Chapter 6 Overview: Applications of Integrals · 2018. 8. 23. · 2 x S, and x S. x y ³b a xªº...

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Chapter 6 Overview: Applications of Integrals

Calculus, like most mathematical fields, began with trying to solve everyday

problems. The theory and operations were formalized later. As early as 270 BC,

Archimedes was working on the problem of problem of finding the volume of a

non-regular shapes. Beyond his bathtub incident that revealed the relationship

between weight volume and displacement, He had actually begun to formalize the

limiting process to explore the volume of a diagonal slice of a cylinder.

This is where Calculus can give us some very powerful tools. In geometry, we can

find lengths of specific objects like line segments or arcs of circles, while in

Calculus we can find the length of any arc that we can represent with an equation.

The same is true with area and volume. Geometry is very limited, while Calculus

is very open-ended in the problems it can solve. On the next page is an illustration

of the difference.

In this chapter, we will investigate what have become the standard applications of

the integral:

Area

Volumes of Rotation

Volume by Cross-Section

Arc Length

Just as AP, we will emphasize how the formulas relate to the geometry of the

problems and the technological (graphing calculator) solutions rather than the

algebraic solutions.

536

Below is an illustration of what we can accomplish with Calculus as opposed to

geometry:

Geometry Calculus

Length

Area

Volume

Calculus can also be used to generate surface areas for odd-shaped solids as well,

but that is out of the scope of this class.

537

6.1 Area Between Two Curves

We have learned that the area under a curve is associated with an integral. But

what about the area between two curves? It turns out that it is a simple

proposition, very similar to some problems we encountered in geometry. In

geometry, if we wanted to know the area of a shaded region that was composed of

multiple figures, like the illustration below, we find the area of the larger and

subtract the area of the smaller.

The area for this figure would be 2

2

2Total square circle

sA A A s

Similarly, since the integral gets us a numeric value for the area between a curve

and an axis, if we simply subtract the “smaller” curve from the “larger” one and

integrate, we can find the area between two curves.

x

y

Objectives:

Find the area of the region between two curves.

538

Unlike before, when we had to be concerned about positives and negatives from a

definite integral messing up our interpretation of area, the subtraction takes care of

the negative values for us (if one curve is under the axis, the subtraction makes the

negative value of the integral into the positive value of the area.

Area Between Two Curves:

The area of the region bounded by the curves y f x and y g x and the lines

x a and x b where f and g are continuous and f g for all x in ,a b is

b

aA f x g x dx

You can also think of this expression as the ‘top’ curve minus the ‘bottom’ curve.

If we associate integrals as ‘area under a curve’ we are finding the area under the

top curve, subtracting the area under the bottom curve, and that leaves us with the

area between the two curves.

The area of the region bounded by the curves x f y , x g y , and the lines

y c and y d where f and g are continuous and f g for all y in ,c d is

d

cA f y g y dy

You can also think of this expression as the ‘right’ curve minus the ‘left’ curve.

Steps to Finding the Area of a Region:

1. Draw a picture of the region.

2. Sketch a Reimann rectangle – if the rectangle is vertical your integral

will have a dx in it, if your rectangle is horizontal your integral will

have a dy in it.

3. Determine an expression representing the length of the rectangle:

(top – bottom) or (right – left).

4. Determine the endpoints the boundaries (points of intersection).

5. Set up an integral containing the limits of integration, which are the

numbers found in Step 4, and the integrand, which is the expression

found in Step 3.

6. Solve the integral.

539

Ex 1 Find the area of the region bounded by siny x , cosy x , 2

x

, and x .

x

y

b

aA f x g x dx

Our pieces are perpendicular to the x – axis,

so our integrand will contain dx

sin cosb

aA x x dx

On our interval sin x is greater than cos x

2

sin cosA x x dx

Our region extends from 2

x

to x

2

cos sinx xA

cos sin cos sin2 2

2A

540

What if the functions are not defined in terms of x but in terms of y instead?

Ex 2 Find the area of the region bounded by yx e , 2 4x y , 1

2y , and

1

2y .

x

y

d

cA f y g y dy

Our pieces are perpendicular to the y – axis,

so our integrand will contain dy

2 4d

y

cA e y dy

On our interval ye is greater than 2 4y

12 2

12

4yA e y dy

Our region extends from 1

2y to

1

2y

4.959A

541

Ex 3 Find the area of the region bounded by siny x , cosy x , 0x , and2

x

.

x

y

4 2

04

cos sin sin cos .828A x x dx x x dx

With this problem, we had to split it into two integrals, because the “top” and

“bottom” curves switch partway through the region. We could have also done this

with an absolute value:

2

0cos sin .828A x x dx

If we are asked this question on an AP test or in a college classroom, they usually

want to see the setup. The absolute value is just an easy way to integrate using the

calculator.

NOTE: Any integral can be solved going dy or dx . It is usually the case though,

that one is “easier” than the other.

542

Ex 4 Sketch the graphs of 2 6x y y and 24x y y . If asked to find the area of

the region bounded by the two curves which integral would you choose – an

integral with a dy and an integral with a dx ? Both will work but one is less time

consuming.

x

y

We can see that if we went dx we would need three integrals (because the tops and

bottoms are different on three different sections), but if we go dy we only need one

integral. I like integrals as much as the next person, but I’ll just do the one integral

if that’s OK right now.

d

cA f y g y dy

5

2 2

04 6A y y y y dy

5

2

0

2 35

0

10 2

25

3

y y dy

y y

1253

A

543

6.1 Homework

Draw and find the area of the region enclosed by the given curves.

1. 21, 9 , 1, 2y x y x x x

2. sin , , 0, 2

xy x y e x x

3. 2 3, 8 , 3, 3y x y x x x

544

4. sin2 , cos , 0, 2

y x y x x x

5. 25 , y x x y x

6. 3

1 , 3

xy yx

545

7. 2, 2, 1, 1yx e x y y y

8. 2 21 , 1x y x y

9. 2

12, , 1, 2y x y x x

x

546

10. 2 22 , 4x y y x y y

Use your grapher to sketch the regions described below. Find the points of

intersection and find the area of the region described region.

