Quiz 4 Practice Problems

Practice problems are similar, both in difficulty and in scope, to the type of

problems you will see on the quiz. Problems marked with a ? are ‘‘for your

entertainment’’ and are not essential.

SUGGESTED REFERENCE MATERIAL:

As you work through the problems listed below, you should reference sections

6.1 and 6.2 of the recommended textbook (or the equivalent section in your

alternative textbook/online resource) and your lecture notes.

EXPECTED SKILLS:

• Be able to find the area between the graphs of two functions over an

interval of interest.

• Know how to find the area enclosed by two graphs which intersect.

• Know how to use the method of disks and washers to find the volume

of a solid of revolution formed by revolving a region in the xy-plane

about the x-axis, the y-axis, or any other horizontal or vertical line.

1

PRACTICE PROBLEMS:

1. Let R be the shaded region shown below.

(a) Set up but do not evaluate an integral (or integrals) in terms of x

that represent(s) the area of R.

(b) Set up but do not evaluate an integral (or integrals) in terms of y

that represent(s) the area of R.

For problems 2-4, compute the area of the shaded region.

2.

2

3.

4.

For problems 5-13, compute the area of the region which is enclosed

by the given curves.

5. y = 4x, y = 6x2

6. y = 2x2, y = x2 + 2

7. y = x2/3, y = x4, in the first quadrant

8. y =1

x, y =

1

x2, x = 4

3

9. y = sinx, y = 2− sinx,π

2≤ x ≤ 5π

2

10. y = e5x, y = e8x, x = 1

11. x = 4− y2, x = y2 − 4

12. y = x4, y = |x|

13. y = x2, y =2

x2 + 1

14. Let R be the region enclosed by y = x, y = 8x, and y = 4.

(a) Compute the area of R by evaluating an integral (or integrals)

in terms of x.

(b) Compute the area of R by evaluating an integral (or integrals)

in terms of y.

15. Use an integral (or integrals) to compute the area of the triangle in the

xy-plane which has vertices (0, 0), (2, 3), and (–1, 6).

16. Consider the 2D ice cream cone topped with a delicious scoop of ice

cream that is enclosed by y = 6|x| and y = 16− x2.

(a) Compute the area enclosed within the ice cream cone (including

the scoop portion).

(b) After a bite is taken from the top, the remaining area is enclosed

by y = 6|x|, y = 16 − x2, and y = x2 + 12. Compute the area of

the remaining portion.

17. For each of the following, set up but do not evaluate an integral (or

integrals) which represent(s) the volume of the solid that results from

revolving the shaded region around the indicated axis

4

(a) Around the x-axis:

(b) Around the x-axis:

(c) Around the y-axis:

(d) Around the y-axis:

5

For problems 18-23, compute the volume of the solid that results

from revolving the region enclosed by the given curves around the

x-axis.

18. y =√4− x2, and y = 0.

19. y =√x− 1, x = 5, and the x-axis

20. y = ex, y = 1, and x = ln 5

21. y = secx, y =1

2, x = –

π

4, and x =

π

4

22. y =√tanx, x =

π

3, and the x-axis.

23. y =1√

x2 + 9, x = –3, x = 3, and the x-axis.

24. A solid called a torus is formed by revolving the circle x2 + (y− 2)2 = 1

around the x-axis.

(a) Set up an integral which represents the volume of this solid.

6

(b) Now compute the volume of the torus by evaluating your integral

from part (a). [Hint : Consider using an appropriate formula from

geometry at some point during your calculation.]

For problems 25-27, compute the volume of the solid that results

from revolving the region enclosed by the given curves around the

y-axis.

25. x = y2 and y = 2x

26. y = x2 and y = 2x+ 3, in the first quadrant

27. y = lnx, x = e, and y = 0

28. By revolving an appropriate region around the x-axis, show that the

volume of a sphere with a radius of R is V = 43πR3.

29. Consider the right circular cone with a height of h and a base-radius of

R, shown below.

By revolving an appropriate region around the y-axis, show that the

volume is V =1

3πR2h.

