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 combinat
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,
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
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
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 evalua
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
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 ord
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
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 rot
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
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
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,
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
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
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
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 h
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 avai
xercises on independence, basis, and dimension
Problem 9.1: (3.5 #2. Introduction to Linear lgebra: Strang) Find the largest
possible number of independent vectors among:
2
3
2
3
2
3
1
1
1
6 1 7
6 0 7
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 equi
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 eac
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 ce
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)
(
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 rot
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 (
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 detai
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
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 ,
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
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 m
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 r