11. 2, , 1xy x y e x

12. 2ln 1 , cosy x y x

547

13. 2, 2xy x y

548

Answers: 6.1 Homework

1. 21, 9 , 1, 2y x y x x x

A=19.5

2. sin , , 0, 2

xy x y e x x

A=2.810

3. 2 3, 8 , 3, 3y x y x x x

A=60.252

4. sin2 , cos , 0, 2

y x y x x x

A=0.5

5. 25 , y x x y x

A=32

3

6. 3

1 , 3

xy yx

A=4

7. 2, 2, 1, 1yx e x y y y

A=5.684

8. 2 21 , 1x y x y

A=8

3

9. 2

12, , 1, 2y x y x x

x

A=1.369

10. 2 22 , 4x y y x y y

A=9

11. 2, , 1xy x y e x

A=0.350

12. 2ln 1 , cosy x y x

A=1.168

13. 2, 2xy x y

A=2.106

549

6.2 Volume by Rotation about an Axis

We know how to find the volume of many objects (remember those geometry

formulas?) i.e. cubes, spheres, cones, cylinders … but what about non-regular

shaped object? How do we find the volumes of these solids? Luckily we have

Calculus, but the basis of the Calculus formula is actually geometry.

If we take a function (like the parabola below) and rotate it around the x-axis on an

interval, we get a shape that does not look like anything we could solve with

geometry.

x

y

When we rotate this curve about the axis, we get a shape like the one below:

x

y

z

550

This is sort of like a cone, but because the surface curves inward its volume would

be less than a cone with the same size base. In fact, we cannot use a simple

geometry formula to find the volume. But if we begin by thinking of the curve as

having Reimann rectangles, and rotate those rectangles, the problem becomes a

little more obvious:

x

y

Each of the rotated rectangles becomes a cylinder. The volume of a cylinder is

2V r h

Note that the radius (r) is the distance from the axis to the curve (which is f x )

and the height is the change in x ( x ), giving us a volume of

2

V f x x

551

To find the total volume, we would simply add up all of the individual volumes on

the interval at which we are looking (let’s say from x = a to x = b).

2b

total

x a

V f x x

Of course, this would only give us an approximation of the volume. But if we

made the rectangles very narrow, the height of our cylinders changes from x to

dx , and to add up this infinite number of terms with these infinitely thin cylinders,

the b

x a

becomes b

a

giving us the formula

2

b

total

a

V f x dx

What you may realize is that the f x is simply the radius of our cylinder, and we

usually write the formula with an r instead. The reason for this will become more

obvious when we rotate about an axis that is not an x- or y-axis.

And since integration works whether you have dx or dy, this same process works

for curves that are rotated around the y-axis and are defined in terms of y as well.

Volume by Disc Method (Part 1): The volume of the solid generated when the

function f x from x = a and x = b, where 0f x , is rotated about the x – axis

[or g y from y = c and y = d, where 0g y , is rotated about the y – axis is

given by

2b

aV f x dx

or

2d

cV g y dy

or

2b

aV r dx

where r is the height (or length if they are horizontal) of your Riemann rectangle.

552

Steps to Finding the Volume of a Solid:

1. Draw a picture of the region to be rotated.

2. Draw the axis of rotation.

3. Sketch a Reimann rectangle – if the piece is vertical your integral will

have a dx in it, if your piece is horizontal your integral will have a dy

in it. Your rectangle should always be sketched perpendicular to the

axis of rotation.

4. Determine an expression representing the radius of the rectangle (in

these cases, the radius is the function value).

5. Determine the endpoints the region covers.

6. Set up an integral containing outside the integrand, the limits of

integration are the numbers found in Step 5, and the integrand is the

expression found in Step 4.

Objective:

Find the volume of a solid rotated when a region is rotated about a given line.

553

Ex 1 Let R be the region bounded by the equations 2y x , the x – axis, 2x ,

and 7x . Find the volume of the solid generated when R is rotated about the x –

axis.

x

y

2b

aV r dx Our pieces are perpendicular to the x – axis, so our

integrand will contain dx 2b

aV y dx We cannot integrate y with respect to x so we will

substitute out for y 2

2b

aV x dx

The expression for y is 2x

272

2V x dx

Our region extends from 2x to 7x

3355V

When the region, R, is rotated as in the example above, the solid generated would

look like this:

554

x

y

z

Note that knowing that this is what the rotated solid looks like has no bearing on

the math. It makes an interesting shape, but being able to draw the image and

being able to generate the volume by integration are two completely separate

things. We are much more concerned with finding the volume.

555

Ex 2 Let R be the region bounded by the curves secy x , the x – axis, 3

4x

,

and5

4x

. Find the volume of the solid generated when R is rotated about the x –

axis.

x

y

2b

aV r dx Our pieces are perpendicular to the x – axis, so our

integrand will contain dx

2b

aV y dx We cannot integrate y with respect to x so we will

substitute out for y 2

secb

aV x dx The expression for y is sec x

5 24

34

secV x dx

Our region extends from

3

4x

to

5

4x

2V

556

Again, when the region, S, is rotated as in the example above, the solid generated

would look like this:

x

y

z

It’s pretty cool-looking, but of no great consequence in terms of the math.

557

What if we had a function defined in terms of y?

Ex 3 Find the volume of the solid obtained by rotating about the y – axis the

region bounded by 22x y y and the y – axis.

x

y

2d

cV r dy Our pieces are perpendicular to the y – axis, so our

integrand will contain dy

2d

cV x dy We cannot integrate x with respect to y so we will

sub out for x 2

22d

cV y y dy

The expression for x is 22y y

222

02V y y dy

Our region extends from 0y to 2y

16

15V

558

Suppose we wanted to find the volume of a region bounded by two curves that was

then rotated about an axis.

