Chapter 7:

Applications of Integrals

! !

428

Chapter 7 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.

429

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.

430

7.1 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

22

2Total square circles

A A A s π ⎛ ⎞= − = − ⎜ ⎟⎝ ⎠

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.

−2 −1 1 2 3

−1

1

2

3

x

y

Objectives: Find the area of the region between two curves.

431

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.

432

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

x π= , and x π= .

1 2 3 4 5 6

−2

−1

1

2

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

( ) ( )2sin cosA x x dx

π

π⎡ ⎤⎣ ⎦= −∫ Our region extends from

2x π= to x π=

( ) ( )2

cos sinx xAπ

π⎡ ⎤⎣ ⎦− −=

( ) ( )( )cos sin cos sin2 2

2A π ππ π ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

− − − − − ==

433

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= − , 12

y = , and

12

y = − .

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−4

−3

−2

−1

1

2

3

4

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 yc

A e y dy⎡ ⎤⎢ ⎥⎣ ⎦

= − −∫ On our interval ye is greater than 2 4y −

( )12 212

4yA e y dy−

⎡ ⎤⎢ ⎥⎣ ⎦

= − −∫ Our region extends from 12

y = − to 12

y =

4.959A=

434

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

x π= .

−0.5 0.5 1.0 1.5 2.0 2.5 3.0

−1.5

−1.0

−0.5

0.5

1.0

1.5

x

y

( ) ( ) ( ) ( )4 20 4cos sin sin cos .828A x x dx x x dx

π π

π⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= − + − =∫ ∫

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:

( ) ( )20cos sin .828A x x dx

π= − =∫

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 ordx. It is usually the case though, that one is “easier” than the other.

435

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 dyand an integral with adx? Both will work but one is less time consuming.

−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5

−3

−2

−1

1

2

3

4

5

6

7

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⎡ ⎤

⎣ ⎦= −∫

To find the boundaries, we need to set the equations equal to each other:

4y− y2 = y2 −6y0 = 2y2 −10y0 = 2y y−5( )y = 0,5

436

( ) ( )5 2 204 6A y y y y dy⎡ ⎤

⎢ ⎥⎣ ⎦= − − −∫

= 10y− 2y2( )dy0

5

∫= 5y2 − 2

3 y3⎤

⎦⎥0

5

=125− 2503

125

3A=

437

7.1 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,

2xy x y e x x π= = = =

3. 2 3, 8 , 3, 3y x y x x x= = − =− = 4. sin 2 , cos , 0,

2y x y x x x π= = = =

5. 25 , y x x y x= − =

6. 31 , 3xy yx += + =

7. 2, 2, 1, 1yx e x y y y= = − =− = 8. 2 21 , 1x y x y= − = −

9. 212, , 1, 2y x y x xx

= + = = =

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= + =

13. 2, 2xy x y= =

438

7.2 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.

1 2 3

−4

−3

−2

−1

1

2

3

4

x

y

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

x

y

z

439

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:

1 2 3

−4

−3

−2

−1

1

2

3

4

x

y

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

2V r hπ=

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 ( xΔ ), giving us a volume of

( )( )2V f x xπ= Δ

440

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

totalx a

V f x xπ=

= Δ∑

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 xΔ 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

( )( )2b

totala

V f x dxπ= ∫

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 dxπ ⎡ ⎤

⎣ ⎦= ∫ or ( ) 2d

cV g y dyπ ⎡ ⎤

⎣ ⎦= ∫

or 2b

aV r dxπ= ∫

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

441

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 π 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.

442

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.

5 10

−10

10

20

30

40

50

x

y

2b

aV r dxπ= ∫ Our pieces are perpendicular to the x – axis, so our

integrand will contain dx 2b

aV y dxπ ⎡ ⎤⎣ ⎦= ∫ We cannot integrate y with respect to x so we will

substitute out for y 22b

aV x dxπ ⎡ ⎤

⎣ ⎦= ∫ The expression for y is 2x 27 2

2V x dxπ ⎡ ⎤

⎣ ⎦= ∫ Our region extends from 2x = to 7x =

3355V π= When the region, R, is rotated as in the example above, the solid generated would look like this:

443

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.

