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6.3b: Volumes of Revolution: Cylindrical Shells OS - Mathematics


Learning Objectives

  • Calculate the volume of a solid of revolution by using the method of cylindrical shells.
  • Compare the different methods for calculating a volume of revolution.

In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.

The Method of Cylindrical Shells

Again, we are working with a solid of revolution. As before, we define a region (R), bounded above by the graph of a function (y=f(x)), below by the (x)-axis, and on the left and right by the lines (x=a) and (x=b), respectively, as shown in Figure (PageIndex{1a}). We then revolve this region around the (y)-axis, as shown in Figure (PageIndex{1b}). Note that this is different from what we have done before. Previously, regions defined in terms of functions of (x) were revolved around the (x)-axis or a line parallel to it.

As we have done many times before, partition the interval ([a,b]) using a regular partition, (P={x_0,x_1,…,x_n}) and, for (i=1,2,…,n), choose a point (x^∗_i∈[x_{i−1},x_i]). Then, construct a rectangle over the interval ([x_{i−1},x_i]) of height (f(x^∗_i)) and width (Δx). A representative rectangle is shown in Figure (PageIndex{2a}). When that rectangle is revolved around the (y)-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in Figure (PageIndex{2}).

To calculate the volume of this shell, consider Figure (PageIndex{3}).

The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius (x_i) and inner radius (x_{i−1}). Thus, the cross-sectional area is (πx^2_i−πx^2_{i−1}). The height of the cylinder is (f(x^∗_i).) Then the volume of the shell is

[ egin{align*} V_{shell} =f(x^∗_i)(π,x^2_{i}−π,x^2_{i−1}) [4pt] =π,f(x^∗_i)(x^2_i−x^2_{i−1}) [4pt] =π,f(x^∗_i)(x_i+x_{i−1})(x_i−x_{i−1}) [4pt] =2π,f(x^∗_i)left(dfrac {x_i+x_{i−1}}{2} ight)(x_i−x_{i−1}). end{align*}]

Note that (x_i−x_{i−1}=Δx,) so we have

[V_{shell}=2π,f(x^∗_i)left(dfrac {x_i+x_{i−1}}{2} ight),Δx.]

Furthermore, (dfrac {x_i+x_{i−1}}{2}) is both the midpoint of the interval ([x_{i−1},x_i]) and the average radius of the shell, and we can approximate this by (x^∗_i). We then have

[V_{shell}≈2π,f(x^∗_i)x^∗_i,Δx.]

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate (Figure (PageIndex{4})).

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height (f(x^∗_i)), width (2πx^∗_i), and thickness (Δx) (Figure). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

[V_{shell}≈f(x^∗_i)(2π,x^∗_i),Δx,]

which is the same formula we had before.

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

[V≈sum_{i=1}^n(2π,x^∗_if(x^∗_i),Δx).]

Here we have another Riemann sum, this time for the function (2π,x,f(x).) Taking the limit as (n→∞) gives us

[V=lim_{n→∞}sum_{i=1}^n(2π,x^∗_if(x^∗_i),Δx)=int ^b_a(2π,x,f(x)),dx.]

This leads to the following rule for the method of cylindrical shells.

Rule: The Method of Cylindrical Shells

Let (f(x)) be continuous and nonnegative. Define (R) as the region bounded above by the graph of (f(x)), below by the (x)-axis, on the left by the line (x=a), and on the right by the line (x=b). Then the volume of the solid of revolution formed by revolving (R) around the (y)-axis is given by

[V=int ^b_a(2π,x,f(x)),dx.]

Now let’s consider an example.

Example (PageIndex{1}): The Method of Cylindrical Shells I

Define (R) as the region bounded above by the graph of (f(x)=1/x) and below by the (x)-axis over the interval ([1,3]). Find the volume of the solid of revolution formed by revolving (R) around the (y)-axis.

Solution

First we must graph the region (R) and the associated solid of revolution, as shown in Figure (PageIndex{5}).

Figure (PageIndex{5}) (c) Visualizing the solid of revolution with CalcPlot3D.

Then the volume of the solid is given by

[ egin{align*} V =int ^b_a(2π,x,f(x)),dx =int ^3_1left(2π,xleft(dfrac {1}{x} ight) ight),dx =int ^3_12π,dx =2π,xigg|^3_1=4π, ext{units}^3. end{align*}]

Exercise (PageIndex{1})

Define R as the region bounded above by the graph of (f(x)=x^2) and below by the (x)-axis over the interval ([1,2]). Find the volume of the solid of revolution formed by revolving (R) around the (y)-axis.

Hint

Use the procedure from Example (PageIndex{1}).

