Calculus Concept Collection - Chapter 6

Area Between Two Curves

Answers

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. The integral associated with this area is 2

01 2 sin( )x x x dx

.

This definite integral can be resolved as follows:

2 2

3

0 0

3

2 2

0 0

2 20

1 2 sin( ) 1 sin( )2

1 1 71) 2 cos( ) | 1) 2( (

33 3

x x dx x x

x

x x

x x

12. The integral associated with this area is 3

2cos( ) cos(2 )x x dx .

This definite integral can be resolved as follows:

3

2

3

2

1 1 1cos( ) cos(2 ) sin( ) sin(2 ) | sin(3) sin(6) sin(2) sin(4)

2 2 2x x dx x x

13. First, we express our functions in terms of y.

3 39 9x yy x

1 1

x yy x

Now, we set up and evaluate the integral.

1

33

4 4

1 1 1

3 31

4

3

1 19

( 9) ( 9

( 9)

3) ( 8

3 3ln( ) | )

4)ln(

4 4

y dy dyy y

y d

yy

y

14. Set them equal to each other and solve for x.

2

2

6

0 4 5 6

0 (4 3)( )

5 4

2

x x

x x

x x

The curves intersect at 2x and3

4x .

15. First find points of intersection.

2 2

2

8

0 8 2

0 2(2 )(2 )

x x

x

x x

They intersect at 2x .

To find the area, integrate over the difference. On the interval, 2 28 xx so integrate

2 22 2 2

2

3

2

2

28 ) ( ) 8( (

2

2 )

864

3 3

x dx x dxx

xx

Volumes by Cross Section

Answers

1. Find a formula for area of a cross-section as a function of height. A good way to do this is to

first find a function for length of a side, since it changes linearly with height. At height zero the

length is 20; at height 15 the length is 0. Create a straight line function between these points:

0 20 420 20

15 0( )

3x xl x

Then area is length squared:

2 2 24 16 160( ) ( ) 20) 400

3 9 3(A x l x x x x

To find volume integrate the area formula over the total height, from 0 to 15.

152

0

16 160( 400) 2000

9 3V x x dx

2. Find a formula for area of a cross-section as a function of height. Start by finding a function

for radius as a function of height. At height zero, radius is 10/2=5. At height 4, radius is 0. The

radius changes linearly with height so make a linear function between these points.

0 5 5( ) 5 5

4 0 4r x x x

Now find cross-sectional area:

2 2 25 25 25( ) ( ) ( 5) ( 25)

4 16 2A x r x x x x

Integrate the area function over the height.

4 4 42 2

0 0 0

25 25 25 25 100( ) ( 25) ( 25)

16 2 16 2 3A x dx x x dx x x dx

3. Cross-sections of the swimming pool are rectangles with sides of lengths 12 and

6 2cos( )20

x

; the area of each rectangle is ( ) 12 6 2cos(( ))20

A x x

. Integrate the area

over the length 15:

15 15 15 15

0 0 0 0

15

0

( ) ( )) 12 ( ) )20 20

40 40 2 240 212 (6 15 sin ( ) 12 (90 )

12 6 2cos( 6 2c

10802 2

os

0

(A x dx x dx dx x dx

x

4. Find area of a cross section as a function of height by first finding side length as a function of

height; side length varies linearly with height, from (0,2) to (2,0): ( ) 2s x x .

The area of an equilateral triangles is 23

4A s .

2 2 23 3 3( ) ( ) (2 ) (4 4

4)

4 4A x s x x x x

Integrate over the height:

2 22

0 0

3 3 8 2 3( ) (4 4 )

4 4 3 3A x dx x x dx

5. The formula for area as a function of length x is

1( ) ( ) 2 ( ) 2(1 ) 2 2

2A x h w x w x x x

Integrate over the total length. 5

02 2 35xdx

6. 1

3 4 2 1

00

|1 1 3

4 2 4xx x xdx

7. 6

6

55

1ln(6) lnln( 5)) (|dx x

x

8. 1

2 2 1

11

1 1cos(co ) sin(

2 2s( ) ) | 0x x u dud xx

9.

4 4

2 2

2 4 2 2

2

ln( )

| ln(2)

ln( ) 1

1 1ln( ) (ln(4 ))

2 2

xdx dx udu

x x

x

x

10.

320

20210

10

3 3

2 2

2cos( ) 1 (cos( ) 1)

3

2 2(cos(20) 1

sin( )

) (cos(10) 1)3 3

|x dxx udu x

11. Two regions between two parallel lines have the same area if and only if every line parallel

to the two bounding lines intersects the two regions with line segments of equal length.