444

Ex 2 Let R be the region bounded by the curves secy x= , the x – axis, 34

x π= ,

and 54

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

axis.

π/2 π 3π/2 2π 5π/2

−4

−3

−2

−1

1

2

3

4

x

y

2b

aV r dxπ= ∫ Our pieces are perpendicular to the x – axis, so our

integrand will contain dx 2b

aV y dxπ= ∫ We cannot integrate y with respect to x so we will

substitute out for y 2sec

b

aV x dxπ ⎡ ⎤⎣ ⎦= ∫ The expression for y is sec x

5 2434secV x dx

π

ππ ⎡ ⎤⎣ ⎦= ∫ Our region extends from 34

x π= to 54

x π=

2V π=

445

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.

446

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.

−2 −1 1 2

−2

−1

1

2

x

y

2d

cV r dyπ= ∫ Our pieces are perpendicular to the y – axis, so our

integrand will contain dy 2d

cV x dyπ= ∫ We cannot integrate x with respect to y so we will

sub out for x 222

d

cV y y dyπ ⎡ ⎤

⎣ ⎦= −∫ The expression for x is 22y y− 22 2

02V y y dyπ ⎡ ⎤⎣ ⎦= −∫ Our region extends from 0y = to 2y =

1615

V π=

447

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 dxπ ⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠= −∫ or ( ) ( )2 2d

cV f y g y dyπ ⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠= −∫

or

( )2 2b

aV R r dxπ= −∫ or ( )2 2d

cV R r dyπ= −∫

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

2 4

x

y

448

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.

xy

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:

449

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.

−0.5 0.5 1.0 1.5

−0.5

0.5

1.0

x

y

R

r

450

( )2 2b

aV R r dxπ= −∫ Our pieces are perpendicular to the x – axis, so our

integrand will contain dx

( ) ( )22 2b

aV x x dxπ ⎡ ⎤

⎢ ⎥⎣ ⎦= −∫ The expression for R1 is x and the expression

for R2 is 2x

( ) ( )21 2 20

V x x dxπ ⎡ ⎤⎢ ⎥⎣ ⎦

= −∫ Our region extends from 0x = to 1x =

.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.

−0.5 0.5 1.0 1.5

−0.5

0.5

1.0

x

y

( )2 2b

aV R r dxπ= −∫ Our pieces are perpendicular to the x – axis, so our

integrand will contain dx

( ) ( )221b

aV x dxπ ⎡ ⎤

⎢ ⎥⎣ ⎦

= −∫ The expression for R1 is 1and the expression

for R2 is x

( ) ( )21 2

01V x dxπ ⎡ ⎤

⎢ ⎥⎣ ⎦

= −∫ Our region extends from 0x = to 1x =

1.571V =

R

r

451

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.

−0.5 0.5 1.0 1.5

−0.5

0.5

1.0

x

y

( )2 2d

cV R r dyπ= −∫ Our pieces are perpendicular to the y – axis, so our

integrand will contain dy

( ) ( )22 31d

cV y dyπ ⎡ ⎤

⎢ ⎥⎣ ⎦

= −∫ The expression for R1 is 1and the expression

for R2 is 3 y

( ) ( )21 2 301V y dyπ ⎡ ⎤

⎢ ⎥⎣ ⎦

= −∫ Our region extends from 0y = to 1y =

1.257V =

R r

452

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.

−1 1 2 3

2

4

6

8

10

x

y

( )( ) ( )

2 2

210 2

51 2

39.270

d

cV R r dy

y dy

π

π ⎡ ⎤⎢ ⎥⎣ ⎦

= −

= − −

=

∫

∫

R r

453

7.2 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= − = 5. 2 , 0, 1; about the -axis.y x x y y= = = 6. sec , 1, , ; about the -axis.

3 3y x y x x xπ π= = = − =

7. 2 , 2 ; about the -axis.y x x y y= = 8. 2

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

454

7.3 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 dxπ ⎡ ⎤

⎣ ⎦= −∫ or ( ) 2d

cV g y h dyπ ⎡ ⎤

⎣ ⎦= −∫

or 2b

aV r dxπ= ∫

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 dxπ ⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠= − − −∫

or

( ) ( )2 2d

cV f y h g y h dyπ ⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠= − − −∫

( )2 2d

cV R r dyπ= −∫

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

455

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 = − .