30. Let V be the volume of the solid which results from revolving the region

enclosed by y = x3, x = 0, and y = k (k > 0) around the y-axis. Find

the value of k such that V = 9π.

7

31. Let R be the region enclosed by y = x3, x = 2, and y = 0. For each

of the following, set up but do not evaluate the integral (or integrals)

which represent the volume of the solid which results from revolving R

around the indicated line.

(a) Revolved around y = –1

(b) Revolved around y = 8

(c) Revolved around x = –2

(d) Revolved around x = 2

32. Let R be the region enclosed by y = x and y =√x. For each of the

following, set up but do not evaluate the integral (or integrals) which

represent the volume of the solid which results from revolving R around

the indicated line.

(a) Revolved around y = 1

(b) Revolved around y = –1

(c) Revolved around x = 1

(d) Revolved around x = –1

? Let O be the origin. The horizontal line y = c intersects the curve y = 2x−x3

at P and Q in the first quadrant and cuts the y-axis at R. Find c so that the

region OPR bounded by the y-axis, the line y = c, and the curve has the

same area as the region between P and Q under the curve and above the line

y = c.

? Let R be the region {(x, y) : 0 ≤ x ≤ 1, 3x − x − 1 ≤ y ≤ x}. Find the

volume of the solid obtained by rotating R around the line y = x.

8

SOLUTIONS

1. (a) The top curve is y = 5− x2 and the bottom curve is y =1

2x. We

are given the left intersection point with x = –5/2. The second

one can be attained by setting 5− x2 = x/2 so that

10− 2x2 = x⇒ 2x2 + x− 10 = 0⇒ (2x+ 5)(x− 2) = 0,

which yields x = 2. Hence

area of R =

∫ 2

–5/4

(5− x2 − 1

2x

)dx.

(b) First rewrite the equations of the curves so that they are functions

of y. The line becomes x = 2y, and solving for x in y = 5−x2 givesx =√5− y to the right of the y-axis and x = –

√5− y to the left.

We are given the bottom intersection point with y = –5/4. The

higher one we found at an x-coordinate of 2. Plugging that into

either one of the curves yield y = 1. The integral must be split

into 2 pieces however since the rightmost curve changes from a

line to a parabola whose vertex is at (0, 5) and so

area of R =

∫ 1

–5/4

(2y − (–√

5− y))dy +∫ 5

1

(√

5− y − (–√5− y))dy

=

∫ 1

–5/4

(2y +√

5− y)dy +∫ 5

1

2√

5− y dy.

2. The area is simply

∫ 1

–2

(x2 + 2− (–x)) dx =

∫ 1

–2

(x2 + x+ 2) dx

9

=

[x3

3+x2

2+ 2x

]1–2

=

(1

3+

1

2+ 2

)−(–8

3+ 2− 4

)=

15

2

3. The top and bottom curves alternate at the points of intersection of

x = π/4 and x = 5π/4. Hence

area =

∫ π/4

0

(cosx− sinx) dx+

∫ 5π/4

π/4

(sinx− cosx) dx+

∫ 2π

5π/4

(cosx− sinx) dx

= [sinx+ cosx]π/40 + [– cosx− sinx]

5π/4π/4 + [sinx+ cosx]2π5π/4

=

(1√2+

1√2

)− (0 + 1) +

(1√2+

1√2

)−(–

1√2− 1√

2

)+ (0 + 1)−

(–

1√2− 1√

2

)=

8√2= 4√2.

4. Since y = x+1 and x = y2 are both functions of y, it is easier to integrate

in terms of y. Note that the equation of the line can be rewritten as

x = y − 1, and so the area is

∫ 2

–2

(y2 − (y − 1)) dy =

∫ 2

–2

(y2 − y + 1) dy

=

[y3

3− y2

2+ y

]2–2

=

(8

3− 4

2+ 2

)−(–8

3− 4

2− 2

)=

28

3.

