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Estimate the Volume if the Region Is Rotated About the Y Axis Again Use the Midpoint Rule With N 4

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6. Applications of Integration

6.3 Volumes of Revolution: Cylindrical Shells

Learning Objectives

  • Summate the book of a solid of revolution by using the method of cylindrical shells.
  • Compare the different methods for computing a volume of revolution.

In this section, nosotros examine the method of cylindrical shells, the final method for finding the book of a solid of revolution. Nosotros can utilize 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 centrality of revolution. With the method of cylindrical shells, we integrate forth 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 reward 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 function of this section, we review all the methods for finding volume that we accept studied and lay out some guidelines to help you determine which method to utilise in a given situation.

The Method of Cylindrical Shells

Once again, we are working with a solid of revolution. Every bit before, we define a region R, bounded above past the graph of a office y=f(x), beneath by the x\text{-axis,} and on the left and right by the lines x=a and x=b, respectively, as shown in (Effigy)(a). We so circumduct this region effectually the y-axis, as shown in (Effigy)(b). Notation that this is different from what we have done before. Previously, regions divers in terms of functions of x were revolved around the x\text{-axis} or a line parallel to it.

As we have washed many times before, sectionalisation the interval \left[a,b\right] using a regular partition, P=\left\{{x}_{0},{x}_{1}\text{,…},{x}_{n}\right\} and, for i=1,2\text{,…},n, choose a point {x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right]. Then, construct a rectangle over the interval \left[{x}_{i-1},{x}_{i}\right] of superlative f({x}_{i}^{*}) and width \text{Δ}x. A representative rectangle is shown in (Figure)(a). When that rectangle is revolved effectually the y-axis, instead of a disk or a washer, nosotros become a cylindrical vanquish, as shown in the post-obit figure.

To calculate the volume of this shell, consider (Figure).

This figure is a graph in the first quadrant. The curve is increasing and labeled
Effigy iii. Computing the volume of the trounce.

The beat out is a cylinder, then its book is the cross-sectional area multiplied by the height of the cylinder. The cantankerous-sections are annuli (band-shaped regions—essentially, circles with a pigsty in the middle), with outer radius {x}_{i} and inner radius {x}_{i-1}. Thus, the cross-sectional area is \pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}. The height of the cylinder is f({x}_{i}^{*}). And so the book of the shell is

\begin{array}{cc}\hfill {V}_{\text{shell}}& =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\hfill \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\hfill \end{array}

Note that {x}_{i}-{x}_{i-1}=\text{Δ}x, and so we take

{V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\text{Δ}x.

Furthermore, \frac{{x}_{i}+{x}_{i-1}}{2} is both the midpoint of the interval \left[{x}_{i-1},{x}_{i}\right] and the average radius of the crush, and nosotros can approximate this by {x}_{i}^{*}. We and so have

{V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\text{Δ}x.

Another way to call up of this is to think of making a vertical cut in the shell and then opening it up to grade a flat plate ((Figure)).

This figure has two images. The first is labeled
Figure 4. (a) Make a vertical cut in a representative beat. (b) Open the shell up to form a apartment plate.

In reality, the outer radius of the shell is greater than the inner radius, and hence the dorsum edge of the plate would exist slightly longer than the front border of the plate. Still, we can approximate the flattened trounce by a flat plate of height f({x}_{i}^{*}), width 2\pi {x}_{i}^{*}, and thickness \text{Δ}x ((Figure)). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the meridian, width, and depth of the plate, we get

{V}_{\text{shell}}\approx f({x}_{i}^{*})(2\pi {x}_{i}^{*})\text{Δ}x,

which is the same formula nosotros had before.

To calculate the volume of the entire solid, nosotros so add the volumes of all the shells and obtain

V\approx \underset{i=1}{\overset{n}{\text{∑}}}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{Δ}x).

Here we have another Riemann sum, this time for the function 2\pi xf(x). Taking the limit as n\to \infty gives u.s.a.

V=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{Δ}x)={\int }_{a}^{b}(2\pi xf(x))dx.

This leads to the post-obit rule for the method of cylindrical shells.

Now let's consider an example.

The Method of Cylindrical Shells 1

Define R as the region bounded in a higher place by the graph of f(x)=1\text{/}x and below by the x\text{-axis} over the interval \left[1,3\right]. Find the volume of the solid of revolution formed by revolving R around the y\text{-axis}.

Solution

First we must graph the region R and the associated solid of revolution, as shown in the post-obit figure.