Answer

(dfrac{15π}{2} , ext{units}^3 )

Example (PageIndex{2}): The Method of Cylindrical Shells II

Define (R) as the region bounded above by the graph of (f(x)=2x−x^2) and below by the (x)-axis over the interval ([0,2]). Find the volume of the solid of revolution formed by revolving (R) around the (y)-axis.

Solution

First graph the region (R) and the associated solid of revolution, as shown in Figure (PageIndex{6}).

Then the volume of the solid is given by

[egin{align*} V =int ^b_a(2π,x,f(x)),dx =int ^2_0(2π,x(2x−x^2)),dx = 2πint ^2_0(2x^2−x^3),dx =2π left. left[dfrac {2x^3}{3}−dfrac {x^4}{4} ight] ight|^2_0 =dfrac {8π}{3}, ext{units}^3 end{align*}]

Exercise (PageIndex{2})

Define (R) as the region bounded above by the graph of (f(x)=3x−x^2) and below by the (x)-axis over the interval ([0,2]). Find the volume of the solid of revolution formed by revolving (R) around the (y)-axis.

Hint

Use the process from Example (PageIndex{2}).

Answer

(8π , ext{units}^3 )

As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the (x)-axis, when we want to integrate with respect to (y). The analogous rule for this type of solid is given here.

Rule: The Method of Cylindrical Shells for Solids of Revolution around the (x)-axis

Let (g(y)) be continuous and nonnegative. Define (Q) as the region bounded on the right by the graph of (g(y)), on the left by the (y)-axis, below by the line (y=c), and above by the line (y=d). Then, the volume of the solid of revolution formed by revolving (Q) around the (x)-axis is given by

[V=int ^d_c(2π,y,g(y)),dy.]

Example (PageIndex{3}): The Method of Cylindrical Shells for a Solid Revolved around the (x)-axis

Define (Q) as the region bounded on the right by the graph of (g(y)=2sqrt{y}) and on the left by the (y)-axis for (y∈[0,4]). Find the volume of the solid of revolution formed by revolving (Q) around the (x)-axis.

Solution

First, we need to graph the region (Q) and the associated solid of revolution, as shown in Figure (PageIndex{7}).

Label the shaded region (Q). Then the volume of the solid is given by

[ egin{align*} V =int ^d_c(2π,y,g(y)),dy =int ^4_0(2π,y(2sqrt{y})),dy =4πint ^4_0y^{3/2},dy =4πleft[dfrac {2y^{5/2}}{5} ight]∣^4_0 =dfrac {256π}{5}, ext{units}^3 end{align*}]

Exercise (PageIndex{3})

Define (Q) as the region bounded on the right by the graph of (g(y)=3/y) and on the left by the (y)-axis for (y∈[1,3]). Find the volume of the solid of revolution formed by revolving (Q) around the (x)-axis.

Hint

Use the process from Example (PageIndex{3}).

Answer

(12π) units3

For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

[egin{align*} V_{shell} =f(x^∗_i)(π,x^2_i−π,x^2_{i−1}) [4pt] =π,f(x^∗_i)(x^2_i−x^2_{i−1}) [4pt] =π,f(x^∗_i)(x_i+x_{i−1})(x_i−x_{i−1}) [4pt] =2π,f(x^∗_i)left(dfrac {x_i+x_{i−1}}{2} ight)(x_i−x_{i−1}).end{align*}]

This was based on a shell with an outer radius of (x_i) and an inner radius of (x_{i−1}). If, however, we rotate the region around a line other than the (y)-axis, we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line (x=−k,) where (k) is some positive constant. Then, the outer radius of the shell is (x_i+k) and the inner radius of the shell is (x_{i−1}+k). Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line (x=−k,) the volume of a shell is given by

[egin{align*} V_{shell} =2π,f(x^∗_i)(dfrac {(x_i+k)+(x_{i−1}+k)}{2})((x_i+k)−(x_{i−1}+k)) [4pt] =2π,f(x^∗_i)left(left(dfrac {x_i+x_{i−2}}{2} ight)+k ight)Δx.end{align*}]

As before, we notice that (dfrac {x_i+x_{i−1}}{2}) is the midpoint of the interval ([x_{i−1},x_i]) and can be approximated by (x^∗_i). Then, the approximate volume of the shell is

[V_{shell}≈2π(x^∗_i+k)f(x^∗_i)Δx.]

The remainder of the development proceeds as before, and we see that

[V=int ^b_a(2π(x+k)f(x))dx.]

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the (x)-term in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

Example (PageIndex{4}): A Region of Revolution Revolved around a Line

Define (R) as the region bounded above by the graph of (f(x)=x) and below by the (x)-axis over the interval ([1,2]). Find the volume of the solid of revolution formed by revolving (R) around the line (x=−1.)