−2.00 −1.50 −1.00 −0.50 0.50 1.00 1.50 2.00 2.50

−1.50

−1.00

−0.50

0.50

1.00

1.50

2.00

2.50

3.00

3.50

x

y

2

27 22

1

3583.333

b

aV r dx

x dx

π

π ⎡ ⎤⎣ ⎦

=

= +

=

∫∫

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= − −

456

x

y

z

457

Ex 2 Let R be the region bounded by the curves secy x= , 2y = , 34

x π= , and

54

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

line 2y = .

π/2 π 3π/2 2π 5π/2

−4

−3

−2

−1

1

2

3

4

x

y

2

5 2434

2

2 sec

48.174

b

aV y dx

x dxπ

π

π

π

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

= −

= −

=

∫∫

R

2 secR x= −

458

Ex 3 Find the volume of the solid obtained by rotating about the y – axis the region bounded by 22x y y= − and the line 3x = − about the line 3x = − .

−4 −3 −2 −1 1 2

−2

−1

1

2

3

4

x

y

2

23 21

3

3 2

961.746

d

cV x dy

y y dy

π

π−

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

= +

= + −

=

∫∫

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= − − −

459

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 = − .

−3 −2 −1 1 2 3 4 5 6 7

−5

5

10

15

20

25

x

y

( )( ) ( )( )

2 2

25 222

2 0 2

2433.478

b

aV R r dx

x dx

π

π ⎡ ⎤⎢ ⎥⎣ ⎦

= −

= + − − −

=

∫∫

R

r

460

Ex 5 Let R be the region bounded by the curves secy x= , the x – axis, 34

x π= ,

and 54

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

the line 2y = .

( )( ) ( )

2 2

5 2 24342 sec 2

28.435

b

aV R r dx

x dxπ

π

π

π ⎡ ⎤⎢ ⎥⎣ ⎦

= −

= − −

=

∫∫

π/2 π 3π/2 2π 5π/2

−4

−3

−2

−1

1

2

3

4

x

y

The rotated figure looks like this:

r

R

461

x

y

z

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

462

Ex 6 Find the volume of the solid obtained by rotating about the y – axis the 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.

−1 1 2 3 4 5 6

−2

−1

1

2

3

4

x

y

Set up the integration:

( )( ) ( )

2 2

22 2 204 4 2

30.159

d

cV R r dy

y y dy

π

π ⎡ ⎤⎢ ⎥⎣ ⎦

= −

= − − +

=

∫∫

R r

463

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

2x π= . Find the volume of the solid obtained by rotating R about the x –

axis.

−π/4 π/4 π/2 3π/4 π

−1.5

−1.0

−0.5

0.5

1.0

x

y

( ) ( )( ) ( ) ( ) ( )

2 2 2 2

2 22 24 20 4

cos sin sin cos

b c

a bV R r dx R r dx

x x dx x x dxπ π

π

π π

π π

π

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − + −

= − + −

=

∫ ∫∫ ∫

464

7.3 Homework Set A 1. Given the curves ( ) lnf x x= , ( ) xg x e−= and x = 4.

1 2 3 4 5

1

2

3

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.

465

2. Let f and g be the functions given by ( ) ( )4 1f x x x= − and

( ) ( )4 2 1g x x x= −

−0.5 0.5 1.0 1.5 2.0

−2

−1

1

2

3

x

y

a. Find the area of the region bounded by the two curves. b. Find the volume of the solid generated by rotating the region around

the line y = 3. c. Find the volume of the solid generated by rotating the region around

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.

466

Find the volume of the solid formed by rotating the described region about the 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 yπ= + = = = = 5. 3, ; about the line 1.y x y x y= = = 6. 3, ; about the line 1.y x y x x= = = Use your grapher to sketch the regions described below. Find the points of intersection and find the volume of the solid formed by rotating the described region about the given line. 7.

y = ln x2 +1( ), y = cosx; about the line y =1.

