xercises on solving
x = b and row reduced form R
Problem 8.1: (3.4 #13.(a,b,d) Introduction to Linear
plain why these are all false:
lgebra: Strang) Ex-
a) The complete solution is any linear combination of x p and xn .
b) The system
c) If
x = b has at mo
Double integration
1. Set up a double integral for the mass of the planar region R with a variable density
(x, y). Use a simple sketch to illustrate the setup.
Note: the problem as stated is abstract, so your solution will also be abstract. The goal
here
V3. Two-dimensional Flux
In this section and the next we give a dierent way of looking at Greens theorem which
both shows its signicance for ow elds and allows us to give an intuitive physical meaning
for this rather mysterious equality between integrals.
Changing Variables in Multiple Integrals
1. Changing variables.
Double integrals in x, y coordinates which are taken over circular regions, or have integrands involving the combination x2 + y 2 , are often better done in polar coordinates:
(1)
f (x, y) dA
18.02 Practice Exam 3 A
1. Let (, y ) be the center of mass of the triangle with vertices at ( 2, 0), (0, 1), (2, 0) and
x
uniform density = 1.
a) (10) Write an integral formula for y . Do not evaluate the integral(s), but write explicitly the
integrand
Fundamental Theorem for Line Integrals
1. Let f = xy + ex .
a) Compute F = rf .
Z
b) Compute
F dr for each of the following paths from (0,0) to (2,1).
C
i) The path consisting of a horizontal segment followed by a vertical segment.
ii) The path consisting
18.02 Exam 3
Problem 1. a) Draw a picture of the region of integration of
Z
0
1 Z 2x
dydx.
x
b) Exchange the order of integration to express the integral in part (a) in terms of integration
in the order dxdy. Warning: your answer will have two pieces.
Pro
18.01 Practice Final Exam
There are 19 problems, totaling 250 points. No books, notes, or calculators. This practice exam should
take 3 hours.
Generally useful trigonometry:
sin2 x =
1 cos 2x
;
2
sec x =
cos2 x =
1
;
cos x
1 + cos 2x
;
2
sec x = ln(sec x
18.01 Practice Exam 4
1/2
Problem 1. (15) Evaluate
0
to change the limits also.
x2
dx by making a trigonometric substitution; remember
1 x2
Problem 2. (15) Find the volume of the solid obtained by rotating about the y-axis (thats the
y-axis) the area unde
Problems: Mass and Average Value
Let R be the quarter of the unit circle in the rst quadrant with density (x, y) = y.
1. Find the mass of R.
Because R is a circular sector, it makes sense to use polar coordinates. The limits of
integration are then 0 r 1
Work and line integrals
Line integrals: (also called path integrals)
Ingredients:
Field F = M i + N j = hM, N i
Curve C: r(t) = x(t)i + y(t)j = hx, yi ) dr = hdx, dyi.
Line integral:
Z
Z
Z
C
F dr =
C
hM, N i hdx, dyi =
M dx + N dy.
C
We need to discuss:
a
18.01 PRACTICE FINAL, FALL 2003
Problem 1 Find the following denite integral using integration by parts.
2
x sin(x)dx.
0
Solution Let u = x, dv = sin(x)dx. The du = dx, v = cos(x). So udv = uv vdu, i.e.,
2
2
x sin(x)dx = (x cos(x)|0 + 02 cos(x)dx =
0
2
Moment of inertia
1. Let R be the triangle with vertices (0, 0), (1, 0), (1,
polar moment of inertia.
p
3) and density
= 1. Find the
Answer: The region R is a 30, 60 , 90 triangle.
y
p
3
r = sec
r
x
1
The polar moment of inertia is the moment of inertia
Work integrals
1. p Let C be the path from (0,0) p (5,5) consisting of the straight line from (0,0)p
to
to
(5 2, 0) followed by the arc from (5 2, 0) to (5,5) that is part of the circle of radius 5 2
centered at the origin.