Solution

First, graph the region (R) and the associated solid of revolution, as shown in Figure (PageIndex{8}).

Note that the radius of a shell is given by (x+1). Then the volume of the solid is given by

[egin{align*} V =int ^2_1 2π(x+1)f(x), dx =int ^2_1 2π(x+1)x , dx=2πint ^2_1 x^2+x , dx =2π left[dfrac{x^3}{3}+dfrac{x^2}{2} ight]igg|^2_1 =dfrac{23π}{3} , ext{units}^3 end{align*}]

Exercise (PageIndex{4})

Define (R) as the region bounded above by the graph of (f(x)=x^2) and below by the (x)-axis over the interval ([0,1]). Find the volume of the solid of revolution formed by revolving (R) around the line (x=−2).

Hint

Use the process from Example (PageIndex{4}).

Answer

(dfrac {11π}{6}) units3

For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

Example (PageIndex{5}): A Region of Revolution Bounded by the Graphs of Two Functions

Define (R) as the region bounded above by the graph of the function (f(x)=sqrt{x}) and below by the graph of the function (g(x)=1/x) over the interval ([1,4]). Find the volume of the solid of revolution generated by revolving (R) around the (y)-axis.

Solution

First, graph the region (R) and the associated solid of revolution, as shown in Figure (PageIndex{9}).

Note that the axis of revolution is the (y)-axis, so the radius of a shell is given simply by (x). We don’t need to make any adjustments to the x-term of our integrand. The height of a shell, though, is given by (f(x)−g(x)), so in this case we need to adjust the (f(x)) term of the integrand. Then the volume of the solid is given by

[egin{align*} V =int ^4_1(2π,x(f(x)−g(x))),dx [4pt] = int ^4_1(2π,x(sqrt{x}−dfrac {1}{x})),dx=2πint ^4_1(x^{3/2}−1)dx [4pt] = 2πleft[dfrac {2x^{5/2}}{5}−x ight]igg|^4_1=dfrac {94π}{5} , ext{units}^3. end{align*}]

Exercise (PageIndex{5})

Define (R) as the region bounded above by the graph of (f(x)=x) and below by the graph of (g(x)=x^2) over the interval ([0,1]). Find the volume of the solid of revolution formed by revolving (R) around the (y)-axis.

Hint

Hint: Use the process from Example (PageIndex{5}).

Answer

(dfrac {π}{6}) units3

Which Method Should We Use?

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure (PageIndex{10}) describes the different approaches for solids of revolution around the (x)-axis. It’s up to you to develop the analogous table for solids of revolution around the (y)-axis.

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

Example (PageIndex{6}): Selecting the Best Method

For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the (x)-axis, and set up the integral to find the volume (do not evaluate the integral).

  1. The region bounded by the graphs of (y=x, y=2−x,) and the (x)-axis.
  2. The region bounded by the graphs of (y=4x−x^2) and the (x)-axis.

Solution

a.

First, sketch the region and the solid of revolution as shown.

Looking at the region, if we want to integrate with respect to (x), we would have to break the integral into two pieces, because we have different functions bounding the region over ([0,1]) and ([1,2]). In this case, using the disk method, we would have

[V=int ^1_0 π,x^2,dx+int ^2_1 π(2−x)^2,dx. onumber]

If we used the shell method instead, we would use functions of y to represent the curves, producing

[V=int ^1_0 2π,y[(2−y)−y] ,dy=int ^1_0 2π,y[2−2y],dy. onumber]

Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.

b.

First, sketch the region and the solid of revolution as shown.

Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then

[V=int ^4_0πleft(4x−x^2 ight)^2,dx onumber]

Exercise (PageIndex{6})

Select the best method to find the volume of a solid of revolution generated by revolving the given region around the (x)-axis, and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of (y=2−x^2) and (y=x^2).

Hint

Sketch the region and use Figure (PageIndex{12}) to decide which integral is easiest to evaluate.

Answer

Use the method of washers; [V=int ^1_{−1}πleft[left(2−x^2 ight)^2−left(x^2 ight)^2 ight],dx onumber]

Key Concepts

  • The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
  • The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.

Key Equations

  • Method of Cylindrical Shells

(displaystyle V=int ^b_aleft(2π,x,f(x) ight),dx)

Glossary

method of cylindrical shells
a method of calculating the volume of a solid of revolution by dividing the solid into nested cylindrical shells; this method is different from the methods of disks or washers in that we integrate with respect to the opposite variable

5.5 Alternating Series

So far in this chapter, we have primarily discussed series with positive terms. In this section we introduce alternating series—those series whose terms alternate in sign. We will show in a later chapter that these series often arise when studying power series. After defining alternating series, we introduce the alternating series test to determine whether such a series converges.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series . For example, the series

are both alternating series.