Volume by Washer Method: The volume of the solid generated when the region

bounded by functions f x and g x , from x = a and x = b, where f x g x [or

f y and g y , from y = c and y = d, where f y g y ], is rotated about the x –

axis is given by

2 2b

aV f x g x dx

or 2 2d

cV f y g y dy

or

2 2b

aV R r dx or 2 2

d

cV R r dy

Where R is the outer radius of your Riemann rectangle and r is the inner radius of

your Riemann rectangle.

x

y

559

Again, think of taking a tiny strip of this region, a Riemann rectangle, and rotating

it about the x – axis. Note that this gives us a cylinder (albeit a very narrow one)

with another cylinder cut out of it. This is why the formula is what we stated

above.

If we took all of those strips for the above example, the solid would look like this.

x

y

z

Notice that overall, this does not look like a “washer” per se, but if you cut a cross

section, you would see washer shapes like the one illustrated below:

Cross section of the solid above:

560

Steps to Finding the Volume of a Solid With the Washer Method:

1. Draw a picture of the region to be rotated.

2. Draw the axis of rotation.

3. Sketch a Riemann rectangle – if the piece is vertical your integral will

have a dx in it, if your piece is horizontal your integral will have a dy

in it. Your rectangle should always be sketched perpendicular to the

axis of rotation.

4. Determine an expression representing the radius of the rectangle – in

the case of washers you will have an expression for the outer radius

and an expression for the inner radius.

5. Determine the boundaries the region covers.

6. Set up an integral containing π outside the integrand, the limits of

integration are the boundaries found in Step 5, and the integrand is

the expression found in Step 4.

Ex 5 Let R be the region bounded by the equations 2y x and y x . Find the

volume of the solid generated when R is rotated about the x – axis.

x

y

R

r

561

2 2b

aV R r dx Our pieces are perpendicular to the x – axis, so our


22 2

b

aV x x dx

The expression for R1 is x and the expression

for R2 is 2x

21 2 2

0V x x dx


.419V

Ex 6 Let R be the region bounded by 2, 1x y y , and 0x . Find the volume of

the solid generated when R is rotated about the x – axis.

x

y

2 2b

aV R r dx Our pieces are perpendicular to the x – axis, so our


22

1b

aV x dx

The expression for R1 is 1and the expression

for R2 is x

21 2

01V x dx


1.571V

R

r

562

Ex 7 Let R be the region bounded by 3, 1y x x , and 0y . Find the volume of

the solid generated when R is rotated about the y – axis.

x

y

2 2d

cV R r dy Our pieces are perpendicular to the y – axis, so our


22

31d

cV y dy

The expression for R1 is 1and the expression

for R2 is 3 y

21 2

3

01V y dy

Our region extends from 0y to 1y

1.257V

R r

563

Ex 8 Let R be the region bounded by 21 , 2y x x , and 10y . Find the volume

of the solid generated when R is rotated about the y – axis.

x

y

2 2

210 2

51 2

39.270

d

cV R r dy

y dy

R

r

564

6.2 Homework

Find the volume of the solid formed by rotating the described region about the

given line.

1. 2, 1, 0; about the axis.y x x y x

2. , 0, 1, 0; about the -axis.xy e x x y x

3. 1

, 1, 2, 0; about the -axis.y x x y xx

4. 24 , 0; about the -axis.y x y x

565

5. 2 , 0, 1; about the -axis.y x x y y

6. sec , 1, , ; about the -axis.3 3

y x y x x x

7. 2 , 2 ; about the -axis.y x x y y

8. 2

3 , 1, 0; about the -axis.y x x y y

566

9. 2 2, 4 , 2 2, 2; about the -axis.y x y x x x x

10. 2 2, ; about the -axis.y x x y x

11. 2, , 1; about the -axis.xy x y e x x

12. 2, 2 ; about the -axis.xy x y x

567


1. 2, 1, 0; about the axis.y x x y x

V=

5

2. , 0, 1, 0; about the -axis.xy e x x y x

V=10.036

3. 1

, 1, 2, 0; about the -axis.y x x y xx

V=

2

4. 24 , 0; about the -axis.y x y x

V=32

3

5. 2 , 0, 1; about the -axis.y x x y y

V=

5

6. sec , 1, , ; about the -axis.3 3

y x y x x x

V=4.303

7. 2 , 2 ; about the -axis.y x x y y

V=64

15

8. 2

3 , 1, 0; about the -axis.y x x y y

V=3

4

9. 2 2, 4 , 2 2, 2; about the -axis.y x y x x x x

V=592

15

10. 2 2, ; about the -axis.y x x y x

V=3

10

11. 2, , 1; about the -axis.xy x y e x x

V=1.207

12. 2, 2 ; about the -axis.xy x y x

V=94.612

568

6.3 Volume by Rotation about a Line not an Axis

In the last section we learned how to find the volume of a non-regular solid. This

section will be more of the same, but with a bit of a twist. Instead of rotating about

an axis, let the region be revolved about a line not the origin.

Volume by Disc Method (Form 2): The volume of the solid generated when the

function f x from x = a and x = b, where 0f x , is rotated about the line y k

[or g y from y = c and y = d, where 0g y , is rotated about the line x h is

given by

2b

aV f x k dx

or

2d

cV g y h dy

or

2b

aV r dx

where r is the Length of your Riemann rectangle.

Volume by Washer Method (Part 2): The volume of the solid generated when

the region bounded by functions f x and g x , from x = a and x = b,

where f x g x [or f y and g y , from y = c and y = d, where f y g y ],

is rotated about the line y k is given by

2 2b

aV f x k g x k dx

or

2 2d

cV f y h g y h dy

2 2d

cV R r dy

where R is the outer radius of the Riemann rectangle and r is the inner radius.

