MATH1014 Tutorial 1

MATH1014 Tutorial 1
General Information
Teaching Assistant
Leung Ho Ming
Office: Rm 3489
Tel: 3469 2017
Email: [email protected]
Tutorial Homepage
All materials used in the tutorials will be found at http://ihome.ust.hk/~malhm
Time and Venue
There are four sessions each week, the materials to be covered will be the same.
You may come to any one of these classes.
T07B Mo 09:30AM - 10:20AM G002, CYT Bldg
T07D Mo 04:30PM - 05:20PM G003, CYT Bldg
T07C Tu 04:30PM - 05:20PM G003, CYT Bldg
T07A We 04:30PM - 05:20PM Rm 3588, Lift 27-28
Math Support Center Hours
http://www.math.ust.hk/~support
Math Support Center (Rm 2612) is opened every weekday from 11am to 7pm. While staff are available there, you
may find me there outside class during the following timeslots.
Mon 14:00PM - 15:00PM
Tue 18:00PM - 19:00PM
Wed 13:00PM - 15:00PM
Thu 11:00AM - 16:00PM, 18:00PM - 19:00PM
Fri 12:00NN - 15:00PM, 16:00 - 19:00PM
Prepared by Leung Ho Ming
Homepage: http://ihome.ust.hk/~malhm
1
6.1
Area between Curves
If f and g are continuous functions on [a, b] where f (x) ≥ g(x) for all x ∈ [a, b], then we can find the area bounded by
the curves y = f (x) and y = g(x) from a to b by
ˆ
b
A=
a
[f (x) − g(x)] dx.
When the condition f (x) ≥ g(x) is not known, or does not hold, the more general expression would be
ˆ
b
A=
a
|f (x) − g(x)| dx,
usually we need to calculate the intervals where each function is larger and split the integral into a sum of several
parts.
1 2
x , y = 2x2 , x + y = 3, x ≥ 0.
4
1. (J. Stewart, section 6.1, ex.28) To find the area enclosed by the curves y =
Solution
First we sketch the curves and locate the intersection points of the curves.
4
y = 14 x2
y = 2x2
y=3−x
3.5
3
y
2.5
2
1.5
1
0.5
0
0
0.5
1
x
1.5
2
Clearly y = 41 x2 and y = 2x2 intersect at (0, 0).
y = 14 x2
Solving the simultaneous equations
gives the point (2, 1).
x+y =3
y = 2x2
Similarly, solving the simultaneous equations
gives the point (1, 2).
x+y =3
Since the “upper side” of the bounded region does not consist of a single curve only, we have to split the region
into two parts at x = 1. Hence the required area is
ˆ 2
ˆ 1
ˆ 1
ˆ 2
ˆ 2
1 2
1 2
1 2
2
2
(3 − x − x ) dx =
2x dx +
(3 − x) dx −
x dx
(2x − x ) dx +
4
4
4
1
0
1
0
0
1 2 2
2 3
1
1 3
=
x
+ 3x − x2 −
x
3
2
12
0
1
0
2 3 2
= + −
3 2 3
3
= .
2
2
2. The figure shows the graph of |x|1/2 + |y|1/2 = 1, find the area enclosed by the graph.
1
|x|1/2 + |y|1/2 = 1
0.8
0.6
0.4
y
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
x
0.5
1
Solution
Note that the graph is symmetric along both the x-axis and y-axis.
Symmetry along the x-axis: (x0 , y0 ) lies on the graph ⇒ |x0 |2 + |y0 |2 = 1 ⇒ | − x0 |2 + |y0 |2 = 1 ⇒ (−x0 , y0 ) lies
on the graph.
Symmetry along the y-axis: (x0 , y0 ) lies on the graph ⇒ |x0 |2 + |y0 |2 = 1 ⇒ |x0 |2 + | − y0 |2 = 1 ⇒ (x0 , −y0 ) lies
on the graph.
Therefore we can simply consider the area in the first quadrant, and multiply the value by 4 to obtain the whole
area enclosed by the graph. Restricted to the first quadrant, x and y are both non-negative, hence the equation
of the graph is equivalent to
x1/2 + y 1/2 = 1
y = (1 − x1/2 )2
y = 1 − 2x1/2 + x.
The area enclosed inside the first quadrant is
ˆ
0
1
1
x2
4
(1 − 2x1/2 + x) dx = x − x3/2 +
3
2 0
4 1
=1− +
3 2
1
= .
6
Hence the total area enclosed by the graph is
1
2
×4= .
6
3
If the “left side” and “right side” of a region are functions of y, where f (y) ≥ g(y), we may also integrate with respect
to y to find the area bounded by the curves x = f (y) and x = g(y) from y = c to y = d by
ˆ
A=
c
d
[f (y) − g(y)] dy,
where x = f (y) is the right-hand curve and x = g(y) is the left hand curve.