12. 2 2R y

We can see this with Pythagorean Theorem. The distance from the center of sphere to a point

on the edge of this cross section is of course R . The distance straight down from this point to

the cross section at the center of the sphere is y by definition. Taking these as two legs of a

triangle, the third leg—which is radius we are looking for—is then 2 2R y .

13.

2

2

2

hR

We can see this with Pythagorean Theorem, again: simply apply the result of the previous

theorem to the top of the cylinder, which coincides with the cross section of height 2

h.

14. We integrate the cross section of the original sphere minus the cross section of the sphere. 2

2 22/2 /22 2 2 2

/2 /2(

2)

2

h h

h h

h hR y yR ydy d

Curiously, this does not rely on the radius of the sphere.

15. Volumes with equal sized cross sections corresponding with each other must necessarily be

equal.

Solids of Revolution: Volumes by Disks

Answers

1. 36

2. 117

5

3. 2 2

4. 2

3

5. 3

5

6. 72

5

7. 2

8. 128

5

9. 128

3

10. 128

7

11. 2

12. 1

12 e

13. Sum infinite solid disks of radius 3x+2 and height dx between x=0 and x=4:

4 4 4

2 2 2

0 0 0( ) (3 2) (9 12 4) 304V r x dx x dx x x dx .

14. Sum solid disks of radius y and height dy between y=0 and y=3.

3 3

2 2

0 0( ) ( ) 9V r y dy y dy .

15. Find boundaries of integration by setting22 12 10y x x equal to y=0:

2

2

0 2 12 10

0 6 5

0 ( 5)( 1)

1,5

x x

x x

x x

x

Then find the sum of infinite disks of radius 22 12 10x x and width dx between these

bounds.

2 2 2

4 3

5 5

1 1

5

1

2

( ) ( 2 12 10)

2048(4 48 184 240 100)

15

V r x dx x x dx

x x x x dx

Solids of Revolution: Volumes by Washers

Answers

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14. Find boundaries of integration by setting the curves equal to each other.

2 2

2

2

7 12 2

2 2

3 3

2

1

1

2

x x

x

x

x

On the interval, 2 27 12 2

2 2x x . Integrate over washes with width dx, outer radius

2 72

2x and inner radius 21

22

x :

1 12 2 2 2 2 2

1 1

7 1( ) ( ) ( 2 ) ( 2) 10

2 2R x r x dx x x dx

.

15. Integrate using washers over y-values. Solve 1y x for x: 2 1 , 0y x y . Since the x-

axis is one boundary, y=0 is the lower bound. For the upper bound set 2 1y x equal to

2x : 2 21 112y y y .The inner radius is 4-2=2; the outer radius is

2 2( ) 4 ( 1) 3R y y y . Then we integrate

1 1 12 2 2 2 2 2 4

0 0 0

12 4

0

( ) (3 ) 2 (9 6 ) 4

16(5 6 )

5

R y r dy y dy y dy

y dyy

y

Solids of Revolution: Volumes by Cylindrical Shells

Answers

1. 2

0

162 ( )

3x x dx

2. 2

0

2 [ ( )] 82

xx x dx

3. 3

0

9 32 ( 3 )

5x x dx

4. 2

0

82 (2 )

3y y dy

5. 3

2

0

92 ( )

2y y dy

6. 3

1

12 ( ) 4x dx

x

7. 2

1

202 [(2 1) ( 2 3)]

3x x x dx

. Lower limit of integration is where slant lines intersect.

8. 8

2

0

42 [ ( 10 16)] 160

9x x x dx

9. 1

0

2 (1 )5

y y dy

10. 1

2

0

2 ( )2

y y dy

11. 1

1/3

0

92 (1 )

14y y dy

12. We sum up shells of thickness dy on the interval 0 14y . Solve 3 2y x for x:

2

3

yx

. So the height as a function of radius y is given by

2 14( ) 4

3 3 3

y yh y

on the

interval 2 14y . On the remaining interval the solid is a right circular cylinder of height 4

and radius 2. The length of each shell is ( ) 2 2l y r y . We integrate over 2 14y :

214 14

2 2

14 142 ( ) 2 ( ) 2 144 288

3 3 3 3

y yy dy y dy

and add the volume of the inner right circular, 2 4 4 16hr .

The total volume is 288 16 304 .