569

Ex 1 Let R be the region bounded by the equations 2y x , 1y , 1x , and

1x . Find the volume of the solid generated when R is rotated about the

line 1y .

x

y

2

272

21

3583.333

b

aV r dx

x dx

As in the previous section, this makes an interesting shape, but is of very little use

to us in actually solving for the volume. You could easily have found the value for

the volume never knowing what the shape actually looked like.

r

2 1r x

570

x

y

z

571

Ex 2 Let R be the region bounded by the curves secy x , 2y , 3

4x

, and

5

4x

. Find the volume of the solid generated when R is rotated about the

line 2y .

x

y

2

5 24

34

2

2 sec

48.174

b

aV y dx

x dx

R

2 secR x

572


region bounded by 22x y y and the line 3x about the line 3x .

x

y

2

232

1

3

3 2

961.746

d

cV x dy

y y dy

What if you were asked to take the region bounded by y x and 2y x and rotate

it about the x – axis? How is this different from the previous problems?

R

22 3R y y

573

Ex 4 Let R be the region bounded by the equations 2y x , the x – axis, 2x , and

5x . Find the volume of the solid generated when R is rotated about the

line 2y .

x

y

2 2

25 22

22 0 2

2433.478

b

aV R r dx

x dx

R

r

574

Ex 5 Let R be the region bounded by the curves secy x , the x – axis, 3

4x

,

and5

4x

. Find the volume of the solid generated when R is rotated about

the line 2y .

2 2

5 2 24

34

2 sec 2

28.435

b

aV R r dx

x dx

x

y

r

R

575

The rotated figure looks like this:

x

y

z

It would be difficult to sketch this by hand, and we definitely do not need it to

solve the problem.

576


region bounded by 22x y y and the y – axis about line 4x .

Given the information above and the sketch of 22x y y below, sketch

your boundaries, the axis of rotation, your Riemann rectangle, and your

inner and outer radii.

x

y

Set up the integration:

2 2

22 2 2

04 4 2

30.159

d

cV R r dy

y y dy

R

r

577

Ex 7 Let R be the region bounded by the curves siny x , cosy x , 0x , and

2x

. Find the volume of the solid obtained by rotating R about the x –

axis.

x

y

2 2 2 2

2 22 24 2

04

cos sin sin cos

b c

a bV R r dx R r dx

x x dx x x dx

578

6.3 Homework Set A

1. Given the curves lnf x x , xg x e and x = 4.

x

y

a. Find the area of the region bounded by three curves in the first

quadrant.

b. Find the volume of the solid generated by rotating the region around

the line y = 2.

c. Find the volume of the solid generated by rotating the region around

the line y = –1.

579

2. Let f and g be the functions given by 4 1f x x x and

4 2 1g x x x

x

y


quadrant.


the line y = 3.


the line y = –2.

d. Set up an integral expression for the curve 4 2 1g x x x and the

curve 1h x x x which is not pictured above that is rotated

around the line y = k, where k >2, in which the volume equals 10. Do

not solve this equation.

580


given line. Sketch the graph so that it is easier for you to apply the Disk/Washer

Methods.

3. 1

, 0, 1, 3; about the line 1.y y x x yx

4. 2 sin , 2, 0, ; about the line 3.y x y x x y

5. 3, ; about the line 1.y x y x y

6. 3, ; about the line 1.y x y x x

581


intersection and find the volume of the solid formed by rotating the described

region about the given line.

7. 2ln 1 , cos ; about the line 1.y x y x y

8. 2, 4; about the line 4.y x y y

9. 2 sin , 2, 0, 2 ; about the line 2.y x y x x y

10. 2 , 1; about the line 1.y x x x

582

6.3 Homework Set B

1. Given the functions 2

1

1f x

x

and 21

6 52

g x x x

, find the area

bounded by the two curves in the first quadrant.

2. Find the volume of the solid generated when the function 3ln 9y x is

rotated around the line 4y on the interval 1,2x

3. Find the volume of the solid generated when the area in the first quadrant

bounded by the curves ln 1x y and 2

2x y is rotated around the y–

axis

583

4. Let R be the region bounded by the graphs sin2

y x

and 3116

4y x x

x

y

a. Find the area of the region R.

b. Find the volume of the solid when the region R is rotated around the

line y = –10.

c. The line y = –3 cuts the region R into two regions. Write and evaluate

an integral to find the area of the region below the line.

R

584

5. Given the functions below are 3

2

yx and 24x y . Let A be the region

bounded by the two curves and the x-axis.

x

y

a. Find the area of region A.

b. Find the value of dx

dy for 24x y at the point where the two curves

intersect.

c. Find the volume of the solid when region A is rotated around the y-

axis.

A

585

Answers: 6.3 Homework Set A

1. Given the curves lnf x x , xg x e and x = 4.


quadrant.

A=2.250


the line y = 2.

V=20.254


the line y = –1.

V=22.156

2. Let f and g be the functions given by 4 1f x x x and

4 2 1g x x x


quadrant.

A=1.089


the line y = 3.

V=19.538


the line y = –2.

V=14.689

d. Set up an integral expression for the curve 4 2 1g x x x and the

curve 1h x x x which is not pictured above that is rotated

around the line y = k, where k >2, in which the volume equals 10. Do

not solve this equation.

1

2

0

210 k g x k h x dx

3. 1

, 0, 1, 3; about the line 1.y y x x yx

V=8.997

4. 2 sin , 2, 0, ; about the line 3.y x y x x y

V=7.632

586

5. 3, ; about the line 1.y x y x y

V=5

14

6. 3, ; about the line 1.y x y x x

V=1.361

7. 2ln 1 , cos ; about the line 1.y x y x y

V=3.447

8. 2, 4; about the line 4.y x y y

V=512

15

9. 2 sin , 2, 0, 2 ; about the line 2.y x y x x y

V=9.870

10. 2 , 1; about the line 1.y x x x

V=16

15

6.3 Homework Set B

1. Given the functions 2

1

1f x

x

and 21

6 52

g x x x

, find the area

bounded by the two curves in the first quadrant.

A=4.798

2. Find the volume of the solid generated when the function 3ln 9y x is

rotated around the line 4y on the interval 1,2x

V=27.286

3. Find the volume of the solid generated when the area in the first quadrant

bounded by the curves ln 1x y and 2

2x y is rotated around the y–

axis

V=124.757

587

4. Let R be the region bounded by the graphs sin2

y x

and 3116

4y x x

a. Find the area of the region R.