Definition

Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form

Series (1), shown in Equation 5.11, is a geometric series. Since | r | = | − 1 / 2 | < 1 , | r | = | − 1 / 2 | < 1 , the series converges. Series (2), shown in Equation 5.12, is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.

To prove this, we look at the sequence of partial sums < S k > < S k >(Figure 5.17).

Proof

Therefore, by the Monotone Convergence Theorem, the sequence < S 2 k > < S 2 k >also converges. Since

It can also be shown that S = ln 2 , S = ln 2 , and we can write

More generally, any alternating series of form (3) (Equation 5.13) or (4) (Equation 5.14) converges as long as b 1 ≥ b 2 ≥ b 3 ≥ ⋯ b 1 ≥ b 2 ≥ b 3 ≥ ⋯ and b n → 0 b n → 0 (Figure 5.18). The proof is similar to the proof for the alternating harmonic series.

Theorem 5.13

Alternating Series Test

An alternating series of the form

This is known as the alternating series test .

We remark that this theorem is true more generally as long as there exists some integer N N such that 0 ≤ b n + 1 ≤ b n 0 ≤ b n + 1 ≤ b n for all n ≥ N . n ≥ N .

Example 5.19

Convergence of Alternating Series

For each of the following alternating series, determine whether the series converges or diverges.

Solution

Checkpoint 5.18

Determine whether the series ∑ n = 1 ∞ ( −1 ) n + 1 n / 2 n ∑ n = 1 ∞ ( −1 ) n + 1 n / 2 n converges or diverges.

Remainder of an Alternating Series

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

Remainders in Alternating Series

Consider an alternating series of the form

In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the N th N th partial sum S N S N is in magnitude at most the size of the next term b N + 1 . b N + 1 .

Example 5.20

Estimating the Remainder of an Alternating Series

Consider the alternating series

Use the remainder estimate to determine a bound on the error R 10 R 10 if we approximate the sum of the series by the partial sum S 10 . S 10 .

Solution

From the theorem stated above,

| R 10 | ≤ b 11 = 1 11 2 ≈ 0.008265 . | R 10 | ≤ b 11 = 1 11 2 ≈ 0.008265 .

Absolute and Conditional Convergence

Definition

Absolute Convergence Implies Convergence

Proof

converges. By using the algebraic properties for convergent series, we conclude that

Example 5.21

Absolute versus Conditional Convergence

For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.

Solution

Checkpoint 5.20

Begin by adding enough of the positive terms to produce a sum that is larger than some real number M > 0 . M > 0 . For example, let M = 10 , M = 10 , and find an integer k k such that

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In Example 5.22, we show how to rearrange the terms to create a new series that converges to 3 ln ( 2 ) / 2 . 3 ln ( 2 ) / 2 . We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number r r however, the proof of that fact is beyond the scope of this text.

Example 5.22

Rearranging Series

to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is 3 ln ( 2 ) / 2 . 3 ln ( 2 ) / 2 .

Solution

Then using the algebraic limit properties of convergent series, since ∑ n = 1 ∞ a n ∑ n = 1 ∞ a n and ∑ n = 1 ∞ b n ∑ n = 1 ∞ b n converge, the series ∑ n = 1 ∞ ( a n + b n ) ∑ n = 1 ∞ ( a n + b n ) converges and

Now adding the corresponding terms, a n a n and b n , b n , we see that

We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since ∑ n = 1 ∞ ( a n + b n ) = 3 ln ( 2 ) / 2 , ∑ n = 1 ∞ ( a n + b n ) = 3 ln ( 2 ) / 2 , we conclude that

Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

Section 5.5 Exercises

State whether each of the following series converges absolutely, conditionally, or not at all.

∑ n = 1 ∞ ( −1 ) n + 1 ( n − 1 n ) n ∑ n = 1 ∞ ( −1 ) n + 1 ( n − 1 n ) n

∑ n = 1 ∞ ( −1 ) n + 1 ( n + 1 n ) n ∑ n = 1 ∞ ( −1 ) n + 1 ( n + 1 n ) n

∑ n = 1 ∞ ( −1 ) n + 1 sin 2 ( 1 / n ) ∑ n = 1 ∞ ( −1 ) n + 1 sin 2 ( 1 / n )

∑ n = 1 ∞ ( −1 ) n + 1 cos 2 ( 1 / n ) ∑ n = 1 ∞ ( −1 ) n + 1 cos 2 ( 1 / n )

