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.

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 \(0 \le a \lt b.\)

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

- If the region is bounded by a curve and \(y-\)axis:

\[V = 2\pi \int\limits_c^d {yf\left( y \right)dy}; \]

- If the region is bounded by two curves:

\[V = 2\pi \int\limits_c^d {y\left[ {f\left( y \right) – g\left( y \right)} \right]dy} .\]

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:

\[{V \text{ = }}\kern0pt{\left\{ {\begin{array}{*{20}{l}} {2\pi \int\limits_a^b {\left( {x – h} \right)f\left( x \right)dx} ,\text{ if } h \le a \lt b}\\ {2\pi \int\limits_a^b {\left( {h – x} \right)f\left( x \right)dx} ,\text{ if } a \lt b \le h} \end{array}} \right.}\]

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

\[{V \text{ = }}\kern0pt{\left\{ {\begin{array}{*{20}{l}} {2\pi \int\limits_c^d {\left( {y – m} \right)f\left( y \right)dy} ,\text{ if } m \le c \lt d}\\ {2\pi \int\limits_c^d {\left( {m – y} \right)f\left( y \right)dy} ,\text{ if } c \lt d \le h} \end{array}} \right.}\]

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 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 \(x = \large{\frac{\pi }{2}}\normalsize\) 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 5

Calculate the volume of a sphere of radius \(R\) using the shell integration.### Example 6

Find the volume of the solid obtained by rotating about the \(y-\)axis the region bounded by the curve \(y = {e^{ – x}},\) where \(0 \le x \lt \infty \) and the horizontal line \(y = 0.\)### Example 7

Find the volume of the solid formed by rotating about the line \(x = -1\) the region bounded by the parabola \(y = {x^2}\) and the lines \(y = 0,\) \(x = 1.\)### Example 8

The region bounded by curves \(y = {x^2}\) and \(y = \sqrt x \) rotates around the \(y-\)axis. Find the volume of the solid of revolution.### Example 9

The region is bounded by the hyperbola \(y = \large{\frac{1}{x}}\normalsize\) and the line \(y = \large{\frac{5}{2}}\normalsize – x.\) Find the volume of the solid formed by rotating the region about the \(y-\)axis.### Example 10

Find the volume of the solid obtained by rotating about the line \(y = 2\) the region bounded by the curve \(x = 1 + {y^2}\) and the lines \(y = 0,\) \(y = 1.\)### 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:

\[{3{x^2} – {x^3} = 0,}\;\; \Rightarrow {{x^2}\left( {3 – x} \right) = 0, }\;\;\Rightarrow {{x_1} = 0,\;}\kern0pt{{x_2} = 3.}\]

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.

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 \(x = \large{\frac{\pi }{2}}\normalsize\) about the \(y-\)axis.Solution.

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.

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,

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

### Example 5.

Calculate the volume of a sphere of radius \(R\) using the shell integration.Solution.

We consider the region which occupies the part of the disk \({x^2} + {y^2} = {R^2}\) in the first quadrant. Taking the integral \(2\pi \int\limits_0^R {x\sqrt {{R^2} – {x^2}} dx} \) and multiplying the result by two, we find the volume of the sphere.

It’s convenient to change the variable. Let \({R^2} – {x^2} = {t^2},\) so \(xdx=-tdt.\) When \(x = 0,\) then \(t = R,\) and when \(x = R,\) then \(t = 0.\) Hence, the integral becomes

\[{V = 4\pi \int\limits_0^R {x\sqrt {{R^2} – {x^2}} dx} }={ 4\pi \int\limits_R^0 {t\left( { – tdt} \right)} }={ 4\pi \int\limits_0^R {{t^2}dt} }={ \left. {\frac{{4\pi {t^3}}}{3}} \right|_0^R }={ \frac{{4\pi {R^3}}}{3}.}\]

### Example 6.

Find the volume of the solid obtained by rotating about the \(y-\)axis the region bounded by the curve \(y = {e^{ – x}},\) where \(0 \le x \lt \infty \) and the horizontal line \(y = 0.\)Solution.

The volume of the solid is expressed by the improper integral

\[V = 2\pi \int\limits_0^\infty {x{e^{ – x}}dx} .\]

To take the integral we use integration by parts:

\[{V = 2\pi \int\limits_0^\infty {x{e^{ – x}}dx} }={ \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = {e^{ – x}}dx}\\ {du = dx}\\ {v = – {e^{ – x}}} \end{array}} \right] }={ 2\pi \left\{ {\left. {\left[ { – x{e^{ – x}}} \right]} \right|_0^\infty – \int\limits_0^\infty {\left( { – {e^{ – x}}} \right)dx} } \right\} }={ 2\pi \left\{ {\left. {\left[ { – x{e^{ – x}}} \right]} \right|_0^\infty + \int\limits_0^\infty {{e^{ – x}}dx} } \right\} }={ 2\pi \left\{ {\left. {\left[ { – x{e^{ – x}}} \right]} \right|_0^\infty – \left. {\left[ {{e^{ – x}}} \right]} \right|_0^\infty } \right\} }={ – 2\pi \left. {\left[ {{e^{ – x}}\left( {x + 1} \right)} \right]} \right|_0^\infty .}\]

We replace the infinite upper limit with a finite value \(b\) and then take the limit as \(b \to \infty.\)

\[{V = – 2\pi \lim\limits_{b \to \infty } \left. {\left[ {{e^{ – x}}\left( {x + 1} \right)} \right]} \right|_0^b }={ – 2\pi \lim\limits_{b \to \infty } \left[ {{e^{ – b}}\left( {b + 1} \right) – {e^0}\left( {0 + 1} \right)} \right] }={ 2\pi \lim\limits_{b \to \infty } \left[ {1 – \frac{{b + 1}}{{{e^b}}}} \right].}\]

By L’Hopital’s Rule,

\[{\lim \limits_{b \to \infty } \frac{{b + 1}}{{{e^b}}} }={\lim \limits_{b \to \infty } \frac{{\left( {b + 1} \right)’}}{{\left( {{e^b}} \right)’}} }={ \lim \limits_{b \to \infty } \frac{1}{{{e^b}}} }={ 0.}\]

So, the volume of the solid is \(V = 2\pi .\)

### Example 7.

Find the volume of the solid formed by rotating about the line \(x = -1\) the region bounded by the parabola \(y = {x^2}\) and the lines \(y = 0,\) \(x = 1.\)Solution.