R=16

b. Find the volume of the solid when the region R is rotated around the

line y = –10.

V=766.489

c. The line y = –3 cuts the region R into two regions. Write and evaluate

an integral to find the area of the region below the line.

23.552799

0.7796156

3116 10 5.775

4V x x dx

5. Given the functions below are 3

2

yx and 24x y . Let A be the region

bounded by the two curves and the x-axis.

a. Find the area of region A.

A=1.609

b. Find the value of dx

dy for 24x y at the point where the two curves

intersect.

3.578dx

dy

c. Find the volume of the solid when region A is rotated around the y-

axis.

V=14.986

588

6.4 Volume by Cross Sections

In the last sections, we have learned how to find the volume of a solid created

when a region was rotated about a line. In this section, we will find volumes of

solids that are not made though revolutions, but cross sections.

Volume by Cross Sections Formula:

The volume of solid made of cross sections with area A is

b

aV A x dx or

d

cV A y dy

Objectives:

Find the volume of a solid with given cross sections.

Steps to Finding the Volume of a Solid by Cross Sections:

1. Draw a picture of the region of the base.

2. Sketch a sample cross section (the base of the sample cross section

will lie on the base sketched in Step 1)– if the piece is vertical your

integral will have a dx in it, if your piece is horizontal your integral

will have a dy in it.

3. Determine an expression representing the area of the sample cross

section (be careful when using semi-circles – the radius is half the

distance of the base).

4. Determine the endpoints the region covers.

5. Set up an integral containing the limits of integration found in Step 5,

and the integrand expression found in Step 4.

589

Ex 1 The base of solid S is 2 2 1x y . Cross sections perpendicular to the x –

axis are squares. What is the volume of the solid?

x

y

z

View of the solid

generated with the

given cross-sections

from the perspective of

the x-axis coming out

of the page.

590

x

y

z

b

aV A x dx Start here

2

sideb

aV dx Area formula for our (square) cross section

1 2

1sideV dx

Endpoints of the region of our solid

1 2

02 sideV dx Simplify the integral

1 2

02 2V y dx Length of the side of our cross section

12

02 4V y dx We cannot integrate y with respect to x so we will

sub out for y

1

2

02 4 1V x dx

V 5.333

View of the solid

generated with the

given cross-sections

from the perspective of

the x-axis coming out

of the page.

591

Ex 2 The base of solid is the region S which is bounded by the graphs

sin2

y x

and 3116

4y x x . Cross sections perpendicular to the x –

axis are semicircles. What is the volume of the solid?

It is easier for set-up to visualize the base region and the cross-section separately

and not to try to imagine the 3D solid.

x

y

d

b

aV A x dx Start here

21

2

b

aV r dx Area formula for our cross section

42

0

1

2V r dx Endpoints of the region of our solid

24

0

1

2 2

dV dx

42

0

1

8V d dx For our cross section y r .

24

3

0

1sin 16

8 2 4V x x x dx

We cannot integrate y with respect to x so

we will substitute out for y.

30.208V

d

592

Here is an illustration of the solid, cut-away so that you can see the cross-section:

x

y

z

Two views of the solid from Example 2

x

y

z

d

593

x

y

594

Ex 3 The base of solid is the region S which is bounded by the graphs

4 1f x x x and 4 2 1g x x x . Cross-sections perpendicular to the

x – axis are equilateral triangles. What is the volume of the solid?

x

y

d

1

0

12

0

1 24

0

1 3

2 2

3

4

34 1 2 1

4

.589

b

aV A x dx

d d dx

d dx

x x x x dx

595

Ex 4 Let R be the region in the first quadrant bounded by cosy x , xy e , and

2x

.

(a) Find the area of region R.

(b) Find the volume of the solid obtained when R is rotated about the

line 1y .

(c) The region R is the base of s solid. For this solid, each cross section

perpendicular to the x – axis is a square. Find the volume of the solid.

(a) 2

0cosxA e x dx

=2.810

(b)

x

y

2 2 2

0V R r dx

2 221 20

2 22

0

1 1

1 1 cos

49.970

x

V y y dx

e x dx

R

596

(c)

x

y

22

0sideV dx

22

0cos

8.045

xV e x dx

V

What you may have noticed is that the base of your cross-section is always

either “top curve” – “bottom curve” or “right curve” – “left curve”.

What this means for us is that all we really need to do is

1) Draw the cross-section and label the base with its length (top – bottom or

right – left).

2) Figure out the area formula for that cross-section.

3) Integrate the area formula on the appropriate interval.

side

cosxe x

cosxe x

597

6.4 Homework


intersection and find the volume of the solid that has the described region as its

base and the given cross-sections.

1. 2, , 1;xy x y e x the cross-sections are semi-circles.

2. 2 ln 1 and cos ;y x y x the cross-sections are squares.

598

3. Given the curves 3 23 2 4f x x x x , 2 4g x x x and in the first

quadrant.

x

y

a. Find the area of the regions R, S, and T.

b. Find the volume of the solid generated by rotating the curve

2 4g x x x around the line y = 8 on the interval 1,3x .

R

S

T

599

c. Find the volume of the solid generated by rotating the region S around

the line y = –1.

d. Find the volume of the solid generated if R forms the base of a solid

whose cross sections are squares whose bases are perpendicular to the

x-axis

600

4. Let f and g be the functions given by 2

2 3x

f x e

and 2 4g x x

x

y

a. Find the area of the region bounded by two curves above.


the line y = 4.

601

c. Find the volume of the solid generated by rotating the area between

g x and the x-axis around the line y = 0.

d. Find the volume of the solid generated if the region above forms the

base of a solid whose cross sections are semicircles with bases

perpendicular to the x-axis.

602

5. Let R and S be the regions bounded by the graphs 2 3x y y and

2y x

x

y

a. Find the area of the regions R and S.

b. Find the volume of the solid when the region S is rotated around the

line x = 3.

