# 1.3: Volume by Cylindrical Shells - Mathematics

## Cylindrical Shells

Consider rotating the region between the curve

[y = x^2, onumber]

the line

[x = 2, onumber]

and the x-axis about the y-axis.

If instead of taking a cross section perpendicular to the y-axis, we take a cross section perpendicular to the x-axis, and revolve it about the y-axis, we get a cylinder. Recall that the area of a cylinder is given by:

[ A = 2pi r h onumber]

where (r) is the radius of the cylinder and (h) is the height of the cylinder. We can see that the radius is the x coordinate of the point on the curve, and the height is the y coordinate of the curve. Hence

[A(x) = 2pi xy = 2pi x(x^2). onumber]

Therefore the volume is given by

[ egin{align*} ext{Volume} &= 2pi int_{0}^{1} x^3 dx &= dfrac{pi}{2}. end{align*} ]

Example 1

Find the volume of revolution of the region bounded by the curves

• (y = x^2 + 2)
• (y = x + 4)
• and the y-axis

Solution

We draw the picture with a cross section perpendicular to the x-axis.

The radius of the cylinder is (x) and the height is the difference of the (y) coordinates:

[h = (x + 4) - (x^2 + 2). onumber]

We solve for (b).

[egin{align*} (x+4)&=(x^2+2) x^2-x-2&=0 (x-2)(x+1)&=0 end{align*} ]

So that (b = 2). Hence the volume is equal to

[egin{align*} 2 pi int_{0}^{2} ig[ (x+4) -(x^2+2) ig] dx &= 2 pi int_{0}^{2}(x^2+2x-x^3) dx &= 2 pi left(dfrac{x^3}{3}+x^2-dfrac{x^4}{4} ight]_{0}^{2} &= 2 pi Big( dfrac{8}{3}+4 -4 Big) &= dfrac{16 pi}{3}. end{align*} ]

## Exercises

1. (y=x^2-3x+2), (y=0 ) about the y-axis
2. (y=x^2-7x+6), (y=0 ) about the y-axis
4. (y=xsqrt{1+x^3}), (y=0), (x=2 ) about the y-axis
7. (y=x^2-2x+1), (y=1) about the line (x=3)

## Use cylindrical shells to find the volume of the solid generated when the region bounded by curves = 4 −

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

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

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.

## 1.3: Description of Strain in the Cylindrical Coordinate System

• Contributed by Tomasz Wierzbicki
• Professor (Mechanical Engineering) at Massachusetts Institute of Technology
• Sourced from MIT OpenCourseWare

In this section the strain-displacement relations will be derived in the cylindrical coordinate system ((r, heta, z)).

The polar coordinate system is a special case with (z = 0). The components of the displacement vector are (, u_z>). There are two ways of deriving the kinematic equations. Since strain is a tensor, one can apply the transformation rule from one coordinate to the other. This approach is followed for example on pages 125-128 of the book on &ldquoA First Course in Continuum Mechanics&rdquo by Y.C. Fung. Or, the expression for each component of the strain tensor can be derived from the geometry. The latter approach is adopted here. The diagonal (normal) components (epsilon_), (epsilon_< heta heta>), and (epsilon_) represent the change of length of an infinitesimal element. The non-diagonal (shear) components describe the change of angles.

Figure (PageIndex<1>): Rectangular and cylindrical coordinate system. Figure (PageIndex<2>): Change of length in the radial direction.

The radial strain is solely due to the presence of the displacement gradient in the (r)-direction

The circumferential strain has two components

The first component is the change of length due to radial displacement, and the second component is the change of length due to circumferential displacement.

From Figure ((PageIndex<3>)) the components (epsilon_< heta heta>^<(1)>) and (epsilon_< heta heta>^<(2)>) are calculated as

Figure (PageIndex<3>): Two deformation modes responsible for the circumferential (hoop) strain.