Then the volume of the solid is given by

\begin{array}{cc}\hfill V& ={\int }_{a}^{b}(2\pi xf(x))dx\hfill \\ & ={\int }_{1}^{3}(2\pi x(\frac{1}{x}))dx\hfill \\ & ={\int }_{1}^{3}2\pi dx={2\pi x|}_{1}^{3}=4\pi {\text{units}}^{3}\text{.}\hfill \end{array}

The Method of Cylindrical Shells 2

Define R equally the region bounded to a higher place by the graph of f(x)=2x-{x}^{2} and below past the x\text{-axis} over the interval \left[0,2\right]. Find the volume of the solid of revolution formed past revolving R around the y\text{-axis}.

Solution

First graph the region R and the associated solid of revolution, as shown in the following figure.

Then the volume of the solid is given by

\begin{array}{cc}\hfill V& ={\int }_{a}^{b}(2\pi xf(x))dx\hfill \\ & ={\int }_{0}^{2}(2\pi x(2x-{x}^{2}))dx=2\pi {\int }_{0}^{2}(2{x}^{2}-{x}^{3})dx\hfill \\ & ={2\pi \left[\frac{2{x}^{3}}{3}-\frac{{x}^{4}}{4}\right]|}_{0}^{2}=\frac{8\pi }{3}{\text{units}}^{3}\text{.}\hfill \end{array}

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

The Method of Cylindrical Shells for a Solid Revolved around the x-axis

For the next case, we await 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 gear up this up, we demand to revisit the development of the method of cylindrical shells. Retrieve that we institute the volume of one of the shells to be given by

\begin{array}{cc}\hfill {V}_{\text{shell}}& =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\hfill \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\hfill \end{array}

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\text{-axis}, we have a unlike outer and inner radius. Suppose, for example, that we rotate the region effectually the line x=\text{−}k, where k is some positive abiding. Then, the outer radius of the crush is {x}_{i}+k and the inner radius of the shell is {x}_{i-1}+k. Substituting these terms into the expression for volume, nosotros run across that when a airplane region is rotated around the line x=\text{−}k, the volume of a vanquish is given by

\begin{array}{cc}\hfill {V}_{\text{shell}}& =2\pi f({x}_{i}^{*})(\frac{({x}_{i}+k)+({x}_{i-1}+k)}{2})(({x}_{i}+k)-({x}_{i-1}+k))\hfill \\ & =2\pi f({x}_{i}^{*})((\frac{{x}_{i}+{x}_{i-2}}{2})+k)\text{Δ}x.\hfill \end{array}

As earlier, we find that \frac{{x}_{i}+{x}_{i-1}}{2} is the midpoint of the interval \left[{x}_{i-1},{x}_{i}\right] and can be approximated past {x}_{i}^{*}. And then, the approximate book of the shell is

{V}_{\text{shell}}\approx 2\pi ({x}_{i}^{*}+k)f({x}_{i}^{*})\text{Δ}x.

The remainder of the evolution gain as before, and we see that

V={\int }_{a}^{b}(2\pi (x+k)f(x))dx.

We could too rotate the region around other horizontal or vertical lines, such equally a vertical line in the right half airplane. In each instance, the volume formula must be adjusted accordingly. Specifically, the x\text{-term} in the integral must be replaced with an expression representing the radius of a shell. To encounter how this works, consider the post-obit example.

A Region of Revolution Revolved around a Line

For our concluding instance 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 ii functions.

A Region of Revolution Bounded by the Graphs of Two Functions

Which Method Should Nosotros Utilize?

Nosotros have studied several methods for finding the volume of a solid of revolution, but how do we know which method to utilize? It often comes down to a choice of which integral is easiest to evaluate. (Figure) describes the unlike approaches for solids of revolution around the x\text{-axis}. Information technology's upwardly to you to develop the analogous table for solids of revolution around the y\text{-axis}.

This figure is a table comparing the different methods for finding volumes of solids of revolution. The columns in the table are labeled
Figure ten.

Permit'south have a look at a couple of additional issues and make up one's mind on the all-time approach to accept for solving them.

Selecting the All-time Method

Solution

  1. 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 accept to interruption the integral into ii pieces, because we accept dissimilar functions bounding the region over \left[0,1\right] and \left[1,2\right]. In this instance, using the disk method, we would have

    V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.

    If we used the trounce method instead, we would use functions of y to stand for the curves, producing

    \begin{array}{cc}\hfill V& ={\int }_{0}^{1}(2\pi y\left[(2-y)-y\right])dy\hfill \\ & ={\int }_{0}^{1}(2\pi y\left[2-2y\right])dy.\hfill \end{array}

    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 become with the shell method in this case.

