**Unformatted text preview: **MATH 27 LECTURE GUIDE
UNIT 3. APPLICATIONS OF THE DEFINITE INTEGRAL
The definite integral has numerous applications especially in the fields of physics,
engineering, and probability and statistics. In this unit, we discuss some applications of the definite
integral. In particular, we deal with problems involving certain geometric quantities such as the area of
a region, the volume and surface area of a solid, the length of an arc, and the center of mass of a
region and of a solid.
Our goals for this unit are as follows. By the end of the unit, you should be able to solve for the area of a plane region;
solve for the volume of the solids of revolution using disk and cylindrical
shell methods;
identify when to use disk or cylindrical shell method;
find centroids of plane area and solids of revolution;
find the length of arc of a curve; and
solve for the area of surfaces of revolution. Note: It is expected that you already know how to evaluate definite integrals. Most examples and
exercise problems will only require setting up the definite integral that will solve for the measure of the
object required. In case you forgot, this is how definite integrals are evaluated:
RECALL: = + , then = − . MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) If is continuous on [, ] and 1 3.1 Area of a Plane Region
Using vertical strips, Using horizontal strips,
= = = = Area: Area: MUST REMEMBER!!!
To get area of a plane region, integrate area of a strip. Use the length of the strip for the length
and or for the width.
Length − ( − ) − ( − ) Vertical strip:
Horizontal strip: Width ILLUSTRATION:
1. Determine the area of the region bounded by = and = ,
Solution: When ≤ ≤ , = lies above = . Using vertical strips, the area of the described
region is given by ≤≤ . = = ∫ ( − ) =∫ − = =( √ √
√ √
+ ) − (−
− ) = √ square units. Note: Using horizontal strips to solve the problem would be difficult. (Why?) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) ⁄
= (− − | ⁄ 2 ILLUSTRATION:
2. Using horizontal strips, set-up the definite integral that will give the area of the region
bounded by = and = , ≤ ≤ .
Solution:
When ≤ ≤ , = (which is equivalent to = ) lies on the right of = (which is
equivalent to = ). Using horizontal strips,
the area of the described region is given by = = = = ∫ ( − ) = = − + TO DO:
1. Using vertical strips, set-up the definite integral that will give the area of the region above.
(Hint: Split the region into two. Solve for the points of intersection of = and = and
of = and = to get the proper bounds of the integral.) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. Evaluate the integrals in the previous item and in + . Compare the results. 3 TRY THIS!
Use definite integral to determine the area of the following regions. Use (a) vertical strips and
(b) horizontal strips.
1.
2.
3.
4.
5.
6.
7. the region bounded by = + and = , ≤ ≤ the region bounded by = √ and = the region bounded by = − and = the region bounded by = − and = − the region bounded by = − + and = − + + the region bounded by = and = , ≤ ≤ the region bounded by = and = − , ≤ ≤ For more exercises, you can refer to:
Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 442-445
James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte.
Ltd., pp. 349-350
For an online tutorial, follow these links:
MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 4 3.2 Volume of a Solid of Revolution
Volume of a cylinder = (area of a cross section) x (height)
Consider the solid on the right.
Let ∈ , and be the area of crosssection of the object at .
Let ∆ be a certain thickness of the cross-section
at . (This cross-section is perpendicular to the −axis.)
Volume of a slice =
Appoximate volume of the solid =
(using several slices)
Volume of solid =
(as a Riemann sum) Volume of solid =
(as a definite integral)
where is an area of a cross-section. If the cross-sections that will be used are perpendicular to the −axis, the volume of a solid will be given by = ⋅ , where is an area of a cross-section and is the thickness of a
slice.
TO DO:
Consider the solid formed by revolving the
region bounded by = √, the −axis, and the line = about the −axis. To get the volume of a solid of revolution using DISKS,
consider a strip of the region that is perpendicular to the
axis of revolution. (Revolving this strip generates a disk.)
If the strips are vertical, = ⋅ where is an area of a cross-section. Hence, = [ ] ⋅ where is the radius of a cross-section. = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) A solid of revolution is generated by revolving a plane region about an axis of revolution that is
either tangent to the region or does not pass through the region. The resulting solid has a circular
cross-section: an area of ⋅ with the center on the axis of revolution. 5 If the strips are horizontal, = [ ] ⋅ where is the radius of a cross-section. ILLUSTRATION
Let be the region in the first quadrant bounded by = , the −axis, and the −axis. Using disks, set-up the definite integral that will solve for the volume of the solid
generated by revolving about (a) the −axis and (b) the −axis.
