,dy=\frac{y^2}{2}. \]
Then
\[ V=\pi\left.\frac{y^2}{2}\right|_0^4. \]
Substitute the limits:
\[ V = \pi\left( \frac{4^2}{2} – \frac{0^2}{2} \right) = 8\pi. \]
Thus,
\[ V=8\pi. \]
Example 4. Find the volume of the solid formed by rotating the region between the graphs
\[
y=2,\qquad y=x,
\]
on the interval
\[
0\leq x\leq2
\]
around the \( x \)-axis.
After rotation, annular cross-sections are formed. On the interval \( [0,2] \), the distance from the \( x \)-axis to the line \( y=2 \) is greater than or equal to the distance to the graph \( y=x \).
Therefore,
\[ R(x)=2, \]
and
\[ r(x)=x. \]
The area of the annular cross-section is
\[ A(x) = \pi\cdot\left(2^2-x^2\right) = \pi\cdot(4-x^2). \]
Then
\[ V=\pi\int_0^2 (4-x^2)\,dx. \]
Let us find the antiderivative:
\[ \int (4-x^2)\,dx = 4x-\frac{x^3}{3}. \]
Therefore,
\[ V = \pi \left. \left( 4x-\frac{x^3}{3} \right) \right|_0^2. \]
Substitute the limits:
\[ V = \pi \left( 8-\frac{8}{3} \right) = \pi\cdot\frac{16}{3}. \]
Thus,
\[ V=\frac{16\pi}{3}. \]
Example 5. Find the volume of the solid formed by rotating around the \( x \)-axis the region bounded by the graphs
\[
y=2\cdot x,\qquad y=x^2.
\]
In this case, the limits of integration are not given, so we first find the points of intersection of the graphs:
\[ 2\cdot x=x^2. \]
Move all terms to one side:
\[ x^2-2\cdot x=0. \]
Factor out \( x \):
\[ x\cdot(x-2)=0. \]
Hence, \( x=0 \) or \( x=2 \). Therefore, the region lies on the interval \( [0,2] \).
On this interval, the line \( y=2\cdot x \) lies above the parabola \( y=x^2 \), and at the points of intersection their values are equal. Therefore, the larger distance from the \( x \)-axis is
\[ R(x)=2\cdot x, \]
while the smaller distance is
\[ r(x)=x^2. \]
The area of the annular cross-section is
\[ A(x) = \pi\cdot \left( (2\cdot x)^2-(x^2)^2 \right). \]
Simplifying, we get
\[ A(x) = \pi\cdot \left( 4\cdot x^2-x^4 \right). \]
Therefore,
\[ V = \pi\int_0^2 \left( 4\cdot x^2-x^4 \right)\,dx. \]
Let us find the antiderivative:
\[ \int \left( 4\cdot x^2-x^4 \right)\,dx = \frac{4\cdot x^3}{3} – \frac{x^5}{5}. \]
Therefore,
\[ V = \pi \left. \left( \frac{4\cdot x^3}{3} – \frac{x^5}{5} \right) \right|_0^2. \]
Substitute the limits:
\[ V = \pi \left( \frac{4\cdot8}{3} – \frac{32}{5} \right). \]
Then
\[ V = \pi \left( \frac{32}{3} – \frac{32}{5} \right). \]
Rewrite the fractions with a common denominator:
\[ V = \pi \left( \frac{160}{15} – \frac{96}{15} \right). \]
Thus,
\[ V=\frac{64\pi}{15}. \]
This example is different because the limits of integration had to be found by solving the equation for the points of intersection of the graphs. After that, the outer and inner radii were determined by their distances from the axis of rotation.
The Next Step in Integral Calculus: Topics for Further Study
After calculating the volumes of solids of revolution, you can move on to other problems in which the definite integral plays a key role. What topics are worth studying next? Here are three logical directions to continue.
- Arc Length of a Curve: Formula Using the Definite Integral — We will look at how to use the definite integral to find the length of a section of a curve between two given points.
- Improper Integrals: Convergence, Divergence, and Examples — We will learn how to work with infinite limits and unbounded functions and determine whether an integral converges.
- Double Integrals: Meaning, Region of Integration, and Examples — We will explore how to integrate functions of two variables, determine the region of integration, and use the main methods of calculation.
Volumes of Solids of Revolution: From a Flowchart to Code
If you enjoy programming, try implementing the algorithm shown in the flowchart in your favorite programming language. It approximately calculates the volume of a solid formed by rotating around the \( x \)-axis the region under the graph of the function
\[ y=x^2+1. \]
To do this, the interval is divided into equal parts, the radius is determined for each part, and the volumes of the thin layers are added together step by step. Try reproducing these steps in code and see how the result changes as the number of parts increases.