5. Note that y = 4x and y = 6x2 intersect when 4x = 6x2 ⇒ 6x2 − 4x =

0⇒ 2x(3x−2) = 0 so that x = 0 or x = 2/3. As the parabola eventually

dominates the line, y = 4x is the curve on top on [0, 2/3].

10

0.1 0.2 0.3 0.4 0.5 0.6

0.5

1

1.5

2

2.5

The area is thus given by

∫ 2/3

0

(4x− 6x2) dx = [2x2 − 2x3]2/30 =

[2

(4

9

)− 2

(8

27

)]=

8

27.

6. Note that y = 2x2 and y = x2 + 2 intersect when 2x2 = x2 + 2⇒ x2 =

2 ⇒ ±√2. At x = 0, 2(0)2 = 0 < 02 + 2 = 2 so that x2 + 2 ≥ 2x2 on

[–√2,√2].

−1 −0.5 0.5 1

1

2

3

4

The area is thus given by

∫ √2–√2

(x2+2−2x2) d = 2

∫ √20

(2−x2) dx = 2

[2x− x3

3

]√20

= 2

(2√2−

(√2)3

3

)=

8√2

3.

11

7. Note that y = x2/3 and y = x4 intersect when x2/3 = x4. We can

eliminate the fractional exponent by raising both sides to the 3/2-power

so that x = (x4)3/2 = x6. Algebraically, this becomes

x = x6 ⇒ x6−x = 0⇒ x(x5−1) = 0⇒ x(x−1)(x4+x3+x2+x+1) = 0.

Because we only care about the first quadrant, so nonnegative x, clearly

x4+x3+x2+x+1 can be ignored since this factor is positive, and thus

nonzero, whenever x > 0. So the only positive points of intersection are

at x = 0 and x = 1. As 2/3 < 4, it follows that x2/3 ≥ x4 on [0, 1].

0.2 0.4 0.6 0.8

0.2

0.4

0.6

0.8

1

The area is thus given by

∫ 1

0

(x2/3 − x4) dx =

[3

5x5/3 − x5

5

]10

=3

5− 1

5=

2

5.

8. Note that y = 1/x and y = 1/x2 intersect when 1/x = 1/x2 ⇒ x = x2

so long as x 6= 0. In this case, we can cancel an x on both sides of the

equation to get that 1 = x. So our region lies between x = 1 and x = 4.

Note that if x > 1, then x2 > x so that 1/x2 < 1/x.

12

1.5 2 2.5 3 3.5 4

0.2

0.4

0.6

0.8

1

The area is thus given by

∫ 4

1

(1

x− 1

x2

)dx =

[ln|x|+ 1

x

]41

=

(ln 4 +

1

4

)− (ln 1+ 1) = ln 4− 3

4.

9. Note that y = sinx and y = 2− sinx intersect when sinx = 2− sinx,

which is when 2 sinx = 2⇒ sinx = 1. The first two points x where this

occurs is at x = π/2 and x = π/2 + 2π = 5π/2. (How fortunate!) To

determine which curve is on top, let us pick an easy point in [π/2, 5π/2]

to test. Trying x = 2π, we recall that sin 2π = 0 whereas 2− sin 2π = 2.

2 3 4 5 6 7

1

2

The area is thus given by

∫ 5π/2

π/2

(2−sinx−sinx) dx =

∫ 5π/2

π/2

(2−sinx) dx = [2x+cosx]5π/2π/2 = (5π+0)−(π+0) = 4π.

13

10. Note that y = e5x and y = e8x intersect when e5x = e8x ⇒ 5x = 8x⇒x = 0. Since 5x < 8x for x > 0, we have e5x ≤ e8x for x in [0, 1].

0.2 0.4 0.6 0.8

500

1,000

1,500

2,000

2,500

The area is thus given by

∫ 1

0

(e8x−e5x) dx =

[1

8e8x − 1

5e5x]10

=

(1

8e8 − 1

5e5)−(1

8− 1

5

)=

1

8e8−1

5e5+

3

40.