13. Here shells have a height y, radius 4-x and thickness dx. In terms of x, height is 1x .

Boundaries of integration are found by the intersection of 1y x with the x-axis ( 1x ) and

by 2x .

Integrate:

2

12 (4 ) 1x x dx

Let u=x-1. Then

12 2 2 2

1 1

/2 3/

1 1

22 (4 ) 1 2 (3 ) 2 (3 ) 2 (3 )u u ux x dx u du u du u duu

8 162

5 5

.

14. Shells have thickness dx, radius x-(-2)=x+2, height 2 2( 1) 4 4 34x x x x .

Boundaries of integration are the intersections of the curves:

2

2

4 1

4 3

( 3)( 1) 0

1,3

4x x

x x

x x

x

Integrate:

23

1

16 322 ( 2)( 4 3) 2

3 3x x x dx

15. Shells have thickness dx, height sin(x), radius x . Boundaries of integration are / 2x

and x . So the integral is:

/22 ( )sin( )x x dx

16. Shells have thickness dx, radius x, height 2 2 2(13 ) ( 8 19) 2 8 6x x x x x . To find

boundaries of integration, set the curves equal to each other to find their points of intersection:

2 2

2

2

13 8 19

2 8 6 0

4 3 0

1, 3

x x x

x x

x x

x x

Then integrate:

23

1

16 322 ( 2 8 6) 2

3 3x x x dx

The Length of a Plane Curve

Answers

1. Yes. The derivative of any polynomial exists:

11

0 0

( ) '( )n n

j j

j j

j j

f x a x f x ja x

Since the derivative is also a polynomial, polynomial are infinitely differentiable.

2. No. While this is a continuous function, it is not differentiable at zero.

3. Yes.

( ) ( ) ( ) '( ) '( ) '( )h x af x bg x h x af x bg x

4. No, because tangent is not defined over the entire real line.

5. 12

6. (17/12)

7. 2

8. 6

9. 2

2

40

41 .

s hL x dx

S

10. The arc length formula is: 21 | ( ) |

b

aL f x dx .

In this case: 1 1

0 0

21 | ( ) | 1 1L f x dx dx

11. We need to take the derivative of the function.

2

2

2 cos( )ln( sin( ))

sin( )

d x x

x xx x

dx

.

In this case: 25

1

2

25

21 | ( ) | 1 ( )sin( )

2 cos( )b

aL f x dx d

x x

xx

x

12. The arc length formula is 21 | ( ) |

b

aL f x dx .

We need to take a derivative: sin( )

ln(cos( )) tan( )cos( )

d xx x

dx x

Now we integrate:

2

1 1

1

2

1sec(1 tan( ) sec( )

) tan(1))

)

ln(sec( ) tan( ) ln(sec(0) tan(0))) | ln(sec(1

x dx

x

L x dx x dx

x

13. The arc length formula is 21 | ( ) |

b

aL f x dx .

We need to take a derivative: 3

22

( 1) 13

dx x

dx

5

0

3 3

52 20

2 21 1 5

3 3|

b

aL x dx xdx x

14. The arc length formula is 21 | ( ) |

b

aL f x dx .

We need to take a derivative: 3 2 2(2 4 1) 6 8 1d

x x x x xdx

.

The function is 3

2 2 2

11 | ( ) | 1 (6 8 1)

b

aL f x dx x x dx

15. The arc length formula is:

2

21 [ ( )] 1 .d d

c c

dxL f y dy dy

dy

2'( ) 1 3f y y .

4

2 2

1

1 [1 3 ]L y dy .

Area of a Surface of Revolution

Answers

1.

2.

3.

4.

5.

6.

7. We can create a sphere of radius by rotating a semicircle of radius around the axis.

The formula for a circle of radius is , and we can solve for to get the

equation for the semicircle: . Then we can also

solve for: and

so . Now we can plug all this into the integral that gives us

the surface area, integrating from :

8. We can create this right circular cone by rotating a line around the axis over the

interval . Two points we know that have to be on the line

are and , which then gives us an equation for the line

of . Then we can determine the derivative of with respect

to , and so . And now we are ready to calculate the integral

that gives the lateral surface area:

9. 5 5 5

0

25

2 2

0 0 01 ( ) 1 (12 2 ) 2 22 22 5

dyy dx x dx xdx x

dx .

10. /2 /2 /2

2 2 2

/4 /4 /41 ( ) cos( ) 1 sin ( ) 2 cos( ) 1 s2 (2 in )

dyy dx x x dx x x dx

dx

11.