8. y = x2, y = 4; about the line y = 4. 9. y = 2+ sin x, y = 2, x = 0, x = 2π; about the line y = 2. 10. y

2 = x, x =1; about the line x =1.

467

7.3 Homework Set B

1. Given the functions ( ) 21

1f x

x=

+ and ( ) ( )21 6 5

2g 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 ( )22x y= − is rotated around the y–axis

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

y xπ⎛ ⎞⎜ ⎟⎝ ⎠

= and ( )31 164

y x x= −

1 2 3 4

−10

−8

−6

−4

−2

2

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

468

8. Given the functions below are 32yx = and 24x y= + . Let A be the region

bounded by the two curves and the x-axis.

1 2 3 4 5

1

2

3

x

y

a. Find the area of region A.

b. Find the value of dxdy

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

469

7.4 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.

470

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 rlπ 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 dxπ= ⋅∫ or ( )2

d

cV y g y dyπ= ⋅∫

** Everything rhymes with the shell method – Shell/Parallel/ 2 rlπ so the disc method must be of the form Disc/Perpendicular/ 2rπ .

471

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.

5 10

−10

10

20

30

40

50

x

y

2

b

aV x ydxπ= ⋅∫ Our pieces are parallel to the y – axis, so our

integrand will contain dx 2

b

aV x ydxπ= ⋅∫ We cannot integrate y with respect to x so we will

substitute out for y

22b

aV x x dxπ= ⋅∫ The expression for y is 2x

3 21

2V x x dxπ= ⋅∫ Our region extends from 1x = to 3x =

40V π=

472

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.

−2 −1 1 2 3 4 5

−2

2

4

6

8

10

12

x

y

( )2 5b

aV x y dxπ= ⋅ −∫ 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 dxπ= ⋅ −∫ The expression for y is 2x .

( )5 20

2 5V x x dxπ= ⋅ −∫ Our region extends from 0x = to 5x =

504V π=

473

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.

10 20 30

−3

−2

−1

1

2

3

4

5

6

x

y

2

d

cV y xdyπ= ⋅∫

4 22

2 2V y y dyπ= ⋅∫ 4 32

4V y dyπ= ∫

240V π=

474

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.

−4 −3 −2 −1 1 2 3 4 5

−3

−2

−1

1

2

3

x

y

2d

cV y xdyπ= ⋅∫

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 3

2y = 3 32

1 230 22 2V y x dy y x dyπ π= ⋅ + ⋅∫ ∫

( )3 32

30 22 2 3V y ydy y y dyπ π= ⋅ + ⋅ − +∫ ∫

27

4V π= The Shell Method can be applied to rotating regions about lines other than the axes, just as the Disk and Washer Methods were.

475

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 = .

4 8 12 16 20 24 28 32

−1

1

2

3

4

5

x

y

2d

cV y xdyπ= ⋅∫ Our pieces are parallel to the x – axis, so our

integrand will contain dy

( ) ( )2 4 7d

cV y x dyπ= − ⋅ −∫ We cannot integrate x with respect to y so

we will substitute out for x

( ) ( )( )22 4 7 3d

cV y y dyπ= − ⋅ − +∫ The expression for x is 2 3y +

( ) ( )( )2 20

2 4 7 3V y y dyπ= − ⋅ − +∫ Our region extends from 0y = to

2y = 108.909V =

476

The Washer Method yields the same result.

( )2 2b

aV R r dxπ= −∫

( ) ( )1

2224

b

ayV y dxπ ⎛ ⎞

⎜ ⎟⎝ ⎠= − −∫

( ) ( )27 2

34 4 3V x dxπ ⎡ ⎤

⎢ ⎥⎣ ⎦

= − − −∫

108.909V =

477

7.4 Homework Set A Find the volume of the solid formed by rotating the described region about the given line. 1. sec , 1, 0, ; about the -axis.

6y x y x x yπ= = = =

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

23

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

= = = =

4. 2

, 0, 2, 2; about the -axis.xy e y x x x−= = =− = 5. ( )21 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 yπ⎡ ⎤

⎣ ⎦= ∈

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

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

478

7.5 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.

479

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

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.

View of the solid generated with the given cross-sections from the perspective of the x-axis coming out of the page.