The total circumferential (hoop) component of the strain tensor is

The strain components in the (z)-direction is the same as in the rectangular coordinate system

The shear strain (epsilon_) describes a change in the right angle.

Figure (PageIndex<4>): Construction that explains change of angles due to radial and circumferential displacement.

From Figure ((PageIndex<4>)) the shear strain over the () plane is

On the () plane, the (epsilon_) shear develops from the respective gradients, see Figure ((PageIndex<5>)).

Figure (PageIndex<5>): Change of angles are () plane.

From the construction in Figure ((PageIndex<4>)), the component (epsilon_) is

Finally, a similar picture is valid on the (tangent) () plane

Figure (PageIndex<6>): Visualization of the strain component (epsilon_< heta z>).

The component (epsilon_< heta z>) of the strain tensor is one half of the change of angles, i.e.

To sum up the derivation, the six components of the infinitesimal strain tensor in the cylindrical coordinate system are

Considerable simplifications are obtained in the case of axial (rotational symmetry for which (u_ < heta>= 0) and (frac [,] = 0)

The application of the above geometrical relations for axi-symmetric loading of circular plates and cylindrical shells will be given in subsequent chapters.

## Volumes by Cylindrical Shells Method

Let’s consider the problem of finding the volume of the solid obtained by rotating about the (x)-axis or parallel to (x)-axis the region, where the core idea of cylindrical shells method for finding volumes. If we slice perpendicular to the (y)-axis, we get a cylinder. But to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation for (x) in terms of (y). The method of cylindrical shells is being used for finding the volume in this case, that is easier to use in such a case. We can see a cylindrical shell with inner radius, outer radius, and height. Its volume is calculated by subtracting the volume of the inner cylinder from the volume of the outer cylinder.

### Worked Example of Finding a Volume by Cylindrical Shells Method

The diagram shows the graph of (displaystyle f(x) = frac <1 + x^2>).

The area bounded by (y = f(x) ), the line (x = 1) and the (x)-axis is rotated about the line (x=1) to form a solid volume. Use the cylindrical shells method to find the volume of the solid.

(egin displaystyle equire
delta V &= pi Big[(1-x + delta x)^2 – (1-x)^2 Big] imes y
&= pi Big[(1-x)^2 + 2(1-x) delta x +(delta x)^2 -(1-x)^2 Big] imes y
&= pi Big[2(1-x) delta x Big] imes frac <1+x^2>&color (delta x)^2 approx 0
&= 2 pi Big[ frac <1 + x^2>Big] delta x
&= 2 pi Big[ frac<1 – (1+x^2) + x> <1 + x^2>Big] delta x
&= 2 pi Big[ frac<1> <1+x^2>– 1 + frac <1+x^2>Big] delta x
V &= 2 pi int_<0>^ <1>Big[ frac<1> <1+x^2>– 1 + frac <1+x^2>Big] dx
&= 2 pi Big[ an^ <-1>x – x + frac<1><2>log_e (1 + x^2) Big]_<0>^ <1>
&= 2 pi Big(frac <4>– 1 + frac<1> <2>log_e <2>Big)
end \ )

## Slicing Spheres

Last week we saw how to compute the area of a circle from first principles. What about spheres?

To compute the volume of a sphere, let’s show that a hemisphere (with radius (r)) has the same volume as the vase shown in the figure below, formed by carving a cone from the circular cylinder with radius and height (r). Why this shape? Here’s why: if we cut these two solids at any height (h) (between 0 and (r)), the areas of the two slices match. Indeed, the slice—usually called cross section—of the sphere is a circle of radius (sqrt), which has area (pi(r^2-h^2)). Similarly, the vase’s cross section is a radius (r) circle with a radius (h) circle cut out, so its area is (pi r^2-pi h^2), as claimed.

When sliced by a horizontal plane at any height (h), the hemisphere and vase have equal cross-sectional areas. (Shown here for (h = 0.4cdot r).) By Cavalieri's Principle, this implies that they have equal volumes.