22 2 2 2

220 0 0

2 20.52 2

2 2 2 00

0.5 0.5 0.5

0.5 0.

0

5 0

0

.5

1 ( ) 1 1 ( ) 1 111

1 11 1

2 2 2

2 2 21 1 1

2

dy x xy dx x dx x dx

dx xx

x xx dx x dx dx x

x x x

12.

4 4 4 4

1 1 1 1

2

4

1

22 2 21 1

1 ( ) 1 ( ) 14

2

12

42

4

dyy dx x dx x dx x dx

dx x xx

x

x dx

Let1

4u x to solve the integral. Then the surface area is

(17 17 5 5) (17 17 5 5)

12 62

13.

5 5 52 2

0 0 01 ( ) 1 (1) 2 2 2 25 22 22 5

dyx dx x dx

dxxdx

14.

/2 /2 /22 2 2

/4 /4 /41 ( ) 12 2 ( ) ( )sin 2 1 sin

dyx dx x dx xx

dx dx

x

15.

0.5 0.52

2 2

220 0 0

2 2

2 2 2 20 0 0

0.5

0.5 0.5 0.5

1 ( ) 1 ( ) 111

1 1

1 1 1 1

2 2 2

2 2 2

dy x xx dx x dx x dx

dx xx

x x xx dx x dx dx

x x x x

Let 21 , 2u x du xdx . Then

11/2

0.5 0.75

2 0.710 5(

1)2 32

1

xdx du

xu du

u

16.

4 4 4

1 1 1

2 2 2

1

2 2

2 2 2 2 2

1 1

2 2 2

2 2 2

1 11 ( ) 1 ( ) 1

42

1 ( ) 1 (2 ) 1 4

dyx dx x dx x dx

dx xx

dxx dy y y dy y y dy

dy

17.

3 3 3

1 1

2 2 2

11 ( ) 1 (2 ) 1 42 2 2

dyx dx x x dx x x dx

dx

let 2 ,1 84 d xu x u dx . Then

2 1/2 3/2 3/23 37

1 5

2 (37 37 5 5)1 4 (37 5 )

8 6 62 u dx x dx u

Applications: Work and Force

Answers

1. 1.471 N

2. 0 N

3. 105 10 J

4.

a. 5 N/mk

b. 8.1 J

5. 3.2 J

6. 15

0

600 50 18(1 ) 2047550

yW dy

ft-lbs

7. 12

0

6 ) 43.94

yW dy ft-lbs

8. 4

2 2

0

30 (5 2)( 25 ) 12440W y y dy ft-lbs. where y is the height of the fluid

9. 6

2

0

2(6 )30(6 1)(2 ) 5280

6

hW h dh

ft-lbs where h is the height of the fluid.

10. 9 3

2

0

240(9 )( ) 5760

3W h h dh

ft-lbs, where h is the height of the fluid.

11. 3

0

40(4 )( 3)10 108002

hW h dh

ft-lbs, where h is the height of the fluid.

12. 3

0

2(4 ) 9

3W h dh ft-lbs.

13. 10

2 2 10

00[ sin( ) 2] [ cos( ) 2 ] | cos(100) 21x x dx x x

14. 5

2

0cos(sin ( )) 4.04x dx

15. 2

3 2 2 2

00

4 31 1 1)

2

1(2 1) | 8

3(1

3 2x x x dx x x x x

Applications: Fluid Forces

Answers

1. 6(1000)(9.8)(8)(10)(16) 1.2544 10 NF PA ghA

2. 8

6

0

10 3.136 10 NF gx dx where x is the depth of the pool.

3. 4( 5)(0.5 1) 2.45 10 NF g .

4. 5 sin

4

5

12.5725 10 N

sin

l

F gx w dx

, where 1l m is the length of the plate,

0.5w m is the width of the plate, is the angle the plate makes with the bottom of the

pool, and x is the depth of the pool.

5. 5 sin

4

5

12.5113 10 N

sin

l

F gx w dx

, where 0.5l m is the length of the plate,

1w m is the width of the plate, is the angle the plate makes with the bottom of the

pool, and x is the depth of the pool.

6. 4

4

0

1(4 ) 9.6975 10 N

sinF g x w dx

, where the integral covers the additional depth

starting at 4 m, and is the angle the bottom of the pool makes with the horizontal ground

plane (2 2

4sin

16 4

).