480

( )b

aV A x dx= ∫ Start here

( )2sideb

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 1 20

2 4V y dx= ∫ We cannot integrate y with respect to x so we will

sub out for y

( )1 20

2 4 1V x dx= −∫

V =5.333

481

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

sin2

y xπ⎛ ⎞⎜ ⎟⎝ ⎠

= and ( )31 164

y 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.

−1 1 2 3 4 5

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

x

y

( )b

aV A x dx= ∫ Start here

212

b

aV r dxπ= ∫ Area formula for our cross section

4 20

12

V r dxπ= ∫ Endpoints of the region of our solid 24

0

12 2

dV dxπ ⎛ ⎞⎜ ⎟⎝ ⎠

= ∫

4 20

18

V d dxπ= ∫ For our cross section y r= .

( )24

3

0

1sin 168 2 4

V x x x dxπ π⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − −∫ We cannot integrate y with respect to x so

we will substitute out for y.

30.208V =

d

482

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

483

x

y

484

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?

−0.5 0.5 1.0 1.5 2.0

−2

−1

1

2

3

x

y

( )

( ) ( )( )

1

0

1 20

1 24

0

1 32 2

343 4 1 2 14.589

b

aV A x dx

d d dx

d dx

x x x x dx

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

=

=

= − − −

=

∫

∫

∫

∫

485

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

2x π= .

(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) ( )20

cosxA e x dxπ

= −∫ =2.810

(b)

−π/4 π/4 π/2

−2

−1

1

2

3

4

5

x

y

( )2 2 20

V R r dxπ

π= −∫

( ) ( )

( ) ( )

2 221 20

2 220

1 1

1 1 cos

49.970

x

V y y dx

e x dx

π

π

π

π

⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎣ ⎦

= + − +

= + − +

=

∫

∫

R

486

(c)

π/4 π/2

−2

−1

1

2

3

4

5

x

y

( )220sideV dx

π= ∫

( )220

cos

8.045

xV e x dx

V

π= −

=∫

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−

487

7.5 Homework Find the volume of the solid described. 1. The base is a circle of radius 4 and the cross-sections are squares. 2. The base is the ellipse 9x2 + 4y2 = 36 and the cross-sections perpendicular to the x-axis are isosceles right triangles with the hypotenuse in the base. 3. The base is the region x2 ≤ y ≤1 and the cross-sections perpendicular to the y-axis are squares. 4. The base is the region x2 ≤ y ≤1 and the cross-sections perpendicular to the x-axis are squares. 5. The base is the region x2 ≤ y ≤1 and the cross-sections perpendicular to the y-axis are equilateral triangles. Use your grapher to sketch the regions described below. Find the points of intersection and find the volume of the solid that has the described region as its base and the given cross-sections. 6. y = x, y = e−2x, x =1; the cross-sections are semi-circles. 7. y = ln x2 +1( ) and  y = cosx; the cross-sections are squares.

8. The infinite region bounded by y = 1x , x =1 in Quadrant I where the cross-

sections are squares. 9. The cross-sectional areas of a felled tree cut into 10-foot sections are given by in table below. Use the Midpoint Riemann Sum with n=5 to estimate the volume of the tree.

x 0 10 20 30 40 50 60 70 80 90 100 Area .68 .65 .64 .61 .58 .59 .53 .55 .52 .50 .48

488

10. Given the curves ( ) 3 23 2 4f x x x x= − + + , ( ) 2 4g x x x= − and in the first

quadrant.

1 2 3 4 5

1

2

3

4

5

6

7

8

x

y

a. Find the area of the regions R, S, and T. b. Find the volume of the solid generated by rotating the region bounded

by ( ) 2 4g x x x= − and the x-axis (that is, the region made up of both Region S and Region T) around the line y = 8 on the interval 1,3x ⎡ ⎤⎣ ⎦∈ .

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

R

S

T

489

11. Let f and g be the functions given by ( )2

2 3x

f x e−

= + and ( ) 2 4g x x= −

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

5

6

x

y

a. Find the area of the region bounded by two curves above. b. Find the volume of the solid generated by rotating the region around

the line y = 4. 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.