  2. Kickoff, 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 role. Therefore, we can dismiss the method of shells. The solid has no cavity in the center, and then we tin use the method of disks. Then

    V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.

Select the all-time method to find the volume of a solid of revolution generated by revolving the given region effectually the x\text{-axis}, and gear up upwards 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}.

Solution

Utilize the method of washers; V={\int }_{-1}^{1}\pi \left[{(2-{x}^{2})}^{2}-{({x}^{2})}^{2}\right]dx

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 considering 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 chief factors in deciding which integration method to use.

Key Equations

  • Method of Cylindrical Shells
    V={\int }_{a}^{b}(2\pi xf(x))dx

For the following exercise, find the book generated when the region between the ii curves is rotated effectually the given centrality. Utilise both the shell method and the washer method. Utilize engineering science to graph the functions and depict a typical slice by hand.

2. [T] Under the curve of y=3x,x=0,\text{ and }x=3 rotated around the y\text{-axis}.

Solution

This figure is a graph in the first quadrant. It is the line y=3x. Under the line and above the x-axis there is a shaded region. The region is bounded to the right at x=3.
54\pi unitsiii

3. [T] Over the bend of y=3x,x=0,\text{ and }y=3 rotated around the x\text{-axis}.

iv. [T] Under the curve of y=3x,x=0,\text{ and }x=3 rotated around the x\text{-axis}.

Solution

This figure is a graph in the first quadrant. It is the line y=3x. Under the line and above the x-axis there is a shaded region. The region is bounded to the right at x=3.
81\pi units3

5. [T] Under the curve of y=2{x}^{3},x=0,\text{ and }x=2 rotated around the y\text{-axis}.

6. [T] Under the bend of y=2{x}^{3},x=0,\text{ and }x=2 rotated around the x\text{-axis}.

Solution

This figure is a graph in the first quadrant. It is the increasing curve y=2x^3. Under the curve and above the x-axis there is a shaded region. The region is bounded to the right at x=2.
\frac{512\pi }{7} units3

For the post-obit exercises, use shells to find the volumes of the given solids. Notation that the rotated regions lie between the curve and the x\text{-axis} and are rotated around the y\text{-axis}.

7. y=1-{x}^{2},x=0,\text{ and }x=1

8. y=5{x}^{3},x=0,\text{ and }x=1

Solution

2\pi unitsthree

9. y=\frac{1}{x},x=1,\text{ and }x=100

10. y=\sqrt{1-{x}^{2}},x=0,\text{ and }x=1

Solution

\frac{2\pi }{3} unitsthree

11. y=\frac{1}{1+{x}^{2}},x=0,\text{ and }x=3

12. y= \sin {x}^{2},x=0,\text{ and }x=\sqrt{\pi }

Solution

2\pi units3

thirteen. y=\frac{1}{\sqrt{1-{x}^{2}}},x=0,\text{ and }x=\frac{1}{2}

14. y=\sqrt{x},x=0,\text{ and }x=1

Solution

\frac{4\pi }{5} unitsiii

xv. y={(1+{x}^{2})}^{3},x=0,\text{ and }x=1

16. y=5{x}^{3}-2{x}^{4},x=0,\text{ and }x=2

Solution

\frac{64\pi }{3} unitsthree

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and y=0 around the x\text{-axis}.

17. y=\sqrt{1-{x}^{2}},x=0,\text{ and }x=1

eighteen. y={x}^{2},x=0,\text{ and }x=2

Solution

\frac{32\pi }{5} units3

19. y={e}^{x},x=0,\text{ and }x=1

xx. y=\text{ln}(x),x=1,\text{ and }x=e

Solution

\pi (e-2) units3

21. x=\frac{1}{1+{y}^{2}},y=1,\text{ and }y=4

22. x=\frac{1+{y}^{2}}{y},y=0,\text{ and }y=2

Solution

\frac{28\pi }{3} unitsthree

23. x= \cos y,y=0,\text{ and }y=\pi

24. x={y}^{3}-4{y}^{2},x=-1,\text{ and }x=2

Solution

\frac{-84\pi }{5} units3

25. x=y{e}^{y}\text{,}x=-1,\text{ and }x=2

26. x= \cos y{e}^{y},x=0,\text{ and }x=\pi

Solution

\text{−}{e}^{\pi }{\pi }^{2} units3

For the post-obit exercises, observe the book generated when the region betwixt the curves is rotated around the given axis.

27. y=3-x,y=0,x=0,\text{ and }x=2 rotated effectually the y\text{-axis}.