Solution:
a. Since the axis of revolution is the −axis, we
need to use vertical strips. For a vertical strip,
the radius of the corresponding cross-section
is given by = . Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ [ ] ⋅ = ∫ b. Since the axis of revolution is the −axis, we
need to use horizontal strips. Note that for the
described region, ≤ ≤ . Hence, the
function = is defined. Now, for a
horizontal
strip,
the
radius
of
the
corresponding cross-section is given by = . Thus, the volume of the
described solid is = ∫ [ ] ⋅ = ∫ TO DO: a. the line = −
=
b. the line = = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) Let be the region bounded by = and
the lines = and = −. Using disks, set-up the
definite integral that will solve for the volume of the
solid generated by revolving about 6 To get the volume of a solid of revolution (with a cavity) using WASHERS, consider a strip of the
region that is perpendicular to the axis of revolution. (Revolving this strip generates a disk.) If the strips are vertical, If the strips are horizontal, = ∫ [( ) − ( ) ] ⋅ = ∫ [( ) − ( ) ] ⋅ where is the outer radius of a crosssection and is the inner radius of a
cross-section. where is the outer radius of a crosssection and is the inner radius of a
cross-section. ILLUSTRATION
Let be the region bounded by = − + and = − + . Using washers, set-up the
definite integral that will solve for the volume of the solid generated by revolving about (a) the −axis and (b) the line = −.
Solution:
a. Since the axis of revolution is the −axis,
we need to use horizontal strips. For a
horizontal strip, we need to determine the
inner and outer radii of the corresponding
cross-section. Note that = − + is
equivalent to = ±√ − , where ≤ ≤
. However, for the described solid, ≤ ≤
. So, we choose = √ − , where ≤ ≤ . Hence, the outer radius of a crosssection is given by = √ − , where ≤ ≤ . On the other hand, the inner
radius of a cross-section is given by =
−
since = − + is equivalent to = − . Thus, the volume of the described solid
is given by = ∫ [( ) − ( ) ] ⋅ − = ∫ [(√ − ) − (
) ] MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 7 ILLUSTRATION (cont’n.)
b. Since the axis of revolution is the line = −, we
need to use vertical strips. For a vertical strip, the
inner and outer radii of the corresponding crosssection are given by = − + − −
and = − + − − , respectively.
Noting that for the described solid, ≤ ≤ , we
obtain its volume to be = ∫ [( ) − ( ) ] ⋅ = [( − + − − ) − ( − + − − ) ] TO DO:
Consider the region described above. Using washers, set-up the definite integral that will
solve for the volume of the solid generated by revolving the region about
1. the −axis
=
2. the line = = TRY THIS!
Using disks or washers, set-up the definite integral that will solve for the volume of the solid of
revolution generated by revolving the described region about a given axis of revolution. If it is
“easy” to evaluate the integral, determine the volume of the solid.
1. the region bounded by = √, the line = , and the −axis revolved about
a. the −axis
b. the line = c. the line = − revolved about 3. the region bounded by = and = , ≤ ≤ revolved about
a. the −axis
b. the −axis
4. the region bounded by = + − and = − revolved about
a. the −axis
b. the line = c. the line = 5. the region bounded by = and = − revolved about
a. the −axis
b. the line = c. the line = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. the region bounded by = and = , ≤ ≤ a. the line = b. the line = − 8 For more exercises, you can refer to:
Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 453-456
James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte.
Ltd., pp. 360-363
For an online tutorial, follow these links:
To get the volume of a solid of revolution using
CYLINDRICAL SHELLS,
Volume of a cylindrical shell = ⋅ ⋅ ∆ ⋅ where = + (or the average radius) and ∆ is the thickness of the shell MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) Consider a strip of the region that is parallel to the axis of revolution. (Revolving this strip generates a
shell.) 9 If the strips are vertical, If the strips are horizontal, = ∫ ⋅ ⋅ = ∫ ⋅ ⋅ where is the distance of a strip from the
axis of revolution, and is the height of a
strip. where is the distance of a strip fromt the
axis of revolution, and is the length of a
strip. ILLUSTRATION
Let be the region bounded by = − + and = − + . Using cylindrical shells,
set-up the definite integral that will solve for the volume of the solid generated by revolving about (a) the −axis and (b) the line = .
Solution:
a. Since the axis of revolution is the −axis,
we need to use vertical strips. For a
vertical strip, its distance from the −axis
is given by = , and its height is given
by = − + − − + . Noting
that for the described solid, ≤ ≤ , we
obtain its volume to be = ∫ ⋅ ⋅ = ∫ [ − + − − + ] b. Since the axis of revolution is the line = ,
we need to use horizontal strips. For a
horizontal strip, we need to determine its
distance from the line = and its length. It
is easy to see that the distance of a horizontal
strip from the line = is given by = − . Thus, the volume of the described solid is
given by = ∫ ⋅ ⋅ = ∫ − (√ − − −
) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) For the length, note that = − + is
equivalent to = ±√ − , where ≤ ≤ .