R

S

603

c. Find the volume of the solid when the region R is rotated around the

line x = –2.

d. Find the volume of the solid if the region S forms the base of a solid

whose cross sections are equilateral triangles with bases perpendicular

to the y-axis. (Hint: the area of an equilateral triangle is 2 3

4

sA ).

604

6. Let h be the function given by 2ln 1h x x as graphed below. Find

each of the following:

a. The volume of the solid generated when h is rotated around the line

2y on the interval 0,2x

b. The volume of the solid formed if the region between the curve and

the lines 1y , 0x , and 2x forms the base of a solid whose

cross sections are rectangles with heights twice their base and whose

bases are perpendicular to the x-axis.

x

y

605


sin4

f x x and xg x e .

Let R be the region in the first quadrant enclosed by the y – axis and the

graphs of f and g, and let S be the region in the first quadrant enclosed by

the graphs of f and g, as shown in the figure below.

x

y

a. Find the area of R

b. Find the area of S

R S

606

c. Find the volume of the solid generated when S is revolved around the

horizontal line 2y

d. The region R forms the base of a solid whose cross-sections are

rectangles with bases perpendicular to the x-axis and heights equal to

half the length of the base. Find the volume of the solid.

607


1. 2, , 1;xy x y e x the cross-sections are semi-circles.

V=0.085

2. 2 ln 1 and cos ;y x y x the cross-sections are squares.

V=0.916

3. Given the curves 3 23 2 4f x x x x , 2 4g x x x and in the first

quadrant.

a. Find the area of the regions R, S, and T.

R=2.463 S=12.470 T=9.835

b. Find the volume of the solid generated by rotating the curve

2 4g x x x around the line y = 8 on the interval 1,3x .

V=48.152

c. Find the volume of the solid generated by rotating the region S around

the line y = –1. V=55.727

d. Find the volume of the solid generated if R forms the base of a solid

whose cross sections are squares whose bases are perpendicular to the

x-axis

V=6.962


2 3x

f x e

and 2 4g x x

a. Find the area of the region bounded by two curves above.

A=27.180


the line y = 4.

V=599.694

c. Find the volume of the solid generated by rotating the area between

g x and the x-axis around the line y = 0.

V=107.233

d. Find the volume of the solid generated if the region above forms the

base of a solid whose cross sections are semicircles with bases

perpendicular to the x-axis.

V=66.842

608

5. Let R and S be the regions bounded by the graphs 2 3x y y and

2y x

a. Find the area of the regions R and S.

R=2.193, S=6.185

b. Find the volume of the solid when the region S is rotated around the

line x = 3.

V=55.345

c. Find the volume of the solid when the region R is rotated around the

line x = –2.

V=45.081

d. Find the volume of the solid if the region S forms the base of a solid

whose cross sections are equilateral triangles with bases perpendicular

to the y-axis. (Hint: the area of an equilateral triangle is 2 3

4

sA ).

V=8.607

6. Let h be the function given by 2ln 1h x x as graphed below. Find each

of the following:

a. The volume of the solid generated when h is rotated around the line

2y on the interval 0,2x

V=4.881

b. The volume of the solid formed if the region between the curve and

the lines 1y , 0x , and 2x forms the base of a solid whose

cross sections are rectangles with heights twice their base and whose

bases are perpendicular to the x-axis.

V=12.840

609


sin4

f x x and xg x e .

Let R be the region in the first quadrant enclosed by the y – axis and the

graphs of f and g, and let S be the region in the first quadrant enclosed by

the graphs of f and g, as shown in the figure below.

a. Find the area of R

R=0.071

b. Find the area of S

S=0.328

c. Find the volume of the solid generated when S is revolved around the

horizontal line 2y

V=5.837

d. The region R forms the base of a solid whose cross-sections are

rectangles with bases perpendicular to the x-axis and heights equal to

half the length of the base. Find the volume of the solid.

V=0.007

610

6.5 Arc Length

Back in your geometry days you learned how to find the distance between two

points on a line. But what if we want to find the distance between to points that lie

on a nonlinear curve?

Objectives:

Find the arc length of a function in Cartesian mode between two points.

Just as the area under a curve can be approximated by the sum of rectangles, arc

length can be approximated by the sum of ever-smaller hypotenuses.

1

0.5

-0.5

-1 1 2

dx2+dy2

dy

dx

1

0.5

-0.5

-1 1 2

Here is our arc length formula:

Arc Length between Two Points:

Let f x

be a differentiable function such that 'f x

is continuous, the length L

of f x over ,a b

is given by

db

a c

dy dxL dx or L dy

dx dy

22

1 1

As with the volume problems, we will use Math 9 in almost all cases to calculate

arc length.

611

Ex 1 Find the length of the curve xy e 1 from x 0 to x 3 .

b

a

dyL dx

dx

2

1 Start here

xL dxe

3 2

0

1 Substitute in for dy

dx

.L 19 528 Math 9

Ex 2 Find the length of the curve x y y 32 from y 1 to y4 .

d

c

dxL dy

dy

2

1

L y dy

2

24

11 2 3

.L 69 083

Ex 3 Use right hand Riemann sums with n = 4 to estimate the arc length of

lnx xy 3 from x 1 to x5 . How does this approximation compare

with the true arc length (use Math 9)?

Let lnf x x 2

1 1 Let f x represent our integrand

ln

. . . .

.

L x dx

f f f f

5

1

21 1

1 2 1 3 1 4 1 5

1 966 2 325 2 587 2 794

9 672

Approximate using rectangles

612

Actual Value:

ln .L x

5 2

1

1 1 9 026

Note: The integral is the arc length. Some people get confused as to how

“rectangle areas” can get us length. Remember that the Riemann rectangles are a

way of evaluating an integral by approximation. Just like the “area under the

curve” could be a displacement if the curve was velocity, in this case, the “area

under the curve

b

a

dydx

dx

2

1 is the length of the arc along the curve y.