28. y={x}^{3},y=0,\text{ and }y=8 rotated effectually the y\text{-axis}.

Solution

\frac{64\pi }{5} units3

29. y={x}^{2},y=x, rotated around the y\text{-axis}.

30. y=\sqrt{x},x=0,\text{ and }x=1 rotated around the line x=2.

Solution

\frac{28\pi }{15} unitsiii

31. y=\frac{1}{4-x},x=1,\text{ and }x=2 rotated around the line x=4.

32. y=\sqrt{x}\text{ and }y={x}^{2} rotated effectually the y\text{-axis}.

Solution

\frac{3\pi }{10} unitsiii

33. y=\sqrt{x}\text{ and }y={x}^{2} rotated around the line x=2.

34. x={y}^{3},y=\frac{1}{x},x=1,\text{ and }y=2 rotated effectually the x\text{-axis}.

Solution

\frac{52\pi }{5} units3

35. x={y}^{2}\text{ and }y=x rotated around the line y=2.

For the following exercises, utilise technology to graph the region. Decide which method you remember would exist easiest to utilize to summate the volume generated when the function is rotated around the specified axis. Then, use your called method to find the volume.

38. [T] y= \cos (\pi x),y= \sin (\pi x),x=\frac{1}{4},\text{ and }x=\frac{5}{4} rotated around the y\text{-axis}.

Solution

This figure is a graph. On the graph are two curves, y=cos(pi times x) and y=sin(pi times x). They are periodic curves resembling waves. The curves intersect in the first quadrant and also the fourth quadrant. The region between the two points of intersection is shaded.
3\sqrt{2} unitsiii

39. [T] y={x}^{2}-2x,x=2,\text{ and }x=4 rotated effectually the y\text{-axis}.

forty. [T] y={x}^{2}-2x,x=2,\text{ and }x=4 rotated around the x\text{-axis}.

Solution

This figure is a graph in the first quadrant. It is the parabola y=x^2-2x. . Under the curve and above the x-axis there is a shaded region. The region begins at x=2 and is bounded to the right at x=4.
\frac{496\pi }{15} units3

41. [T] y=3{x}^{3}-2,y=x,\text{ and }x=2 rotated around the x\text{-axis}.

42. [T] y=3{x}^{3}-2,y=x,\text{ and }x=2 rotated effectually the y\text{-axis}.

Solution

This figure is a graph in the first quadrant. There are two curves on the graph. The first curve is y=3x^2-2 and the second curve is y=x. Between the curves there is a shaded region. The region begins at x=1 and is bounded to the right at x=2.
\frac{398\pi }{15} units3

44. [T] x={y}^{2},x={y}^{2}-2y+1,\text{ and }x=2 rotated around the y\text{-axis}.

Solution

This figure is a graph. There are two curves on the graph. The first curve is x=y^2-2y+1 and is a parabola opening to the right. The second curve is x=y^2 and is a parabola opening to the right. Between the curves there is a shaded region. The shaded region is bounded to the right at x=2.
15.9074 units3

For the following exercises, use the method of shells to guess the volumes of some common objects, which are pictured in accompanying figures.

45. Use the method of shells to find the volume of a sphere of radius r.

This figure has two images. The first is a circle with radius r. The second is a basketball.

46.Utilise the method of shells to find the volume of a cone with radius r and acme h.

This figure has two images. The first is an upside-down cone with radius r and height h. The second is an ice cream cone.

Solution

\frac{1}{3}\pi {r}^{2}h unitsthree

47.Use the method of shells to find the book of an ellipse ({x}^{2}\text{/}{a}^{2})+({y}^{2}\text{/}{b}^{2})=1 rotated around the x\text{-axis}.

This figure has two images. The first is an ellipse with a the horizontal distance from the center to the edge and b the vertical distance from the center to the top edge. The second is a watermelon.

48.Use the method of shells to notice the volume of a cylinder with radius r and acme h.

This figure has two images. The first is a cylinder with radius r and height h. The second is a cylindrical candle.

Solution

\pi {r}^{2}h units3

49.Utilise the method of shells to find the volume of the donut created when the circle {x}^{2}+{y}^{2}=4 is rotated around the line x=4.

This figure has two images. The first has two ellipses, one inside of the other. The radius of the path between them is 2 units. The second is a doughnut.

Glossary

method of cylindrical shells
a method of computing the book 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 contrary variable

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Source: https://opentextbc.ca/calculusv1openstax/chapter/volumes-of-revolution-cylindrical-shells/

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