However, for the described solid, ≤ ≤ .
So, we choose = √ − , where ≤ ≤ .
On the other hand, = − + is equivalent
−
to =
. Hence, the length of the strip is given by
− = √ − −
, where ≤ ≤ . 10 TO DO:
Consider the region described above. Using cylindrical shells, set-up the definite integral
that will solve for the volume of the solid generated by revolving the region about
1. the −axis Answer: = 2. the line = Answer: = TRY THIS!
Using cylindrical shells, set-up the definite integral that will solve for the volume of the solid of
revolution generated by revolving the described region about a given axis of revolution. If it is
“easy” to evaluate the integral, determine the volume of the solid.
1. the region bounded by = √, the line = , and the −axis revolved about
a. the −axis
b. the line = c. the line = − 2. the region bounded by = and = , ≤ ≤ a. the −axis
b. the line = revolved about 3. the region bounded by = and = , ≤ ≤ revolved about
a. the −axis
b. the −axis
4. the region bounded by = + − and = − revolved about
a. the −axis
b. the line = c. the line = 5. the region bounded by = and = − revolved about
a. the −axis
b. the line = c. the line = For more exercises, you can refer to:
Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 462-465
James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte.
Ltd., pp. 366-368
For an online tutorial, follow these links:
MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 11 3.3 Centroid of a Plane Region
Center of mass along a line (discrete case)
Consider several weights distributed along a line. 0 1 2 3 4 5 6 7 8 9 10 Let be the total mass and be the moment of mass which is a product of a mass and its
̅ is the coordinate of the center of mass along the line, then
directed distance from a point. If ̅= ∑= = ∑= Center of mass of a rod
In a homogeneous rod, the mass is directly proportional to the length: = ⋅ where is the
constant linear density and is the length of the rod.
In a non-homogeneous rod, the mass is directly proportional to the length: = ⋅ where is the linear density at point on the rod.
If is the linear density of a rod, then the total mass is =
mass is = ⋅ and the moment of ⋅ ⋅ . TO DO: Center of mass of a plane region
Consider a region on the −plane.
Assumptions: the region is a lamina (the object is a “thin” sheet) and of constant area density, In effect, the center of mass will just be a function of the area of the region and its distances from the −axis and the −axis. (No “mass” in here!) In this case, the center of mass is a centroid of the
plane region. MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) The length of a rod is meters and the linear density of the rod at a point meters from
one end is + kg/m. Determine the center of mass of the rod. 12 Consider the region bounded by = and = , ≤ ≤ . Suppose the area density is .
(Simplification: constant area density) = Total mass: = ⋅ Moments of mass: = ⋅ ⋅ − = ⋅ ⋅ − = ̅, ̅ , where
Center of mass: ̅= ̅= and Using vertical strips, ⋅ [ − ] ⋅ = ⋅ ⋅ [ − ] ⋅ + = ⋅
⋅ [ − = ⋅ In general, using horizontal or vertical strip, ⋅ ⋅ = ⋅ = ⋅ = − ⋅ ⋅ − ⋅ ⋅ , where : .
ILLUSTRATION
1. Set-up the definite integrals that will give the coordinates of the centroid of the region bounded by = and = , ≤ ≤ . Solution:
We use vertical strips. (Why?) When ≤ ≤ , = lies above = . Hence, we have = = = = ⋅ [ − ] ⋅ ⋅ ⋅ [ − ] ⋅ = − ⋅ − + + ⋅ [ − ] ⋅ − . ̅, ̅ =(
Thus, the centroid of the described region is at , ). MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) = 13 ILLUSTRATION
2. Using horizontal strips, set-up the definite integrals that will give the coordinates of the
centroid of the region bounded by = and = . Solution:
The curves = and = intersect at
the points , and , , and for the
described region, ≤ ≤ and ≤ ≤ .
Now, note that = is equivalent to = ±√.
However, we choose = √ since ≤ ≤ for the described region. On the other hand, = is equivalent to = √. Hence, using
horizontal strips, we have
=
= ⋅ [ ( √ − =
= =
= = = − ] ⋅ √) + ⋅
⋅ [ √+√ ( √ − − ] ⋅ √) ⋅ ⋅ [ − ( √ − √). ⋅ ̅, ̅ =(
Thus, the centroid of the described region is at , ). TRY THIS!
Set-up the definite integrals that will give the coordinates of the centroid of the following
regions. If it is “easy” to evaluate the integrals, determine the coordinates of the centroid of the
region.
1.
2.
3.
4.
5. the region bounded by = − + , = , and the −axis
the region bounded by = and = − + the region bounded by = and = , ≤ ≤ the circle defined by the equation + = , where is a positive constant
the semi-circle bounded by = √ − (where is a positive constant) and the −axis Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 494-496
James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte.