Z
Compute
F dr for the followin
V4.1-2 Greens Theorem in Normal Form
C
n
1. Greens theorem for ux.
R
Let F = M i +N j represent a two-dimensional ow eld, and C a simple
closed curve, positively oriented, with interior R.
n
According to the previous section,
(1)
ux of F across C =
M dy N
V2.2-3 Gradient Fields and Exact Dierentials
2. Finding the potential function.
In example 2 in the previous reading we saw that
F=
xi + yj
= ln
x2 + y 2
x2 + y 2 = ln r.
This raises the question of how we found the function 1 ln(x2 + y 2 ). More generall
Limits in Iterated Integrals
For most students, the trickiest part of evaluating multiple integrals by iteration is to
put in the limits of integration. Fortunately, a fairly uniform procedure is available which
works in any coordinate system. You must al
V2.1 Gradient Fields and Exact Dierentials
1. Criterion for gradient elds.
Let F = M (x, y) i + N (x, y) j be a two-dimensional vector eld, where M and N are
continuous functions. There are three equivalent ways of saying that F is conservative, i.e.,
a g
V1. Plane Vector Fields
1. Vector elds in the plane; gradient elds.
We consider a function of the type
(1)
F(x, y) = M (x, y) i + N (x, y) j .
where M and N are both functions of two variables. To each pair of values (x0 , y0 ) for
which both M and N are
Mass and average value
Center of Mass
Example 1: For two equal masses, the center of mass is at the midpoint between them.
m1 = 1
m2 = 1
xcm = x1 +x2 .
2
xcm
x1
x2
Example 2: For unequal masses the center of mass is a weighted average of their positions.
Changing Variables in Multiple Integrals
3. Examples and comments; putting in limits.
If we write the change of variable formula as
(18)
f (x, y) dx dy =
R
g(u, v)
R
(x, y)
du dv ,
(u, v)
where
(19)
(x, y)
=
(u, v)
xu
yu
xv
yv
,
g(u, v) = f (x(u, v), y(u,
Moment of Inertia
Moment of inertia
We will leave it to your physics class to really explain what moment of inertia means. Very
briey it measures an objects resistance (inertia) to a change in its rotational motion. It is
analogous to the way mass measure
Integration in polar coordinates
Polar Coordinates
Polar coordinates are a dierent way of describing points in the plane. The polar coordinates
(r, ) are related to the usual rectangular coordinates (x, y) by by
x = r cos ,
y = r sin
The gure below shows
Denition of double integration
In this note we will work abstractly, dening double integration as a sum, technically a limit
of Riemann sums. It is best to learn this rst before getting into the details of computing
the value of a double integral we will
V5. Simply-Connected Regions
1. The Extended Greens Theorem.
In the work on Greens theorem so far, it has been assumed that the region R has as
its boundary a single simple closed curve. But this isnt necessary. Suppose the region
has a boundary composed
Fundamental Theorem for Line Integrals
Gradient elds and potential functions
Earlier we learned about the gradient of a scalar valued function
rf (x, y) = hfx , fy i.
For example, rx3 y 4 = h3x2 y 4 , 4x3 y 3 i.
Now that we know about vector elds, we reco
Changing Variables in Multiple Integrals
y
dA
2. The area element.
rd
dr
In polar coordinates, we found the formula dA = r dr d for the area element by
drawing the grid curves r = r0 and = 0 for the r, -system, and determining (see
the picture) the innit
Greens Theorem
Greens Theorem
We start with the ingredients for Greens theorem.
(i) C a simple closed curve (simple means it never intersects itself)
(ii) R the interior of C.
We also require that C must be positively oriented, that is, it must be travers
Double integration in polar coordinates
1.
Compute
ZZ
R
f (x, y) dx dy, where f (x, y) = p
1
x2
+ y2
and R is the region inside the
circle of radius 1, centered at (1,0).
Answer: First we sketch the region R
y
r = 2 cos
x
1
Both the integrand and the reg