Calculus

Applications of Integrals

Applications of Integrals Logo

Volume of a Solid of Revolution: Cylindrical Shells

Sometimes finding the volume of a solid of revolution using the disk or washer method is difficult or impossible.

For example, consider the solid obtained by rotating the region bounded by the line \(y = 0\) and the curve \(y = {x^2}-{x^3}\) about the \(y-\)axis.

Solid obtained by rotating the region bounded by the cubic curve y=x^2-x^3 around the y-axis.
Figure 1.

The cross section of the solid of revolution is a washer. However, in order to use the washer method, we need to convert the function \(y = {x^2} - {x^3}\) into the form \(x = f\left( y \right),\) which is not easy.

In such cases, we can use the different method for finding volume called the method of cylindrical shells. This method considers the solid as a series of concentric cylindrical shells wrapping the axis of revolution.

With the disk or washer methods, we integrate along the coordinate axis parallel to the axes of revolution. With the shell method, we integrate along the coordinate axis perpendicular to the axis of revolution.

As before, we consider a region bounded by the graph of the function \(y = f\left( x \right),\) the \(x-\)axis, and the vertical lines \(x = a\) and \(x = b,\) where \(0 \le a \lt b.\)

Finding volume of a solid using the method of cylindrical shells
Figure 2.

The volume of the solid obtained by rotating the region about the \(y-\)axis is given by the integral

\[V = 2\pi \int\limits_a^b {xf\left( x \right)dx},\]

where \(2\pi x\) means the circumference of the elementary shell, \({f\left( x \right)}\) is the height of the shell, and \(dx\) is its thickness.

If a region is bounded by two curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) on an interval \(\left[ {a,b} \right],\) where \(0 \le g\left( x \right) \le f\left( x \right),\) then the volume of the solid obtained by rotating the region about the \(y-\)axis is expressed by the integral of the difference of two functions:

\[V = 2\pi \int\limits_a^b {x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} .\]

We can easily modify these formulas if a solid is formed by rotating around the \(x-\)axis. The two formulas listed above become

Suppose now that the region bounded by a curve \(y = f\left( x \right)\) and the \(x-\)axis on the interval \(\left[ {a,b} \right]\) is rotating around the vertical line \(x = h.\) In this case, we can apply the following formulas for finding the volume of the solid of revolution:

\[V = \left\{ {\begin{array}{*{20}{l}} {2\pi \int\limits_a^b {\left( {x - h} \right)f\left( x \right)dx} ,\text{ if } h \le a \lt b}\\ {2\pi \int\limits_a^b {\left( {h - x} \right)f\left( x \right)dx} ,\text{ if } a \lt b \le h} \end{array}} \right..\]

Similarly, if the region bounded by a curve \(x = f\left( y \right)\) and the \(y-\)axis on the interval \(\left[ {c,d} \right]\) is rotating around the horizontal line \(y = m,\) then the volume of the obtained solid is given by

\[V = \left\{ {\begin{array}{*{20}{l}} {2\pi \int\limits_c^d {\left( {y - m} \right)f\left( y \right)dy} ,\text{ if } m \le c \lt d}\\ {2\pi \int\limits_c^d {\left( {m - y} \right)f\left( y \right)dy} ,\text{ if } c \lt d \le h} \end{array}} \right..\]

Now let's return to the example mentioned above and find the volume of the solid using the shell method.

The cubic curve \(y = {x^2} - {x^3}\) intersects the \(x-\)axis at the points \(x = 0\) and \(x = 1.\) These will be the limits of integration. Then, the volume of the solid is

\[V = 2\pi \int\limits_a^b {xf\left( x \right)dx} = 2\pi \int\limits_0^1 {x\left( {{x^2} - {x^3}} \right)dx} = 2\pi \int\limits_0^1 {\left( {{x^3} - {x^4}} \right)dx} = 2\pi \left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = 2\pi \left( {\frac{1}{4} - \frac{1}{5}} \right) = \frac{\pi }{{10}}.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the volume of the solid obtained by rotating about the \(y-\)axis the region bounded by the curve \[y = 3{x^2} - {x^3}\] and the line \(y = 0.\)

Example 2

Find the volume of the solid obtained by rotating the sine function between \(x = 0\) and \(x = \pi\) about the \(y-\)axis.

Example 3

Calculate the volume of the solid obtained by rotating the cosine function between \(x = 0\) and \(x = \frac{\pi }{2}\) about the \(y-\)axis.

Example 4

The region bounded by the parabola \[x = {\left( {y - 1} \right)^2}\] and coordinate axes rotates around the \(x-\)axis. Find the volume of the obtained solid of revolution.

Example 1.

Find the volume of the solid obtained by rotating about the \(y-\)axis the region bounded by the curve \[y = 3{x^2} - {x^3}\] and the line \(y = 0.\)

Solution.