∑ n = 1 ∞ ( −1 ) n + 1 ln ( 1 + 1 n ) ∑ n = 1 ∞ ( −1 ) n + 1 ln ( 1 + 1 n )

∑ n = 1 ∞ ( −1 ) n + 1 n 2 1 + n 4 ∑ n = 1 ∞ ( −1 ) n + 1 n 2 1 + n 4

∑ n = 1 ∞ ( −1 ) n + 1 n e 1 + n π ∑ n = 1 ∞ ( −1 ) n + 1 n e 1 + n π

∑ n = 1 ∞ ( −1 ) n + 1 ( ln ( n + 1 ) − ln n ) ∑ n = 1 ∞ ( −1 ) n + 1 ( ln ( n + 1 ) − ln n )

∑ n = 1 ∞ ( −1 ) n + 1 ( ( n + 1 ) 2 − n 2 ) ∑ n = 1 ∞ ( −1 ) n + 1 ( ( n + 1 ) 2 − n 2 )

∑ n = 1 ∞ ( −1 ) n + 1 ( 1 n − 1 n + 1 ) ∑ n = 1 ∞ ( −1 ) n + 1 ( 1 n − 1 n + 1 )

∑ n = 1 ∞ sin ( n π / 2 ) sin ( 1 / n ) ∑ n = 1 ∞ sin ( n π / 2 ) sin ( 1 / n )

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

1 + 1 2 − 1 3 − 1 4 + 1 5 + 1 6 − 1 7 − 1 8 + ⋯ 1 + 1 2 − 1 3 − 1 4 + 1 5 + 1 6 − 1 7 − 1 8 + ⋯

1 + 1 2 − 1 3 + 1 4 + 1 5 − 1 6 + 1 7 + 1 8 − 1 9 + ⋯ 1 + 1 2 − 1 3 + 1 4 + 1 5 − 1 6 + 1 7 + 1 8 − 1 9 + ⋯

Show that the alternating series 1 − 1 2 + 1 2 − 1 4 + 1 3 − 1 6 + 1 4 − 1 8 + ⋯ 1 − 1 2 + 1 2 − 1 4 + 1 3 − 1 6 + 1 4 − 1 8 + ⋯ does

not converge. What hypothesis of the alternating series test is not met?

[T] Plot the series ∑ n = 1 100 sin ( 2 π n x ) n ∑ n = 1 100 sin ( 2 π n x ) n for 0 ≤ x < 1 0 ≤ x < 1 and comment on its behavior

[T] Plot the series ∑ n = 1 100 cos ( 2 π n x ) n 2 ∑ n = 1 100 cos ( 2 π n x ) n 2 for 0 ≤ x < 1 0 ≤ x < 1 and describe its graph.

[T] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If a n ≥ 0 a n ≥ 0 is such that a n → 0 a n → 0 as n → ∞ n → ∞ but ∑ n = 1 ∞ a n ∑ n = 1 ∞ a n diverges, then, given any number A A there is a sequence s n s n of ± 1 's ± 1 's such that ∑ n = 1 ∞ a n s n → A . ∑ n = 1 ∞ a n s n → A . Show this for A > 0 A > 0 as follows.

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    • Authors: Gilbert Strang, Edwin “Jed” Herman
    • Publisher/website: OpenStax
    • Book title: Calculus Volume 2
    • Publication date: Mar 30, 2016
    • Location: Houston, Texas
    • Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
    • Section URL: https://openstax.org/books/calculus-volume-2/pages/5-5-alternating-series

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    The volume of the solid shell between two different cylinders, of the same height, one of radius and the other of radius r^2 > r^1 is π(r_2^2 –r_1^2) h = 2π r_2 + r_1 / 2 (r_2 – r_1) h = 2 πr △rh, where, r = ½ (r_1 + r_2) is the radius and △r = r_2 – r_1 is the change in radius.

    If a profile b = f(a), for (a) between x and y is rotated about the y quadrant, then the volume can be approximated by the Riemann sum method of cylinders:

    Every cylinder at the position x* is the width △a and height b = f(a*): so every component of the Riemann sum has the form 2π x* f(x*) △a.
    In the limit when the value of cylinders goes to infinity, the Riemann sum becomes an integral representation of the volume V:

    $ V = ∫_a^b 2 π x y (dx) = V = ∫_a^b 2 π x f (x) dx $

    To construct the integral shell method calculator find the value of function y and the limits of integration.
    If the area between two different curves b = f(a) and b = g(a) > f(a) is revolved around the y-axis, for x from the point a to b, then the volume is:

    Now, the cylindrical shell method calculator computes the volume of the shell by rotating the bounded area by the x coordinate, where the line x = 2 and the curve y = x^3 about the y coordinate.