3
3. Repeat example 1 by integrating with respect to y.
Solution
The curves y = 14 x2 , y = 2x2 , x + y = 3 can be written as x =
√
4y, x =
py
2,x
= 3 − y respectively.
Splitting the region at y = 1 into two parts, then the area can be calculated as
r
r
ˆ 1 p
ˆ 2
ˆ 1p
ˆ 2
ˆ 2r
y
y
y
( 4y −
) dy +
(3 − y −
) dy =
4y dy +
(3 − y) dy −
dy
2
2
2
0
1
0
1
0
2
1 2 4 3/2
1 2
2 3/2
=
y
+ 3y − y
− √ y
3
2
3 2
0
0
1
4 3 4
= + −
3 2 3
3
= .
2
6.2
Volumes
In this section, we study in particular volume by slicing.
Suppose a solid has integrable cross-sectional area (perpendicular to the x-axis) A(x) from x = a to x = b, then the
volume of the solid can be found by
ˆ b
V =
A(x) dx.
a
Similar formula holds when x is replaced by y.
4. (J. Stewart, section 6.2, ex.56) Find the volume of the solid S if the base of S is the triangular region with
vertices (0, 0), (1, 0), and (0, 1). Cross-sections perpendicular to the y-axis are equilateral triangles.
Solution
First we will find the cross-sectional area in terms of y. The side length of the equilateral triangle for each y is
1 − y. By Pythagoras’s theorem, the altitude has length
s
r
√
2
1−y
1
3
2
(1 − y) −
= (1 − y) 1 − =
(1 − y).
2
4
2
Hence the cross-sectional area is
√
√
1
3
3
A(y) = · (1 − y) ·
(1 − y) =
(1 − y)2 .
2
2
4
Thus the required volume is
ˆ
1
ˆ
√
3
(1 − y)2 dy
4
0
√ ˆ 1
3
=
(y 2 − 2y + 1) dy
4 0
√ 3
1
3 y
2
−y +y
=
4
3
0
√
3
=
.
12
A(y) dy =
0
1
Remark By changing the orientation, you may observe that the solid is indeed a triangular pyramid whose base
is an equilateral triangle of unit side length and the height of the pyramid is equal to 1. Therefore we may check
that the volume is
√
√
1
1
3
3
× base area × height = ·
=1=
.
3
3 4
12
4
When a plane region is revolved about a line, we obtain a solid of revolution. Depending on whether the cross-section
obtained is a disk or an annulus. If the cross-section is a disk, we use
2
A = π(radius) ,
else if the cross-section is an annulus, we use
2
2
2
2
A = π(outer radius) − π(inner radius) = π[(outer radius) − (inner radius) ].
5. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified
line.
(a) (J. Stewart, section 6.2, ex.6) y = ln x, y = 1, y = 2, x = 0; about the y-axis
Solution
The cross-section perpendicular to the y-axis is a disk of radius x = ey . Therefore the volume of the required
solid is
2y 2
ˆ 2
πe
π
2y
πe dy =
= (e4 − e2 ).
V =
2
2
1
1
A picture of the solid:
2
1.8
1.6
1.4
1.2
1
5
5
0
0
−5
−5
(b) (J. Stewart, section 6.2, ex.12) y = e−x , y = 1, x = 2; about y = 2
Solution
The cross-section perpendicular to the x-axis is an annulus of outer radius 2−e−x and inner radius 2−1 = 1.
Therefore the volume of the required solid is
ˆ 2
V =
π[(2 − e−x )2 − 12 ] dx
0
ˆ
=π
0
2
(3 − 4e−x + e−2x ) dx
2
1
= π 3x + 4e−x − e−2x
2
0 1 −4
1
−2
= π 6 + 4e − e
− 4−
2
2
1
5
+ 4e−2 − e−4 .
=π
2
2
5
6. (J. Stewart, section 6.2, ex.64) Find the volume common to two circular cylinders, each with radius r, if the axes
of the cylinders intersect at right angles.
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
1
−1
1
0.5
0.5
1
0
−0.5
−1
1
0
0
0
−0.5
−1
−1
−1
Solution
Choose the coordinate system such that the x-axis is the axis of one of the cylinders, and the other one has the
circle x2 + y 2 = r2 as its cross-section. By symmetry, the cross-section perpendicular to the y-axis is a square,
where the side length is the distance
between the left and the right of the circle x2 + y 2 = r2 . Using Pathogoras’s
p
2
theorem, this side length is 2 r − y 2 , so that the cross-sectional area is
A = 4(r2 − y 2 ).
Hence the required volume is
ˆ
ˆ
r
2
V =
−r
2
r
(r2 − y 2 ) dy
r
1 3
2
=8 r y− y
3
0
16 3
=
r .
3
4(r − y ) dy = 8
0
Remark You may google “Steinmetz Solid” for more about this solid.
6