11. Note that x = 4− y2 and x = y2 − 4 are functions of y this time, and

they intersect when 4− y2 = y2− 4⇒ 2y2 = 8⇒ y2 = 4⇒ y = ±2. Todetermine which curve is rightmost, let us pick an easy point in [–2, 2]

to test. Trying y = 0, we have that 4− 02 = 4 whereas 02 − 4 = –4.

y

x

−2

2

−4 4

The area is thus given by

∫ 2

–2

(4− y2 − (y2 − 4)) dy =

∫ 2

–2

(8− 2y2) dy

14

= 2

∫ 2

0

(8− 2y2) dy

= 2

[8y − 2

3y3]20

= 2

[16− 2

3(8)

]=

64

3.

12. Note that the graphs of y = x4 and y = |x| are symmetric about the

y-axis, so we only need to consider the behavior of the two curves in

the first quadrant, in which case |x| = x. These two curves intersect

in the first quadrant when x4 = x ⇒ x4 − x = 0 ⇒ x(x3 − 1) = 0 ⇒x(x− 1)(x2 + x+ 1) = 0. So the two real roots are x = 0 and x = 1.

Observe that x ≥ x4 for x in [0, 1].

−1 −0.5 0.5

0.2

0.4

0.6

0.8

1

The area is thus given by

∫ 1

–1

(|x| − x4) dx = 2

∫ 1

0

(x− x4) dx = 2

[x2

2− x5

5

]10

= 2

(1

2− 1

5

)=

3

5.

13. Note that y = x2 and y =2

x2 + 1intersect when x2 =

2

x2 + 1⇒

x4 + x2 = 2 ⇒ x4 + x2 − 2 = 0. This equation is quadratic in form.

Letting u = x2 changes it to u2+u−2 = 0⇒ (u+2)(u−1) = 0⇒ u = –2

or u = 1⇒ x2 = –2 or x2 = 1. As the square of a real number is positive,

15

we can ignore x2 = –2. So x2 = 1⇒ x = ±1. To determine which curve

is on top, let us pick an easy point in [–1, 1] to test. Trying x = 0, we

have that 02 = 0 whereas 2/(02 + 1) = 2.

−1 −0.5 0.5

0.5

1

1.5

The area is thus given by

∫ 1

–1

(2

x2 + 1− x2

)dx = 2

∫ 1

0

(2

x2 + 1− x2

)dx

= 2

[2 tan–1 x− x3

3

]10

= 2

(2 · π

4− 1

3

)= π − 2

3.

14. Note that y = x and y = 8x intersect when x = 8x⇒ 7x = 0⇒ x = 0.

Clearly y = 8x is on top. The upper bound of y = 4 gives the following.

y

x

4

4

(a) The line up top is y = 8x until x = 1/2 where it becomes y = 4.

This gives the area of R as

16

∫ 1/2

0

(8x− x) dx+∫ 4

1/2

(4− x) dx =

∫ 1/2

0

7x dx+

[4x− x2

2

]41/2

=7x2

2

∣∣∣∣1/20

+

(16− 16

2

)−(2− 1/4

2

)=

7(1/4)

2+ 49/8 = 7.

(b) Both lines are functions of y. Indeed, rewrite them as x = y/8 and

x = y, the rightmost of the two. This gives the area of R as

∫ 4

0

(x− x

8

)dx =

7

8

∫ 4

0

x dx =7

8· x

2

2

∣∣∣∣40

=7

8· 162

= 7.

15. Here is the triangle below.

y

x

The equation of the line connecting (0, 0) and (2, 3) is y = 32x. The

equation of the line connecting (2, 3) and (–1, 6) is y = 6−3–1−2(x−2)+3 =

–x+ 5. The equation of the line connecting (–1, 6) and (0, 0) is y = –6x.

We can now integrate to find the area, which is

∫ 0

–1

(–x+ 5− (–6x)) dx+

∫ 2

0

(–x+ 5− 3

2x

)dx =

∫ 0

–1

(5x+ 5) dx+

∫ 2

0

(5− 5

2x

)dx

17

=

[5x2

2+ 5x

]0–1

+

[5x− 5

4x2]20

= −(5

2− 5

)+

(10− 5

4· 4)

=15

2.