Ex 4 Find the length of the curve x

f x t dt 3

21 on t 2 7 .

.

b

a

b

a

dyL dx

dx

x x dx

2

22 3

1

1 3 1

4235 511

613

6.5 Homework

Find the arc length of the curve.

1. 3

21 6 on 0,1 y x x

2. 21 1ln on 2, 4

2 4y x x x

3. 1

3 on 1, 93

x y y y

4. ln sec on 0, 4

y x x

614

5. 3

2ln on 1, 3y x x

6. on 0,1 xy e x

7. Find the perimeter of each of the two regions bounded by 2y x and 2xy .

8. Find the length of the arc along 0cos

xf x t dt

on 0 ,

2x

.

615

9. Find the length of the arc along 4

23 1

xf x t dt

on 2, 1x .

616


1. 3

21 6 on 0,1 y x x

L=6.103

2. 21 1ln on 2, 4

2 4y x x x

L=6.499

3. 1

3 on 1, 93

x y y y

L=10.177

4. ln sec on 0, 4

y x x

L=0.881

5. 3

2ln on 1, 3y x x

L=1.106

6. on 0,1 xy e x

L=2.003

7. Find the perimeter of each of the two regions bounded by 2y x and 2xy .

L=34.553

8. Find the length of the arc along 0cos

xf x t dt

on 0 ,

2x

.

L=2

9. Find the length of the arc along 4

23 1

xf x t dt

on 2, 1x .

L=1.337

617

Volume Test

1. The area of the region enclosed by 2 4y x and 4y x is given by

(a) 1

2

0

x x dx

(b) 1

2

0

x x dx

(c) 2

2

0

x x dx

(d) 2

2

0

x x dx

(e) 4

2

0

x x dx

2. Which of the following integrals gives the length of the graph tany x

between x=a to x=b if 02

a b

?

(a) 2 2tanb

a

x xdx (b) tanb

a

x xdx

(c) 21 secb

a

xdx (d) 21 tanb

a

xdx

(e) 41 secb

a

xdx

618

3. Let R be the region in the first quadrant bounded by 2x

y e , 1y and

ln3x . What is the volume of the solid generated when R is rotated about

the x-axis?

(a) 2.80 (b) 2.83 (c) 2.86 (d) 2.89 (e) 2.92

4. The base of a solid is the region enclosed by cosy x and the x axis for

2 2x

. If each cross-section of the solid perpendicular to the x-axis is

a square, the volume of the solid is

(a) 4

(b)

2

4

(c)

2

(d)

2

2

(e) 2

5. A region is bounded by 1

yx

, the x-axis, the line x m , and the line

2x m , where 0m . A solid is formed by revolving the region about the x-

axis. The volume of the solid

(a) is independent of m.

(b) increases as m increases.

(c) decreases as m increases.

(d) increases until 1

2m , then decreases.

(e) is none of the above

619

6. Let R be the region in the first quadrant bounded by 1siny x , the y-axis,

and 2

y

. Which of the following integrals gives the volume of the solid

generated when R is rotated about the y-axis?

(a) 2

2

0

y dy

(b) 1

21

0

sin x dx

(c) 2 2

1

0

sin x dx

(d) 2

2

0

sin y dy

(e) 1

2

0

sin y dy

620

x

y

7. Let R be the region bounded by the graphs 212 4f x x x x and

3sin3

g x x

pictured above .

(a) Find the area of R.

(b) Find the volume of the figure if R is rotated around the line y = –4.

621

(c) The region, R, is the base of a solid whose cross-sections are squares

perpendicular to the x-axis. Find the volume of this solid.

(d) Find the volume of the solid if only 212 4f x x x x is rotated around the

x-axis.

622

x

y

8. Let S be the region bounded by the curves 1

3f x x and 2 5g x x x .

(a) Find the area between the two curves.

(b) Find the volume formed by rotating the region S around the line y = 7.

623

(c) Find the perimeter of the region S.

(d) Find the volume formed by rotating the region S around the x- axis.

624

Answers: Volume Test

1. The area of the region enclosed by 2 4y x and 4y x is given by

(a) 1

2

0

x x dx

(b) 1

2

0

x x dx

(c) 2

2

0

x x dx

(d) 2

2

0

x x dx

(e) 4

2

0

x x dx

2. Which of the following integrals gives the length of the graph tany x

between x=a to x=b if 02

a b

?

(a) 2 2tanb

a

x xdx (b) tanb

a

x xdx

(c) 21 secb

a

xdx (d) 21 tanb

a

xdx

(e) 41 secb

a

xdx

625

3. Let R be the region in the first quadrant bounded by 2x

y e , 1y and

ln3x . What is the volume of the solid generated when R is rotated about

the x-axis?

(a) 2.80 (b) 2.83 (c) 2.86 (d) 2.89 (e) 2.92

4. The base of a solid is the region enclosed by cosy x and the x axis for

2 2x

. If each cross-section of the solid perpendicular to the x-axis is

a square, the volume of the solid is

(a) 4

(b)

2

4

(c)

2

(d)

2

2

(e) 2

5. A region is bounded by 1

yx

, the x-axis, the line x m , and the line

2x m , where 0m . A solid is formed by revolving the region about the x-

axis. The volume of the solid

(a) is independent of m.

(b) increases as m increases.

(c) decreases as m increases.

(d) increases until 1

2m , then decreases.

(e) is none of the above

626

6. Let R be the region in the first quadrant bounded by 1siny x , the y-axis,

and 2

y

. Which of the following integrals gives the volume of the solid

generated when R is rotated about the y-axis?

(a) 2

2

0

y dy

(b) 1

21

0

sin x dx

(c) 2 2

1

0

sin x dx

(d) 2

2

0

sin y dy

(e) 1

2

0

sin y dy

627

x

y

7. Let R be the region bounded by the graphs 212 4f x x x x and

3sin3

g x x

pictured above .

(a) Find the area of R.

R=54.595

(b) Find the volume of the figure if R is rotated around the line y = –4.

