Home → Calculus → Applications of Integrals → 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.

You are watching: Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.

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.

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 = fleft( y

ight),) 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 = fleft( x

ight),) the (x-)axis, and the vertical lines (x = a) and (x = b,) where (0 le a lt b.)

Figure 2.

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

where (2pi x) means the circumference of the elementary shell, ({fleft( x

ight)}) is the height of the shell, and (dx) is its thickness.

If a region is bounded by two curves (y = fleft( x

ight)) and (y = gleft( x

ight)) on an interval (left< {a,b}
ight>,) where (0 le gleft( x

ight) le fleft( x

ight),) 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:

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:

If the region is bounded by two curves:

Suppose now that the region bounded by a curve (y = fleft( x

ight)) and the (x-)axis on the interval (left< {a,b}
ight>) 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:

Similarly, if the region bounded by a curve (x = fleft( y

ight)) and the (y-)axis on the interval (left< {c,d}
ight>) is rotating around the horizontal line (y = m,) then the volume of the obtained solid is given by

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

## 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

### 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

### Example 1.

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Find the volume of the solid obtained by rotating about the (y-)axis the region bounded by the curve

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} ight) = 0, ;;Rightarrow {x_1} = 0,;{x_2} = 3.>

By the shell method,

### 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.

Figure 3.

Using the shell method and integrating by parts, we have

ight|_0^pi – intlimits_0^pi {left( { – cos x}

ight)dx} }

ight} = 2pi left{ {left. {left< { - xcos x}
ight>}

ight|_0^pi + intlimits_0^pi {cos xdx} }

ight} = 2pi left{ {left. {left< { - xcos x}
ight>}

ight|_0^pi + left. {left< {sin x}
ight>}

ight|_0^pi }

ight} = 2pi left. {left< {sin x - xcos x}
ight>}

ight|_0^pi = 2pi left< {0 - pi cdot left( { - 1}
ight) - 0}
ight> = 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.

Figure 4.

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

ight|_0^{frac{pi }{2}} – intlimits_0^{frac{pi }{2}} {sin xdx} }

ight} = 2pi left{ {left. {left< {xsin x}
ight>}

ight|_0^{frac{pi }{2}} + left. {left< {cos x}
ight>}

ight|_0^{frac{pi }{2}}}

ight} = 2pi left. {left< {xsin x + cos x}
ight>}

ight|_0^{frac{pi }{2}} = 2pi left< {left( {frac{pi }{2} cdot 1 + 0}
ight) - left( {0 + 1}
ight)}
ight> = pi left( {pi – 2}

ight).>

### Example 4.

The region bounded by the parabola

Solution.

Figure 5.

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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

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

In our case, (c = 0,) (d = 1,) (xleft( y

ight) = {left( {y – 1}

ight)^2}.) Hence,

ight|_0^1 = 2pi left( {frac{1}{4} – frac{2}{3} + frac{1}{2}}

ight) = frac{pi }{6}.>

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