If we imagine the hemisphere and vase as being made from lots of tiny grains of sand, then we just showed, intuitively, that the two solids have the same number of grains of sand in every layer. So there should be the same number of grains in total, i.e., the volumes should match. This intuition is exactly right:

Cavalieri’s Principle: any two shapes that have matching horizontal cross sectional areas also have the same volume.

So the volumes are indeed equal, and all that’s left is to compute the volume of the vase. But we can do this! Recall that the cone has volume (frac<1> <3>( ext) ( ext) = frac<1><3>pi r^3) (better yet, prove this too! Hint: use Cavalieri’s Principle again to compare to a triangular pyramid). Likewise, the cylinder has volume (( ext) ( ext) = pi r^3), so the vase (and hemisphere) have volume (pi r^3 – frac<1> <3>pi r^3 = frac<2><3>pi r^3). The volume of the whole sphere is thus (frac<4><3>pi r^3). Success!

The following visualization illustrates what we have shown, namely $ext + ext = ext.$ The “grains of sand” in the hemisphere are being displaced horizontally by the stabbing cone, and at the end we have exactly filled the cylinder.

Stabbing a cone into a hemisphere made of horizontally-moving "sand particles" exactly fills a cylinder.

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## Solids of Revolution and the Shell Method

Briefly, a solid of revolution is the solid formed by revolving a plane region around a fixed axis.

We defined solids of revolution in a previous article, AP Calculus Review: Disk and Washer Methods. So you might want to read up before continuing.

### Shells, Shells, and more Shells…

Suppose you need to find the volume of a solid of revolution. First we have to decide how to slice the solid. If you wanted to slice perpendicular to the axis of revolution, then you would get slabs that look like thin cylinders (disks) or cylinders with circles removed (washers). However, the Shell Method requires a different kind of slicing.

Imagine that your solid is made of cookie dough. And you have a set of circular cookie cutters of various sizes. Starting with the smallest cookie cutter and progressing to larger ones, let’s slice through the dough in concentric rings.

Making sure to slice in the same direction as the axis of revolution, you will get a clump of nested shells, or thin hollow cylindrical objects.

### Approximating the Volume

Now let’s take a closer look at a single shell.

As long as the thickness is small enough, the volume of the shell can be approximated by the formula:

Note that the volume is simply the circumference (2&pir) times the height (h) times the thickness (w). In fact, you can think of cutting the shell along its height and “unrolling” it to produce a thin rectangular slab. Then the volume is simply length × height × width as in any rectangular solid.

Now suppose we have a solid of revolution with generating region being the area under a function y = f(x) between x = a and x = b. And suppose that the y-axis as its axis of symmetry. (This is the easiest case).

f(x) = x 2 + 1, between 2 and 6, revolved around the y-axis, generating a solid of revolution using shell method” width=�″ height=�″ wp-image-12021″ />

Imagine what happens to a thin vertical strip of the region as it revolves around the y-axis. Because the axis is also vertical, the strip will sweep out a cylindrical shell. Furthermore, we have the following info about each shell:

• Its radius is r = x (distance from a typical point to the yaxis).
• The height is h = y = f(x).
• Its thickness is a small change in x, which we label as &Deltax or dx.

Therefore, the approximate volume of a typical shell is:

### Integration

But remember, that’s only a single shell. The solid consists of shells that were sliced at various positions xk along the x-axis. So we should add them up to get the approximate volume of the entire solid.

Finally, after taking the limit as n &rarr &infin (so that we have infinitely many shells to fill out the solid), we get the exact formula.

### Example 1

Find the volume of the solid generated by revolving the region under f(x) = x 2 + 1, where 2 &le x &le 6, around the y-axis.

#### Solution

It might help to sketch a figure. Fortunately, this is exactly what’s pictured in the figure above.

First identify the dimensions of a typical shell.

In addition, we use a = 2 and b = 6 because we have 2 &le x &le 6.