16. Notice that y = 6|x| and y = 16−x2 are both even functions (symmetric

about the y-axis), so finding points of intersection in the first quadrant

suffices. So 6x = 16 − x2 ⇒ x2 + 6x − 16 = 0 ⇒ (x + 8)(x − 2) = 0.

Since we only care about the positive root, we have that the curves

intersect at x = 2. They thus also intersect at x = –2. To determine

which curve is on top, let us pick an easy point in [–2, 2]. Trying x = 0,

we have that 6|0| = 0 whereas 16− 02 = 16.

−2 −1 1 2

5

10

15

(a) The area is, due to symmetry, thus

∫ 2

–2

(16− x2 − 6|x|) dx = 2

∫ 2

0

(16− x2 − 6x) dx

= 2

[16x− x3

3− 3x2

]20

= 2

[32− 8

3− 3(4)

]=

104

3.

(b) The curve y = x2 + 12 intersects y = 16 − x2 when x2 + 12 =

16 − x2 ⇒ 2x2 = 4 ⇒ x2 = 2 ⇒ x = ±√2, which is certainly

between –2 and 2, and so y = x2 + 12 cannot intersect y = 6|x| byIVT. It must lie above it. Here is our new region:

18

y

x

14

−2 2

The area is, due to symmetry, thus

2

∫ √20

(x2 + 12− 6x) dx+ 2

∫ 2

√2

(16− x2 − 6x) dx

= 2

[x3

3+ 12x− 3x2

]√20

+ 2

[16x− x3

3− 3x2

]2√2

= 2

[23/2

3+ 12√2− 3(2)

]+2

[32− 8

3− 3(4)

]−2[16√2− 23/2

3− 3(2)

]= 2

[2

3

√2 + 12

√2

]+

104

3− 2

[16√2− 2

3

√2

]=

104

3− 16

3

√2.

17. (a) The region is touching the axis of revolution, so we use disk method.

Here r(x) =1

2x2 + 1, and so the volume is

∫ 2

0

A(x) dx =

∫ 2

0

π[r(x)]2 dx = π

∫ 2

0

(1

2x2 + 1

)2

dx.

(b) The region is not touching the axis of revolution, so we use washer

method. Here r1(x) = 3− x2 and r2(x) = 2x, and so the volume is

∫ 1

0

A(x) dx =

∫ 1

0

π([r1(x)]2−[r2(x)]2) dx = π

∫ 1

0

[(3−x2)2−(2x)2] dx.

19

(c) The region is not touching the axis of revolution, so we use washer

method. Since the axis is vertical, we integrate with respect to

y and so rewrite y = 1/x2 as a function of y. Doing so yields

x = 1/√y. Also note that y = 1/x2 intersects the line x = 4 when

y = 1/42 = 1/16. Hence r1(y) = 4 and r2(y) = 1/√y, and so the

volume is

∫ 4

1/16

A(y) dy =

∫ 4

1/16

π([r1(y)]2−[r2(y)]2) dy = π

∫ 4

1/16

(16− 1

y

)dy.

(d) The region is touching the axis of revolution, so we use disk

method. Since the axis is vertical, we integrate with respect to

y and so rewrite y = 2x and y = 3− x2 as functions of y. Doing

so yields x = y/2 and x =√3− y, respectively. Also note that

the volume must be written as the sum of two integrals since the

radius function r(x) changes at y = 2. Thus the volume is

π

∫ 2

0

y2

4dy + π

∫ 3

2

(3− y) dy.

18. Here is a sketch of the region

y

x

2

2 2

y =√4− x2

The region is touching the axis of revolution, so we use the disk method.

Here r(x) =√4− x2, and so the volume is

20

∫ 2

–2

A(x) dx = 2

∫ 2

0

π[r(x)]2 dx = 2π

∫ 2

0

(4−x2) dx = 2π

[4x− x3

3

]20

= 2π

(8− 8

3

)=

32

3π.