Ltd., pp. 585-586
For an online tutorial, follow this link:
MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) For more exercises, you can refer to: 14 3.4 Centroid of a Solid of Revolution Point in the three-dimensional space: , , where
: directed distance from the −plane
: directed distance from the −plane
: directed distance from the −plane
Consider a region on the −plane revolved about a
certain axis of revolution on the −plane. Let be the
(constant) density of the solid of revolution. Total mass: = ⋅ Moments of mass: = ⋅ ⋅ − = ⋅ ⋅ − = ⋅ ⋅ − ̅, ̅, ̅ , where
Centroid of the solid: ̅= Remark: ̅= ̅ = The centroid of a solid of revolution always lies on the axis of revolution.
̅ or ̅ is dependent on the axis of revolution.
Hence, ̅ = , and either Caution! Be careful in the choice of strips: either vertical or horizontal. ILLUSTRATION
1. Let be the region in the first quadrant bounded by = , the −axis, and the −axis. Set-up the definite integrals that will give the coordinates of the centroid of the
solid generated by revolving about the line = −. Solution: We first determine the volume of the solid. The
inner and outer radii of the corresponding crosssection of a vertical strip are given by = −
− = and = − − = + ,
respectively. Noting that for the described solid, ≤ ≤ , we obtain its volume to be MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) By default, ̅ = . And since the axis of revolution
̅ = −. We are left with ̅,
is the line = −, then i.e., we need to determine the total mass and the
moment of mass with respect to the −plane .
For this problem, it is easier to use vertical strips;
hence, we will do so. Equivalently, we will use
washers to determine and . 15 ILLUSTRATION (cont’n.) = ∫ ⋅ [( ) − ( ) ] ⋅ = ∫ [ + − ] . Thus, = ⋅ = ∫ ⋅ ⋅ [( ) − ( ) ] ⋅ = ∫ [ + − ] . and = ⋅ − ⋅ = ∫ ⋅ ⋅ ( ⋅ [( ) − ( ) ] ⋅ ) (Note: The directed distance from the −plane is .) = ∫ [ + − ] . ̅, ̅, ̅ = (
Therefore, the centroid of the described solid is at , −, ). Remark: We can also use horizontal strips (cylindrical shells) to solve the problem.
However, we should note that the directed distance from the −plane should be in terms of
. In this case, the directed distance from the −plane is the average of = (which is the
equation of the −axis) and = (which is the equation equivalent to = ). That is, it is
. TO DO:
Let be the region in the first quadrant bounded by = , the −axis, and the −axis.
Set-up the definite integrals that will give the coordinates of the centroid of the solid generated by
revolving about the line = . ILLUSTRATION
− Let be the region bounded by = √ − and =
. Using horizontal strips, set-up the definite integrals that will give the coordinates of the centroid of the solid generated
by revolving about the line = . Solution:
̅ = . We are
By default, ̅ = . And since the axis of revolution is the line = , then ̅, i.e., we need to determine the total mass and the moment of mass with respect
left with to the −plane . For this problem, we are required to use horizontal strips. Note that
the strips are perpendicular to the axis of revolution. Thus, we will use washers to determine and . MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. 16 ILLUSTRATION (cont’n.)
We first determine the volume of the solid.
The inner and outer radii of the corresponding
cross-section of a horizontal strip are given by
− = − √ − and = −
, respectively. Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ ⋅ [( ) − ( ) ] ⋅ = ∫ [( − − ) − ( − √ − ) ] Thus, = ⋅ = ∫ ⋅ ⋅ [( ) − ( ) ] ⋅ = ∫ [( − − ) − ( − √ − ) ] . and = ⋅ − ⋅ = ∫ ⋅ ⋅ ( ⋅ [( ) − ( ) ] ⋅ ) (Note: The directed distance from the −plane is .) = ∫ [( − − ) − ( − √ − ) ] . ̅, ̅, ̅ = (,
Therefore, the centroid of the described solid is at , ). Remark: We can also use vertical strips (cylindrical shells) to solve the problem. However,
we should note that the directed distance from the −plane should be in terms of . In this
case, the directed distance from the −plane is the average of = − + where ≥ (which is the equation equivalent to = √ − ) and = − + (which is the equation
equivalent to = − ). That is, it is (− +)+ −+ . − Let be the region bounded by = √ − and =
. Set-up the definite integrals that will give the coordinates of the centroid of the solid generated by revolving about the line = −. MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) TO DO: 17 TRY THIS!
Set-up the definite integrals that will give the coordinates of the centroid of the solid of
revolution generated by revolving the described region about a given axis of revolution. If it is
“easy” to evaluate the integrals, determine the coordinates of the cent...

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