Now set up the Shell Method integral and evaluate to find the volume.

Thus the volume is equal to 672&pi cubic units.

Recall that an oblique cylinder is one that 'leans over' - where the top center is not over the base center point. In the figure above check "allow oblique' and drag the top orange dot sideways to see an oblique cylinder.

It turns out that the volume formula works just the same for these. You must however use the perpendicular height in the formula. This is the vertical line to left in the figure above. To illustrate this, check 'Freeze height'. As you drag the top of the cylinder left and right, watch the volume calculation and note that the volume never changes.

## Volume of Cone Derivation Proof

To derive the volume of a cone formula, the simplest method is to use integration calculus. The mathematical principle is to slice small discs, shaded in yellow, of thickness delta y, and radius x. If we were to slice many discs of the same thickness and summate their volume then we should get an approximate volume of the cone.

The derivation usually begins by taking one such disc of thickness delta y, at a distance y from the vertex of a right circular cone. The radius of the disc is x, however, there will be a small error, shaded in red, due to the thickness delta y. As the thickness reduces to zero then so does the error.

Proportional triangles: As you can see, there are two right-angled triangles shaded in yellow. Their sides are proportional which is useful for building the expressions shown below.

Since the sides are proportional, we can write them like this. Nothing amazing or complicated happening here as you can see. A greater explanation is not required, as the key is to notice the proportional triangles.

Here, I am simply rearranging the expression to give x, which is the radius of the disc. At this stage, you may have figured out that this expression will be of use in another formula to calculate the volume of a disc.

Here, I am using the standard formula for calculating the volume of a disc. The volume of a disc is the same as the volume of a cylinder with a short height. In this case the disc has a radius x and a height delta y. Delta y is just another way of saying that it is a very small distance.

Substituting the previously found equation for radius into the standard equation for the volume of a disc gives us this expression.

Here we have just expanded out the power term, with simple algebra.

At this stage, the integration principle involves adding the volume of the discs to give us the volume of the cone. The summation symbol simply means that if we add the volume of all such discs between y = zero, and y = h, then we should get an approximate value of the volume of a cone. This is still an approximate value due to the error caused by the thickness of the disc delta y. The animated diagram above shows this by showing many discs sliced at various points along the y-axis.

As delta y approaches zero, then so does the error, and therefore we replace the delta y by dy and perform a definite integration operation. All we have done is replace the summation symbol with the integral operator. I normally leave PI behind the integral as it is a constant term throughout the expression and keeps it out of the way. Here we are integrating with respect to y of course.

After integrating y, the result is in the large square brackets. This makes mathematicians look clever, but it just means that you will be performing the same operation twice. Once with y = h, and then again with y = 0.

Here are the two terms expanded out and within their own square brackets. When you are calculating volume by integration, you always subtract the contents of the second brackets with the contents of the first.

The contents of the second bracket becomes zero because y = 0 and everything there is multiplied by zero. In addition, h² cancels out h³ leaving a spare h.

Removing all the un-needed terms and simplifying, it boils down to this familiar term for the volume. This is the simplest proof that Mr Smith taught me back around 1986 in Carshalton College. He was the best teacher I ever had.

## VOLUME OF 3D SHAPES WORKSHEET WITH ANSWERS

(1)  A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.           Solution

(2)  A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the raise of the water in the glass?          Solution

(3)  If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.          Solution

(4)   A conical container is fully filled with petrol. The radius is 10 m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.          Solution

(5)  A right angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.           Solution

(6)  The volumes of two cones of same base radius are 3600 cm 3  and 5040 cm 3 . Find the ratio of heights.           Solution

(7)   If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.           Solution

(8)  A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volume isਃ  √ 3 : 4 .           Solution

(9)  The outer and the inner surface areas of a spherical copper shell are 576 π c m 2 ਊnd  324 π ਌m 2  respectively. Find the volume of the material required to make the shell.           Solution

(10)  A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of  ₹ 40 per litre.           Solution

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