19. Here is a sketch of the region

y

x

2

1 5

y =√x− 1

The region is touching the axis of revolution, so we use disk method.

Here r(x) =√x− 1, and so the volume is

∫ 5

1

A(x)dx =

∫ 5

1

π[r(x)]2 dx

= π

∫ 5

1

(x− 1) dx

= π

[x2

2− x]51

= π

[(25

2− 5

)−(1

2− 1

)]= 8π.

20. Here is a sketch of the region

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

2

3

4

21

The region is not touching the axis of revolution, so we use washer

method. Here r2(x) = ex and r1(x) = 1, and so the volume is

∫ ln 5

0

A(x) dx =

∫ ln 5

0

π([r2(x)]2 − [r1(x)]

2) dx

= π

∫ ln 5

0

(e2x − 1) dx

= π

[1

2e2x − x

]ln 5

0

= π

[(52

2− ln 5

)− 1

2

)= (12− ln 5)π.

21. Here is a sketch of the region

−0.6−0.4−0.2 0.2 0.4 0.6

0.6

0.8

1

1.2

1.4

The region is not touching the axis of revolution, so we use washer

method. Here r2(x) = sec x and r1(x) = 1/2, and so the volume is

∫ π/4

–π/4

A(x) dx =

∫ π/4

–π/4

π([r2(x)]2 − [r1(x)]

2) dx

= π

∫ π/4

–π/4

(sec2x− 1

4

)dx

= 2π

∫ π/4

0

(sec2x− 1

4

)dx (by symmetry)

22

= 2π

[tanx− 1

4x

]π/40

= 2π

(tan

π

4− 1

4· π4

)= 2π − 2π2

8.

22. This is a tough one. Chances are that you do not know what the graph

of y =√tanx looks like. However, one can see that

√tanx = 0 when

x = 0 and that it is an increasing function on [0, π/3] since f(x) = tanx

is itself increasing. So the region is simply what is under the curve

y =√tanx in the first quadrant over [0, π/3]. Such a region demands

that we use disk method with r(x) =√tanx. So the volume is

∫ π/3

0

A(x) dx =

∫ π/3

0

π[r(x)]2 dx = π

∫ π/3

0

tanx dx = πln|secx|

∣∣∣∣∣π/3

0

= π ln 2.

(For those who curious, the sketch of the region is below.)

y

x

4√3

π

3

y =√tanx

23. Here is a sketch of the region

−3 −2 −1 1 2 3

0.1

0.2

0.3

23

The region is touching the axis of revolution, so we use disk method.

Here r(x) =1√

x2 + 9, and so the volume is

∫ 3

–3

A(x) dx =

∫ 3

–3

π[r(x)]2 dx

= π

∫ 3

–3

1

x2 + 9dx

= 2π

∫ 3

0

1

x2 + 9dx (by symmetry)

= 2π

∫ 3

0

1

9

1

x2/9 + 1dx

=2

9π

∫ 3

0

1

(x/3)2 + 1dx

=2

3π tan–1

(x3

)∣∣∣∣30

=2

3π tan–11 =

π2

6.

24. First note that x2+(y−2)2 = 1⇒ (y−2)2 = 1−x2 ⇒ |y−2| =√1− x2,

and so the interior of the circle can be instead interpreted as the region

bounded by y = 2 +√1− x2 and y = 2−

√1− x2.

(a) Since the circle does not touch the axis of revolution, we use washer

method. Here r2(x) = 2 +√1− x2 and r1(x) = 2−

√1− x2, and

so the volume is

∫ 1

–1

A(x) dx = 2

∫ 1

0

π([r2(x)]2 − [r1(x)]

2) dx

= 2π

∫ 1

0

[(2 +√1− x2

)2−(2−√1− x2

)2]dx

= 2π

∫ 1

0

[(4 + 4

√1− x2 + (1− x2)

)−(4− 4

√1− x2 + (1− x2)

)]dx

24

= 2π

∫ 1

0

8√1− x2 dx = 16π

∫ 1

0

√1− x2 dx

(b) Recall that the graph of y =√1− x2 over [0, 1] is the quarter of

the unit circle lying in the first quadrant, and so the integral from

part (a), and thus the volume of the torus, is

16π · π4= 4π2.