490

12. Let R and S be the regions bounded by the graphs ( )2 3x y y= − and 2y x= − +

−2 −1 1 2 3 4 5

−2

−1

1

2

3

4

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. 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 34sA= ).

R

S

491

13. 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.

1 2

1

2

x

y

492

14. Let f and g be the functions given by ( ) ( )1 sin4

f x xπ= + 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.

1

1

x

y

a. Find the area of R b. Find the area of S 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.

R S

493

7.6 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 dydx dy

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⌠⌠⎮⎮

⎮ ⎮⌡ ⌡= + = +

22

1 1

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

494

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

b

a

dyL dxdx

⎛ ⎞⎜ ⎟⎝ ⎠

⌠⎮⎮⌡

= +2

1 Start here

( )xL dxe⌠⎮⌡

= +3 2

01 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 y=4 .

d

c

dxL dydy

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⌠⎮⎮⌡

= +2

1

( )L y dy⌠⎮⌡

= + +22

4

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 x = 5 . How does this approximation compare with the true arc length (use Math 9)?

Let ( ) ( )lnf x x= + + 21 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 51 966 2 325 2 587 2 7949 672

Approximate using rectangles

495

Actual Value:

( )ln .L x⌠⎮⌡

= + + =5 2

11 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

dy dxdx

⎛ ⎞⎜ ⎟⎝ ⎠

⌠⎮⎮⌡

+2

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

Ex 4 Find the length of the curve ( ) xf x t dt= +∫

3

21 on t≤ ≤2 7 .

( ).

b

ab

a

dyL dxdx

x x dx

⎛ ⎞⎜ ⎟⎝ ⎠

=

⌠⎮⎮⌡

⌠⎮⌡

= +

+ +

=

2

22 3

1

1 3 1

4235 511

496

7.6 Homework Set A Find the arc length of the curve. 1.

y =1+ 6x

32 on x ∈ 0, 1⎡⎣ ⎤⎦

2. y = 1

2x2 + 1

4ln x on x ∈ 2, 4⎡⎣ ⎤⎦

3. x = 1

3y y − 3( ) on y ∈ 1, 9⎡⎣ ⎤⎦

4. y = ln sec x( ) on x ∈ 0, π

4⎡

⎣⎢

⎤

⎦⎥

5.

y = ln x

32 on x ∈ 1, 3⎡

⎣⎢⎤⎦⎥

6.

y = ex on x ∈ 0, 1⎡⎣ ⎤⎦

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

8. Find the length of the arc along ( ) 0 cos xf x t dt⌠⌡= on 0 ,

2x π⎡ ⎤

⎢ ⎥⎣ ⎦

∈ .

9. Find the length of the arc along ( ) 4

23 1 xf x t dt

−= −∫ on 2, 1x ⎡ ⎤⎣ ⎦∈ − .

497

7.6 Homework Set B Find the lengths of the following curves on the given intervals: 1. 5 3cosy x= + , 0, 2x π⎡ ⎤⎣ ⎦∈ 2. xy e= , 0,1 x ⎡ ⎤⎣ ⎦∈

3. ( )ln secy x= , 0, 4

x π⎡ ⎤⎢ ⎥⎣ ⎦

∈

4. ( )1 33

y x x= − , 1, 9x ⎡ ⎤⎣ ⎦∈

5. 2y x= , 0, 2x ⎡ ⎤⎣ ⎦∈

6. 2

31

6 10xy

x= + , 1, 2x ⎡ ⎤⎣ ⎦∈

Find the perimeter of the regions enclosed by the two curves: 7. 2 lny x= + , 21x y= − 8. 2xy e−= , 2 16y x= − 9. 2 7 6y x x= − + , 3y = 10. 3 9y x x= − , 23y x x= − , 0, 3x ⎡ ⎤⎣ ⎦∈

Find the length of the following curves: 11. 2 3 2 3 1x y+ = 12. 2 2 25x y+ = 13. ( )22 4 2 36x y+ − =

498

Volume Test

1. The area of the region enclosed by y = x2 − 4 and y = x − 4 is given by

(a)

x − x2( )dx0

1⌠⌡ (b)

x2 − x( )dx

0

1⌠⌡ (c)

x − x2( )dx

0

2⌠⌡

(d)

x2 − x( )dx0

2⌠⌡ (e)

x2 − x( )dx

0

4⌠⌡

2. Which of the following integrals gives the length of the graph y = tan x

between x=a to x=b if 0 < a < b <

π2

?