7. 50

9

0

(1000)(9.8) 600 7.35 10 NF x dx .

8. 5 2

2 2 2 2

2 5

( 2) 2 5 ( 2) 2 5 163,417 F g y y dy g y y dy N

, where depth

is taken along the negative y-axis, and the . ( 2)y factor in the integrand is the fluid depth.

The integral was determined using a calculator. The problem can be reformulated as follows

with fluid depth at 0y by moving the center of the tank to (0,2):

3

2

0

( ) 2 25 ( 2) 163,417 NF g y y dy

.

9. 2 1

2 2

1 2

9 9( 1) 2 9 ( 1) 2 9 14,808

4 4F g y y dy g y y dy N

, where depth

is taken along the negative y-axis, and the ( 1)y factor in the integrand is the fluid depth.

The integral was determined using a calculator. The problem can be reformulated as follows

with fluid depth at 0y by moving the center of the tank to (0,1):

2

2

1

9( ) 2 9 ( 1) 14,808 N

4F g y y dy

10. Yes. To explain look at how and are mathematically related. The force is proportional

to

w , or the mass density .

11. 1 1

0 0

1(5 ) 4900 (5 ) 22,050 (100 N

20 9.8) y dy y dy

12. 10 6

2 2 5

6 10

2 4 ( 8) 2 4 ( 8) 9.8520 10 F gy y dy gy y dy N

, where

depth is taken along the negative y-axis, and 22 4 ( 8)y dy in the integrand is the strip

of area on the circular plate [center at (0,-8), radius 2] at depth y .

13. 15 14

4

14 15

3 32 (15 ) 2 (15 ) 8.1098 10

3 3F gy y dy gy y dy N

, where

depth is taken along the negative y-axis, and 22 4 ( 8)y dy in the integrand is a strip

of area on the triangular plate at depth y .

14. 18 17

5

17 18

(1) (1) 1.7150 10 F gy dy gy dy N

, where depth is taken along the

negative y-axis, and (1)dy in the integrand is a strip of area on the square plate at depth

y .

15. 2 5

0

4

)](30 )[2 sin( 5.32186 10 Ny y y dg y , where depth is taken along the positive y-axis

(y=0 is at 30 m depth; y=30 is the surface).

Applications: Probability and Probability Density

Functions

Answers

1. 1

14r : since

7

7

1rdx

, so that 7

7( ) 1r x

2. 6

125r : since

5

0

( 5) 1rx x dx , so that

53 2

0

( 5 ) 13 2

x xr

3. 26

1.0425

r : since 25

20

1(1 )

rdx

x

, so that

25

0

11

r

x

4. 1r : since /2

0

sin 1r xdx

, so that /2

0cos 1r x

5. 1

(7)5

g : since 10

0

1( ) 1 10 (7)

2g x dx g

6. Yes. 10

0

4( 0.9) 1

25x dx

7. a. The probability that a randomly chosen light bulb will have a lifetime

between 1000 and 5000 hours.

b. The probability that a randomly chosen light bulb will have a lifetime of at least 3000

8. a. 1

2

1

1 13( 1 1) ( 9) 0.48

36 27P x x dx

9. ( 2.2 2.2) 0.9P x : using 21( ) ( 9) 0.9

36

T

T

P T x T x dx

Solve the equation 3 27 48.6 0T T to find the only value of T in the interval.

10. a. 3

0.125

0

(0 3) 0.125 0.31tP t e dt

b. 0.125

10

(10 ) 0.125 0.29tP t e dt

11. a. 63.4 63.4

2 2 2 2( ) /(2 ) ( 63.4) /(2(3.2) )

0 0

1 1( ) 0.50

2 3.2 2

x xdx dxP x e e

b. 65

2 2( 63.4) /(2(3.2) )

63

1( ) 0.24

3.2 2

xP x dxe

c. 2 2( 63.4) /(2(3.2) )

72

1( ) 0.0036

3.2 2

x dxP x e

d. (72) 0P

12. M=0: 21( 9) 0.5

36

M

x dx

means 3

3

1( 9 ) 0.5

36 3

M

xx

, or 2( 27) 0M M .

13. ln 2

5.5450.125

M : 0.1250.125 0.5M

te dt

means 0.125

00.125 0.5

Mte

.

14. M=0:

34 23

2 2

33

1( 9) ( 9) ( 9 )

36 36 36 4 2

x x x xx dx x dx

15. 1

80.125

Mean : 0.125 0.125

0

0.125 0.125t tx e dt x e dt