V=3179.787

(c) The region, R, is the base of a solid whose cross-sections are squares

perpendicular to the x-axis. Find the volume of this solid.

V=549.698

(d) Find the volume of the solid if only 212 4f x x x x is rotated around the

x-axis.

V=1900.035

628

x

y

8. Let S be the region bounded by the curves 1

3f x x and 2 5g x x x .

(a) Find the area between the two curves.

S=16.948

(b) Find the volume formed by rotating the region S around the line y = 7.

V=430.435

(c) Find the perimeter of the region S.

P=7.232

(d) Find the volume formed by rotating the region S around the x- axis.

V=314.549

629

6.__ Bonus Section: Volume by the Shell Method

There is another method we can use to find the volume of a solid of rotation. With

the Disc/Washer method we drew our sample pieces perpendicular to the axis of

rotation. Now we will look into calculating volumes when our sample pieces are

drawn parallel to the axis of rotation (parallel and shell rhyme – that’s how you

can remember they go together).

Objectives:

Find the volume of a solid rotated when a region is rotated about a given line.

Take the following function, let’s call it f x , from a to b and rotate it about the y

– axis.

630

How do we begin to find the volume of this solid?

Imagine taking a tiny strip of the function and rotating this little piece around the y

– axis. What would that piece look like once it was rotated?

What is the surface area formula for a cylindrical shell? 2 rl

Now would the surface area of just this one piece give you the volume of the entire

solid?

How would we add up all of the surface areas of all these little pieces?

We arrive at our volume formula.

Volume by Shell Method: The volume of the solid generated when the region

bounded by function f x , from x = a and x = b, where 0f x [or g y , from

y = c and y = d, where 0g y ], is rotated about the y – axis is given by

2b

aV x f x dx or 2

d

cV y g y dy

** Everything rhymes with the shell method – Shell/Parallel/ 2 rl so the disc

method must be of the form Disc/Perpendicular/ 2r .

631

Ex 1 Let R be the region bounded by the equations 2y x , 0y , 1x , 3x .

Find the volume of the solid generated when R is rotated about the y – axis.

x

y

2 b

aV x ydx Our pieces are parallel to the y – axis, so our


2 b

aV x ydx We cannot integrate y with respect to x so we will

substitute out for y

22b

aV x x dx The expression for y is 2x

32

12V x x dx Our region extends from 1x to 3x

40V

632

Ex 2 Let R be the region bounded by 2, 5y x y , and 0x . Find the volume of

the solid generated when R is rotated about the y – axis.

x

y

2 5b

aV x y dx Our pieces are parallel to the y – axis, so our

integrand will contain dx . We cannot integrate y

with respect to x so we will substitute out for y.

22 5b

aV x x dx The expression for y is 2x .

5

2

02 5V x x dx Our region extends from 0x to 5x

504

V

633

Ex 3 Let R be the region bounded by 22 , 1x y x , 2y and 4y . Find the

volume of the solid generated when R is rotated about the x – axis.

x

y

2d

cV y xdy

4

2

22 2V y y dy

43

24V y dy

240V

634

Ex 4 Let R be the region bounded by , 3y x x y , and 0x . Find the volume

of the solid generated when R is rotated about the x – axis.

x

y

2d

cV y xdy

Our pieces are parallel to the x – axis, so our integrand will contain dy . But

we will need two integrals as our region of rotation is bounded by different

curves. The dividing line is 32

y

3 32

1 2302

2 2V y x dy y x dy

3 3

2

302

2 2 3V y ydy y y dy

274

V

The Shell Method can be applied to rotating regions about lines other than the

axes, just as the Disk and Washer Methods were.

635

Ex 5 Let R be the region bounded by the equations 3y x , the x – axis,

and 7x . Find the volume of the solid generated when R is rotated about

the line 4y .

x

y

2d

cV y xdy Our pieces are parallel to the x – axis, so our


2 4 7d

cV y x dy We cannot integrate x with respect to y so

we will substitute out for x

22 4 7 3d

cV y y dy The expression for x is 2 3y

2

2

02 4 7 3V y y dy Our region extends from 0y to

2y

108.909V

636

The Washer Method yields the same result.

2 2b

aV R r dx

1

22

24b

ayV y dx

27 2

34 4 3V x dx

108.909V

637

6.__ Homework Set A


given line.

1. sec , 1, 0, ; about the -axis.6

y x y x x y

2. 3, 0, 8; about the -axis.y x x y y

3. 2

3

1, 1, 8, 0; about the -axis.y x x y y

x

4. 2

, 0, 2, 2; about the -axis.xy e y x x x

638

5. 2

1 and the -axis; about the -axis.y x x x y

6. 2sin and the -axis on 0, ; about the -axis.y x x x y


intersection and find the volume of the solid formed by rotating the described

region about the given line.

7. 2, , 1; about the line 1.xy x y e x x

8. 2ln 1 , cos ; about the line 2.y x y x x

639

9. 2, 2 ; about the line 1.xy x y x

640

Answers: 6.4 Homework Set A

1. sec , 1, 0, ; about the -axis.6

y x y x x y

V=0.064

2. 3, 0, 8; about the -axis.y x x y y

V=60.319

3. 2

3

1, 1, 8, 0; about the -axis.y x x y y

x

V=70.689

4. 2

, 0, 2, 2; about the -axis.xy e y x x x

V=1.969

5. 2

1 and the -axis; about the -axis.y x x x y

V=0.209

6. 2sin and the -axis on 0, ; about the -axis.y x x x y

V=2

7. 2, , 1; about the line 1.xy x y e x x

V=0.554

8. 2ln 1 , cos ; about the line 2.y x y x x

V=14.676

9. 2, 2 ; about the line 1.xy x y x

V=55.428

Chapter 6 Overview: Applications of Integrals · 2018. 8. 23. · 2 x S, and x S. x y ³b a xªº...

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Transcript of Chapter 6 Overview: Applications of Integrals · 2018. 8. 23. · 2 x S, and x S. x y ³b a xªº...