    Here y = x^3 and the limits are x = [0, 2].
    The integral is:

    $ ∫_0^2 2 π x y dx = ∫_0^2 2 π x (x^3)dx $

    $ = 2π∫_0^2x^4 = 2π [ x^5 / 5]_0^2 = 2π 32/5 = 64/5 π$
    The curves meet at the point x = 0 and at the point x = 1, so the volume is:

    $= 2 π[ 2/5 x^ <5/2>– x^4 / 4]_0^1$
    $= 2 π (2 / 5 – 1 / 4) = 3 / 10 π$


    6.2 Determining Volumes by Slicing

    In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for finding these volumes, depending on the characteristics of the solid.

    Volume and the Slicing Method

    Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: V = l w h . V = l w h . The formulas for the volume of a sphere ( V = 4 3 π r 3 ) , ( V = 4 3 π r 3 ) , a cone ( V = 1 3 π r 2 h ) , ( V = 1 3 π r 2 h ) , and a pyramid ( V = 1 3 A h ) ( V = 1 3 A h ) have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

    We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.

    We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 6.11 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: V = A · h . V = A · h . In the case of a right circular cylinder (soup can), this becomes V = π r 2 h . V = π r 2 h .

    If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid S shown in Figure 6.12, extending along the x -axis . x -axis .


    6.3b: Volumes of Revolution: Cylindrical Shells OS - Mathematics

    In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.

    The Method of Cylindrical Shells

    Again, we are working with a solid of revolution. As before, we define a region R , R , bounded above by the graph of a function y = f ( x ) , y = f ( x ) , below by the x -axis, x -axis, and on the left and right by the lines x = a x = a and x = b , x = b , respectively, as shown in [link](a). We then revolve this region around the y-axis, as shown in [link](b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of x x were revolved around the x -axis x -axis or a line parallel to it.

    (a) A region bounded by the graph of a function of x . x . (b) The solid of revolution formed when the region is revolved around the y -axis . y -axis .

    (a) A representative rectangle. (b) When this rectangle is revolved around the y -axis , y -axis , the result is a cylindrical shell. (c) When we put all the shells together, we get an approximation of the original solid.

    To calculate the volume of this shell, consider [link].

    Calculating the volume of the shell.

    The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius x i x i and inner radius x i − 1 . x i − 1 . Thus, the cross-sectional area is π x i 2 − π x i − 1 2 . π x i 2 − π x i − 1 2 . The height of the cylinder is f ( x i * ) . f ( x i * ) . Then the volume of the shell is

    Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ([link]).

    (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate.

    In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height f ( x i * ) , f ( x i * ) , width 2 π x i * , 2 π x i * , and thickness Δ x Δ x ([link]). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

    which is the same formula we had before.

    To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

    Here we have another Riemann sum, this time for the function 2 π x f ( x ) . 2 π x f ( x ) . Taking the limit as n → ∞ n → ∞ gives us


    Contents

    Two common methods for finding the volume of a solid of revolution are the disc method and the shell method of integration. To apply these methods, it is easiest to draw the graph in question identify the area that is to be revolved about the axis of revolution determine the volume of either a disc-shaped slice of the solid, with thickness δx , or a cylindrical shell of width δx and then find the limiting sum of these volumes as δx approaches 0, a value which may be found by evaluating a suitable integral. A more rigorous justification can be given by attempting to evaluate a triple integral in cylindrical coordinates with two different orders of integration.

    Disc method Edit

    The disc method is used when the slice that was drawn is perpendicular to the axis of revolution i.e. when integrating parallel to the axis of revolution.

    The volume of the solid formed by rotating the area between the curves of f(x) and g(x) and the lines x = a and x = b about the x -axis is given by

    If g(x) = 0 (e.g. revolving an area between the curve and the x -axis), this reduces to:

    The method can be visualized by considering a thin horizontal rectangle at y between f(y) on top and g(y) on the bottom, and revolving it about the y -axis it forms a ring (or disc in the case that g(y) = 0 ), with outer radius f(y) and inner radius g(y) . The area of a ring is π(R 2 − r 2 ) , where R is the outer radius (in this case f(y) ), and r is the inner radius (in this case g(y) ). The volume of each infinitesimal disc is therefore πf(y) 2 dy . The limit of the Riemann sum of the volumes of the discs between a and b becomes integral (1).

    Assuming the applicability of Fubini's theorem and the multivariate change of variables formula, the disk method may be derived in a straightforward manner by (denoting the solid as D):

    Cylinder method Edit

    The cylinder method is used when the slice that was drawn is parallel to the axis of revolution i.e. when integrating perpendicular to the axis of revolution.