The curve is rotating about the line \(x = -1,\) which is on the left from the interval of integration \(\left[ {0,1} \right].\) Therefore, we use the following formula for calculating the volume:

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

where \(h = -1,\) \(a = 0,\) \(b = 1.\) So, we have

\[{V = 2\pi \int\limits_0^1 {\left( {x + 1} \right){x^2}dx} }={ 2\pi \int\limits_0^1 {\left( {{x^3} + {x^2}} \right)dx} }={ 2\pi \left. {\left[ {\frac{{{x^4}}}{4} + \frac{{{x^3}}}{3}} \right]} \right|_0^1 }={ 2\pi \left( {\frac{1}{4} + \frac{1}{3}} \right) }={ \frac{{7\pi }}{6}.}\]

### Example 8.

The region bounded by curves \(y = {x^2}\) and \(y = \sqrt x \) rotates around the \(y-\)axis. Find the volume of the solid of revolution.Solution.

Obviously, both curves intersect at the points \(x = 0\) and \(x = 1,\) so using the shell method we will integrate from \(0\) to \(1\).

The region here is bounded by two curves. Therefore the integration formula is written in the form similar to the washer method:

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

Substituting \(a = 0,\) \(b= 1,\) \(f\left( x \right) = \sqrt x,\) \(g\left( x \right) = {x^2},\) we obtain:

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

Compare the result with Example \(3\) on the page Volume of a Solid of Revolution: Disks and Washers.

### Example 9.

The region is bounded by the hyperbola \(y = \large{\frac{1}{x}}\normalsize\) and the line \(y = \large{\frac{5}{2}}\normalsize – x.\) Find the volume of the solid formed by rotating the region about the \(y-\)axis.Solution.

First we determine the points of intersection of both curves:

\[{\frac{1}{x} = \frac{5}{2} – x,}\;\; \Rightarrow {{x^2} – \frac{5}{2}x + 1 = 0,}\;\; \Rightarrow {2{x^2} – 5x + 2 = 0,}\;\; \Rightarrow {D = {\left( { – 5} \right)^2} – 4 \cdot 2 \cdot 2 = 9,}\;\; \Rightarrow {{x_{1,2}} = \frac{{5 \pm \sqrt 9 }}{4} }={ \frac{1}{2},\,2.}\]

We use the shell method for finding the volume. As the region is bounded by two curves, the integration formula is given by

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

Here \(a = \large{\frac{1}{2}}\normalsize,\) \(b = 2,\) \(f\left( x \right) = \large{\frac{5}{2}}\normalsize – x,\) \(g\left( x \right) = \large{\frac{1}{x}}\normalsize.\) So the volume of the solid is

\[{V = 2\pi \int\limits_{\frac{1}{2}}^2 {x\left[ {\frac{5}{2} – x – \frac{1}{x}} \right]dx} }={ 2\pi \int\limits_{\frac{1}{2}}^2 {\left[ {\frac{{5x}}{2} – {x^2} – 1} \right]dx} }={ 2\pi \left. {\left[ {\frac{{5{x^2}}}{4} – \frac{{{x^3}}}{3} – x} \right]} \right|_{\frac{1}{2}}^2 }={ 2\pi \left[ {\left( {5 – \frac{8}{3} – 2} \right) }\right.}-{\left.{ \left( {\frac{5}{{16}} – \frac{1}{{24}} – \frac{1}{2}} \right)} \right] }={ 2\pi \left[ {\frac{7}{2} – \frac{8}{3} – \frac{5}{{16}} + \frac{1}{{24}}} \right] }={ 2\pi \cdot \frac{9}{{16}} }={ \frac{{9\pi }}{8}}\]

### Example 10.

Find the volume of the solid obtained by rotating about the line \(y = 2\) the region bounded by the curve \(x = 1 + {y^2}\) and the lines \(y = 0,\) \(y = 1.\)Solution.

The given figure rotates around the horizontal axis \(y = m = 2.\) To determine the volume of the solid of revolution, we apply the shell integration formula in the form

\[V = 2\pi \int\limits_c^d {\left( {m – y} \right)x\left( y \right)dy} .\]

In this case, \(m = 2,\) \(c = 0,\) \(d = 1,\) \(x\left( y \right) = 1 + {y^2}.\) This yields

\[{V = 2\pi \int\limits_0^1 {\left( {2 – y} \right)\left( {1 + {y^2}} \right)dy} }={ 2\pi \int\limits_0^1 {\left( {2 – y + 2{y^2} – {y^3}} \right)dy} }={ 2\pi \left. {\left[ {2y – \frac{{{y^2}}}{2} + \frac{{2{y^3}}}{3} – \frac{{{y^4}}}{4}} \right]} \right|_0^1 }={ 2\pi \left[ {2 – \frac{1}{2} + \frac{2}{3} – \frac{1}{4}} \right] }={ \frac{{23\pi }}{6}.}\]