By solving the equation \(3{x^2} - {x^3} = 0\) we find the limits of integration:

\[3{x^2} - {x^3} = 0,\;\; \Rightarrow {x^2}\left( {3 - x} \right) = 0, \;\;\Rightarrow {x_1} = 0,\;{x_2} = 3.\]

By the shell method,

\[V = 2\pi \int\limits_a^b {xf\left( x \right)dx} = 2\pi \int\limits_0^3 {x\left( {3{x^2} - {x^3}} \right)dx} = 2\pi \int\limits_0^3 {\left( {3{x^3} - {x^4}} \right)dx} = 2\pi \left. {\left( {\frac{{3{x^4}}}{4} - \frac{{{x^5}}}{5}} \right)} \right|_0^3 = 2\pi \cdot {3^5}\left( {\frac{1}{4} - \frac{1}{5}} \right) = \frac{{243\pi }}{{10}}.\]

Example 2.

Find the volume of the solid obtained by rotating the sine function between \(x = 0\) and \(x = \pi\) about the \(y-\)axis.

Solution.

Solid obtained by rotating the sine half-wave around the y-axis.
Figure 3.

Using the shell method and integrating by parts, we have

\[V = 2\pi \int\limits_0^\pi {x\sin xdx} = \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = \sin xdx}\\ {du = dx}\\ {v = - \cos x} \end{array}} \right] = 2\pi \left\{ {\left. {\left[ { - x\cos x} \right]} \right|_0^\pi - \int\limits_0^\pi {\left( { - \cos x} \right)dx} } \right\} = 2\pi \left\{ {\left. {\left[ { - x\cos x} \right]} \right|_0^\pi + \int\limits_0^\pi {\cos xdx} } \right\} = 2\pi \left\{ {\left. {\left[ { - x\cos x} \right]} \right|_0^\pi + \left. {\left[ {\sin x} \right]} \right|_0^\pi } \right\} = 2\pi \left. {\left[ {\sin x - x\cos x} \right]} \right|_0^\pi = 2\pi \left[ {0 - \pi \cdot \left( { - 1} \right) - 0} \right] = 2{\pi ^2}.\]

Example 3.

Calculate the volume of the solid obtained by rotating the cosine function between \(x = 0\) and \(x = \frac{\pi }{2}\) about the \(y-\)axis.

Solution.

Solid obtained by rotating the cosine function between x=0 and x=pi/2 about the y-axis.
Figure 4.

Using the method of cylindrical shells and integrating by parts, we get

\[V = 2\pi \int\limits_0^{\frac{\pi }{2}} {x\cos xdx} = \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = \cos xdx}\\ {du = dx}\\ {v = \sin x} \end{array}} \right] = 2\pi \left\{ {\left. {\left[ {x\sin x} \right]} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {\sin xdx} } \right\} = 2\pi \left\{ {\left. {\left[ {x\sin x} \right]} \right|_0^{\frac{\pi }{2}} + \left. {\left[ {\cos x} \right]} \right|_0^{\frac{\pi }{2}}} \right\} = 2\pi \left. {\left[ {x\sin x + \cos x} \right]} \right|_0^{\frac{\pi }{2}} = 2\pi \left[ {\left( {\frac{\pi }{2} \cdot 1 + 0} \right) - \left( {0 + 1} \right)} \right] = \pi \left( {\pi - 2} \right).\]

Example 4.

The region bounded by the parabola \[x = {\left( {y - 1} \right)^2}\] and coordinate axes rotates around the \(x-\)axis. Find the volume of the obtained solid of revolution.

Solution.

Solid obtained by rotating about the x-axis the region bounded by the parabola and two coordinate axes.
Figure 5.

We can use the shell method to calculate the volume of the given solid. The region here rotates around the \(x-\)axis, so the volume is defined by the formula

\[V = 2\pi \int\limits_c^d {yx(y)dy} ,\]

assuming the variable \(y\) varies from \(c\) to \(d.\)

In our case, \(c = 0,\) \(d = 1,\) \(x\left( y \right) = {\left( {y - 1} \right)^2}.\) Hence,

\[V = 2\pi \int\limits_0^1 {y{{\left( {y - 1} \right)}^2}dy} = 2\pi \int\limits_0^1 {y\left( {{y^2} - 2y + 1} \right)dy} = 2\pi \int\limits_0^1 {\left( {{y^3} - 2{y^2} + y} \right)dy} = 2\pi \left. {\left[ {\frac{{{y^4}}}{4} - \frac{{2{y^3}}}{3} + \frac{{{y^2}}}{2}} \right]} \right|_0^1 = 2\pi \left( {\frac{1}{4} - \frac{2}{3} + \frac{1}{2}} \right) = \frac{\pi }{6}.\]

See more problems on Page 2.

Page 1 Page 2