    The volume of the solid formed by rotating the area between the curves of f(x) and g(x) and the lines x = a and x = b about the y -axis is given by

    If g(x) = 0 (e.g. revolving an area between curve and y -axis), this reduces to:

    The method can be visualized by considering a thin vertical rectangle at x with height f(x) − g(x) , and revolving it about the y -axis it forms a cylindrical shell. The lateral surface area of a cylinder is 2πrh , where r is the radius (in this case x ), and h is the height (in this case f(x) − g(x) ). Summing up all of the surface areas along the interval gives the total volume.

    This method may be derived with the same triple integral, this time with a different order of integration:


    Calculus: The Notebook

    The study of Calculus enables us to solve problems and articulate abstract concepts far beyond the theoretical reach of Algebra. The power of Calculus derives from the ingenuity and simplicity of its notation. This mathematical language allows the mathematician the freedom and immense versatility to accurately describe physical problems and the tools to solve them. This book consists of lectures on all the topics of a 3-course series on Calculus: Calculus I, II, and III. It can be used as a textbook, or as a supplement to other texts. Both students and instructors will find it helpful in elucidating the ideas and methods of Calculus.

    Preface
    About the Author
    Formulas
    Introduction

    Part 1 Calculus I
    Chapter 1 Functions and Limits

    1.1 Functions, Transformations
    1.2 Tangent and Velocity Functions
    1.3 Limit of a Function, Limit Laws
    1.4 Formal Definition of a Limit
    1.5 Limit Laws
    1.6 Continuity

    Chapter 2 Derivatives
    2.1 Derivatives and Rates of Change
    2.2 Derivative as a Function, Differentiation Formulas
    2.3 Derivatives of Trigonometric Functions
    2.4 Linear Approximations and Differentials
    2.5 Chain Rule
    2.6 Implicit Differentiation
    2.7 Related Rates

    Chapter 3 Applications of Differentiation
    3.1 Mean Value Theorem
    3.2 Maximum and Minimum Values
    3.3 Optimization Problems
    3.4 Derivatives and Curve Sketching
    3.5 Limits at Infinity
    3.6 Newton’s Method

    Chapter 4 Integrals
    4.1 Antiderivatives
    4.2 Areas and Distances
    4.3 Definite Integral
    4.4 Fundamental Theorem of Calculus

    Chapter 5 Applications of Integration
    5.1 Areas Between Curves
    5.2 Volumes of Solids of Revolution: Slices
    5.3 Volumes of Solids of Revolution: Cylindrical Shells
    5.4 Average Value of a Function
    5.5 Improper Integrals

    Part 2 Calculus II
    Chapter 6 Special Functions, Indeterminate Forms

    6.1 Logs and Exponents
    6.2 Exponential Growth and Decay
    6.3 Inverse Trig Functions
    6.4 L’Hospital’s Rule

    Chapter 7 Techniques of Integration
    7.1 U-Substitution
    7.2 Integration by Parts
    7.3 Trig Integrals
    7.4 Trigonometric Substitution
    7.5 Partial Fractions
    7.6 Numerical Integration

    Chapter 8 Applications of the Integral
    8.1 Arclength
    8.2 Surface Areas of Revolution
    8.3 Mass, Work

    Chapter 9 Introduction to Differential Equations
    9.1 Direction Fields and Euler’s Method
    9.2 Separable Differential Equations
    9.3 Linear Differential Equations

    Chapter 10 Conics, Polar Coordinates, and Parametric Equations
    10.1 Conics
    10.2 Polar Coordinates
    10.3 Parametric Equations

    Chapter 11 Sequences and Series
    11.1 Sequences
    11.2 Series
    11.3 Convergence Tests
    11.4 Power Series

    Part 3 Calculus III
    Chapter 12 Geometry in 3 Dimensions

    12.1 3D coordinates
    12.2 Vectors
    12.3 The Dot Product
    12.4 The Cross Product
    12.5 Vector Equations of Lines, Planes
    12.6 Vector Valued Functions
    12.7 Calculus of Vector Valued Functions
    12.8 Arclength
    12.9 Tangent, Normal, and Binormal Vectors
    12.10 Curvature
    12.11 Motion in Space: Velocity and Acceleration

    Chapter 13 Functions of Several Variables
    13.1 Introduction
    13.2 Quadratic & Cylindrical Surfaces
    13.3 Limits, Continuity
    13.4 Partial Derivatives
    13.5 Chain Rule
    13.6 Directional Derivatives, Gradients
    13.7 Tangent Planes, Normal Lines
    13.8 Local Maxima, Local Minima, Saddle Points
    13.9 Global Maxima and Minima
    13.10 Lagrange Multipliers

