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
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