25. Note that x = y2 and y = 2x intersect when

x = (2x)2 = 4x2 ⇒ 4x2 − x = 0⇒ x(4x− 1) = 0⇒ x = 0 or x = 1/4.

When x = 0, y = 2(0) = 0, and when x = 1/4, y = 2(1/4) = 1/2. Here

is a sketch of the region

5 · 10−2 0.1 0.15 0.2

0.1

0.2

0.3

0.4

0.5

Since the axis is vertical, we integrate with respect to y. Since the region

is not touching the axis of revolution, we use washer method. Here

r1(y) = y/2 and r2(y) = y2, and so the volume is

∫ 1/2

0

A(y) dy =

∫ 1/2

0

π([r1(y)]2 − [r2(y)]

2) dy

25

= π

∫ 1/2

0

(y2

4− y4

)dy

= π

[y3

12− y5

5

]1/20

= π

(1/8

12− 1/32

5

)=

π

240.

26. Note that y = x2 and y = 2x+ 3 intersect when

x2 = 2x+3⇒ x2−2x−3 = 0⇒ (x−3)(x+1) = 0⇒ x = 3 or x = –1.

Since we are only interested in the first quadrant, we disregard x = –1.

When x = 3, y = 32 = 9. Here is a sketch of the region

0.5 1 1.5 2 2.5 3

2

4

6

8

Observe that the region touches the axis of revolution only from y = 0

until y = 3. This suggests that the volume is the sum of two integrals,

one obtained from disk method and the other from washer method.

Since the axis is vertical, we integrate with respect to y and so rewrite

y = x2 and y = 2x+ 3 as functions of y. Doing so yields x =√y and

x = y/2− 3/2, respectively. The volume is thus

π

∫ 3

0

(√y)2 dy + π

∫ 9

3

[(√y)2 −

(y

2− 3

2

)2]dy = π

∫ 3

0

y dy + π

∫ 9

3

(–y2

4+

5y

2− 9

4

)dy

26

= πy2

2

∣∣∣∣30

+ π

[–y3

12+

5y2

4− 9y

4

]93

=45π

2

27. Here is a sketch of the region

y

x

1

1 e

y = lnx

The region does not touch the axis of revolution, so we use washer

method. Since the axis is vertical, we integrate with respect to y and so

rewrite y = lnx as a function of y. Doing so yields x = ey. So r1(y) = e

and r2(y) = ey, and so the volume is

∫ 1

0

A(y) dy =

∫ 1

0

π([r1(y)]2 − [r2(y)]

2) dy

= π

∫ 1

0

(e2 − e2y) dy

= π

[e2x− 1

2e2y]10

= π

[(e2 − 1

2e2)+

1

2

]=πe2

2+π

2.

28. Let us revolve the semicircular region bounded by y =√R2 − x2 and

the x-axis around the x-axis. The region touches the axis of revolution,

so we use disk method. The volume is thus

∫ R

–R

A(x) dx = 2

∫ R

0

π[r(x)]2 dx = 2π

∫ R

0

(R2−x2) dx = 2π

[R2x− x3

3

]R0

=4

3πR3.

27

29. Let us revolve the triangular region bounded by the y-axis, y = R, and

the line through (0, 0) and (R, h). The equation of this line is y = hRx.

The region touches the axis of revolution, so we use disk method. Since

the axis is vertical, we integrate with respect to y and so rewrite y = hRx

as a function of y. Doing so yields x = Rhy. The volume is thus

∫ h

0

A(y) dy =

∫ h

0

π[r(y)]2 dy

= π

∫ h

0

R2y2

h2dy

= πR2y3

3h2

∣∣∣∣h0

= πR2h3

3h2=

1

3πR2h.