    Chapter 14 Multiple Integrals
    14.1 Double Integrals
    14.2 Double Integrals in Polar Coordinates
    14.3 Triple Integrals
    14.4 Triple Integrals in Cylindrical Coordinates
    14.5 Triple Integrals in Spherical Coordinates

    Chapter 15 Vector Calculus
    15.1 Vector Fields
    15.2 Line Integrals
    15.3 Fundamental Theorem of Line Integrals
    15.4 Green’s Theorem
    15.5 Line Integrals Made Easy
    15.6 Surface Integrals
    15.7 Flux Integrals
    15.8 Stokes Theorem
    15.9 Divergence Theorem


    Calculus Volume 1 Textbook, Test Bank

    Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning.
    The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them.
    Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
    Volume 1 covers functions, limits, derivatives, and integration.

    * Complete Textbook by OpenStax
    * Multiple Choices Questions (MCQ)
    * Essay Questions Flash Cards
    * Key-Terms Flash Cards

    Powered by QuizOver.com the leading online quiz creator
    https://www.quizover.com

    1. Functions and Graphs
    Introduction
    1.1. Review of Functions
    1.2. Basic Classes of Functions
    1.3. Trigonometric Functions
    1.4. Inverse Functions
    1.5. Exponential and Logarithmic Functions
    2. Limits
    Introduction
    2.1. A Preview of Calculus
    2.2. The Limit of a Function
    2.3. The Limit Laws
    2.4. Continuity
    2.5. The Precise Definition of a Limit
    3. Derivatives
    Introduction
    3.1. Defining the Derivative
    3.2. The Derivative as a Function
    3.3. Differentiation Rules
    3.4. Derivatives as Rates of Change
    3.5. Derivatives of Trigonometric Functions
    3.6. The Chain Rule
    3.7. Derivatives of Inverse Functions
    3.8. Implicit Differentiation
    3.9. Derivatives of Exponential and Logarithmic Functions
    4. Applications of Derivatives
    Introduction
    4.1. Related Rates
    4.2. Linear Approximations and Differentials
    4.3. Maxima and Minima
    4.4. The Mean Value Theorem
    4.5. Derivatives and the Shape of a Graph
    4.6. Limits at Infinity and Asymptotes
    4.7. Applied Optimization Problems
    4.8. L’Hôpital’s Rule
    4.9. Newton’s Method
    4.10. Antiderivatives
    5. Integration
    Introduction
    5.1. Approximating Areas
    5.2. The Definite Integral
    5.3. The Fundamental Theorem of Calculus
    5.4. Integration Formulas and the Net Change Theorem
    5.5. Substitution
    5.6. Integrals Involving Exponential and Logarithmic Functions
    5.7. Integrals Resulting in Inverse Trigonometric Functions
    6. Applications of Integration
    Introduction
    6.1. Areas between Curves
    6.2. Determining Volumes by Slicing
    6.3. Volumes of Revolution: Cylindrical Shells
    6.4. Arc Length of a Curve and Surface Area
    6.5. Physical Applications
    6.6. Moments and Centers of Mass
    6.7. Integrals, Exponential Functions, and Logarithms
    6.8. Exponential Growth and Decay
    6.9. Calculus of the Hyperbolic Functions
    Table of Integrals
    Table of Derivatives
    Review of Pre-Calculus


    What is the Shell method ?

    It is a technique to find solids' capacity of revolutions, which considers vertical sides being integrated rather than horizontal ones to simplify some unique problems where the vertical sides are more easily described. Generally, the solid density is the measurement or standard of how much space an object takes up concerning the XYZ axis plane. As unit cubes are used to fill the solid, the volume of solid is measured by the number of cubes.

    To solve the problem using the cylindrical method, choose the region or area in the XYZ plane, which is distributed into thin vertical strips. Each vertical strip is revolved around the y-axis, and then the different object of a revolution is obtained which looks like a cylindrical shell. Mostly, It follows the rotation of rectangles about the y-axis.


    Solids of Revolution

    If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

    Figure 5. (a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.

    Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

    Use an online integral calculator to learn more.

    Using the Slicing Method to find the Volume of a Solid of Revolution

    Use the slicing method to find the volume of the solid of revolution bounded by the graphs of and rotated about the

    Solution

    Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval as shown in the following figure.

    Figure 6. A region used to produce a solid of revolution.

    Next, revolve the region around the -axis, as shown in the following figure.

    Figure 7. Two views, (a) and (b), of the solid of revolution produced by revolving the region in (Figure) about the

    Since the solid was formed by revolving the region around the the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by Use the formula for the area of the circle:

    The volume, then, is (step 3)

    The volume is

    Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function and the over the interval around the See the following figure.


    Watch the video: AP Calc Volumes by Cylindrical Shells (December 2021).