(a)

x2 + tan2 x dx

a

b∫ (b)

x + tan x dx

a

b∫

(c)

1+ sec2 x dx

a

b∫ (d)

1+ tan2 x dx

a

b∫

(e)

1+ sec4 x dx

a

b∫

3. Let R be the region in the first quadrant bounded by y = e

x2 , y =1 and

x = ln3 . 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

499

4. A region is bounded by y =

1x

, the x-axis, the line x = m , and the line x = 2m ,

where m > 0 . 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 m =

12

, then decreases.

(e) is none of the above 5. Let R be the region in the first quadrant bounded by y = sin−1 x , the y-axis,

and y =

π2

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

generated when R is rotated about the y-axis?

(a) π y2 dy

0

π2∫ (b)

π sin−1 x( )2 dx

0

1⌠⌡⎮

(c) π sin−1 x( )2 dx

0

π2⌠

⌡⎮ (d) π sin y( )2 dy

0

π2⌠

⌡

(e) π sin y( )2 dy

0

1⌠⌡

6. The base of a solid is the region enclosed by y = cos x for −π2< x <

π2

. If each

cross-section of the solid perpendicular to the x-axis is a square, the volume of the solid is

(a)

π4

(b)

π 2

4 (c)

π2

(d)

π 2

2 (e) 2

500

7. Let R be the region bounded by the graphs ( ) 212 4f x x x x= + − and

( ) 3sin3

g x xπ⎛ ⎞= − ⎜ ⎟⎝ ⎠

pictured below .

−3 −2 −1 1 2 3 4 5 6 7 8 9

−3

−2

−1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

x

y

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

501

8. Let S be the region bounded by the curves ( ) 13

f x x= and ( ) 2 5g x x x=− + .

−3 −2 −1 1 2 3 4 5 6 7 8 9

−3

−2

−1

1

2

3

4

5

6

7

8

x

y

(a) Find the area between the two curves. (b) Find the volume formed by rotating the region S around the line y = 7. (c) Find the perimeter of the region S. (d) Find the volume formed by rotating the region S around the x- axis.

502

7.1 Homework 1. 19.5 2. 2.810 3. 60.252 4. .5

5. 323 6. 4 7. 5.684 8. 8

3

9. 1.369 10. 9 11. .350 12. 1.168 13. 2.106 7.2 Homework

1. π5 2. 10.036 3. π

2 4. 32π3

5. π5 6. 4.303 7. 64π

15 8. 3π4

9. 592π15 10. 3π

10 11. 1.207 12. 94.612

7.3 Homework 1a. 2.250 1b. 20.254 1c. 22.156 2a. 1.089 2b. 19.538 2c. 4.690

2d. V =π k − 2x4 x−1( )( )0

1⌠⌡⎮

2− k − 4x 1− x( )( )2 dx =10

3. 8.997 4. 7.632 5. 5π14 6. 1.361

7. 3.447 8. 512π15 9. 9.870 10. 16π

15

503

7.4 Homework 1. .064 2. 60.319 3. 70.689 4. 1.969 5. .209 6. 2π 7. .554 8. 14.676 9. 55.428 7.5 Homework

1. 341.333 2. 163 3. 2 4. 3

2

5. 1615 6. . 085 7. 916 8. DNE 9. 58

10a. 2.463, 1.551, 15.515 10b. 184.308 10c. 5.752 10d. 6.515 11a. 27.180 11b. 599.644 11c. 118.903 11d. 21.347 12a. R=2.193, S=6.185 12b. 23.345 12c. 45.081 12d. 8.607 13a. 8.607 13b. 12.840 14a. .071 14b. .328 14c. 5.837 14d. .007 7.6 Homework 1. 6.103 2. 6.499 3. 10.177 4. .881 5. 1.106 6. 2.003 7. 34.553 8. 2 9. 5.196