30. By design, the region touches the axis of revolution, and so we use disk

method. Since the axis is vertical, we integrate with respect to y and so

rewrite y = x3 as a function of y. Doing so yields x = y1/3. The volume

is thus

∫ k

0

A(y) dy =

∫ k

0

π[r(y)]2 dy

= π

∫ k

0

y2/3 dy

=3π

5y5/3

∣∣∣∣k0

=3π

5k5/3.

Because we want this to equal 9π, we see that

3π

5k5/3 = 9π ⇒ k5/3 = 15⇒ k = 153/5.

28

31. Here is a sketch of the region

y

x

8

2

y = x3

Note also that the curve y = x3 can be rewritten as x = y1/3.

(a) The region does not touch the axis of revolution, and so we use

washer method. Since the axis is horizontal, we integrate with

respect to x. Here r1(x) = x3+1 and r2(x) = 1, and so the volume

is

∫ 2

0

A(x) dx =

∫ 2

0

π([r1(x)]2−[r2(x)]2) dx = π

∫ 2

0

[(x3+1)2−1] dx.

(b) The region does not touch the axis of revolution, and so we use

washer method. Since the axis is horizontal, we integrate with

respect to x. Here r1(x) = 8 and r2(x) = 8−x3, and so the volume

is

29

∫ 2

0

A(x) dx =

∫ 2

0

π([r1(x)]2−[r2(x)]2) dx = π

∫ 2

0

[64−(8−x3)2] dx.

(c) The region does not touch the axis of revolution, and so we use

washer method. Since the axis is vertical, we integrate with respect

to y. Here r1(y) = 4 and r2(x) = y1/3 + 2, and so the volume is

∫ 8

0

A(y) dy =

∫ 8

0

π([r1(y)]2−[r2(y)]2) dy = π

∫ 8

0

[16−(y1/3+2)2] dy.

(d) The region touches the axis of revolution, and so we use disk

method. Since the axis is vertical, we integrate with respect to y.

Here r(y) = 2− y1/3, and so the volume is

∫ 8

0

A(y) dy =

∫ 8

0

π[r(y)]2 dy = π

∫ 8

0

(2− y1/3)2 dy.

32. Note that y = x and y =√x intersect when x =

√x ⇒ x2 = x ⇒

x2 − x = 0 ⇒ x(x − 1) = 0 so that the intersection points are x = 0

and x = 1. Also note that x ≤√x for x in [0, 1]. Here is a sketch of the

region

0.2 0.4 0.6 0.8

0.2

0.4

0.6

0.8

1

Note also that the curves y = x and y =√x can be rewritten as x = y

and x = y2, respectively. In each part below, the region does not touch

the axis of revolution, so each uses washer method.

30

(a) Since the axis is horizontal, we integrate with respect to x. Here

r2(x) = 1− x and r1(x) = 1−√x, and so the volume is

∫ 1

0

A(x) dx =

∫ 1

0

π([r2(x)]2−[r1(x)]2) dx = π

∫ 1

0

[(1− x)2 −

(1−√x)2]

dx.

(b) Since the axis is horizontal, we integrate with respect to x. Here

r2(x) =√x+ 1 and r1(x) = x+ 1, and so the volume is

∫ 1

0

A(x) dx =

∫ 1

0

π([r2(x)]2−[r1(x)]2) dx = π

∫ 1

0

[(√x+ 1

)2 − (x+ 1)2]dx.

(c) Since the axis is vertical, we integrate with respect to y. Here

r2(y) = 1− y2 and r1(y) = 1− y, and so the volume is

∫ 1

0

A(y) dy =

∫ 1

0

π([r2(y)]2−[r1(y)]2) dy = π

∫ 1

0

[(1−y2)2−(1−y)2] dy.

(d) Since the axis is vertical, we integrate with respect to y. Here

r2(y) = y + 1 and r1(y) = y2 + 1, and so the volume is

∫ 1

0

A(y) dy =

∫ 1

0

π([r2(y)]2−[r1(y)]2) dy = π

∫ 1

0

[(y+1)2−(y2+1)2] dy.

31