MATH 324F
Final Exam
March 19, 2013
Student ID #
Name
Your exam should consist of this cover sheet, followed by 7 problems. Check that you have
a complete exam.
Unless otherwise indicated, show all your work and justify your answers.
Unless otherwise i
15. Line Integrals of Vector Fields II. February 13, 2013
15.1. Calculation Example
Find
F dr, where F(x, y) = ex1 i + xyj, C : r(t) = t2 i + t3 j, 0 t 1
C
This is one of the practice problems. We have: x(t) = t2 , y(t) = t3 , and
2 1
r (t) =< 2t, 3t2 >,
20. Stokes Theorem I. March 6, 2013
20.1. Formulation
This is a 3D version of Greens Theorem. Consider an oriented surface with an orientation given
by a normal vector eld n. Assume its boundary is a simple closed curve C which is traversed in the
followi
19. Greens Theorem. March 4, 2013
19.1. Formulation
Here, we have only two dimensions: x and y. Consider a curve C in R2 . Assume it is closed: it starts
and ends at the same point. Assume it is traversed counterclockwise. Let D be the region inside this
18. Seven Integrals and Three Theorems. March 1, 2013
18.1. Seven Types of Integrals.
So far, we have studied the following types of integrals:
b
1. One-variable integral. a f (x) dx. It has two interpretations: (1) in 1D: total mass of a
thin wire with l
21. Stokes Theorem II. March 8, 2013
21.1. Independence of Surface
Note that if S1 and S2 are two oriented surfaces with the same boundary curve C, and their orientations agree with the orientation of C, then
curl F dS =
F dr =
S1
C
curl F dS
S2
This fact
22. The Divergence Theorem I. March 11, 2013
22.1. Formulation
Assume that E is a bounded region in R3 , and is its boundary surface, oriented outwards.
If F is a 3D eld, then
F dS =
div FdV
E
Recall the deiniton of the divergence:
div F =
P
Q R
+
+
=
x
y
1. Double Integrals. January 7, 2013
Double Integrals over Rectangles
We dene a single-variable integral over an interval by splitting it into small subintervals and picking
up a point in each of them. A double integral can be dened similarly. Assume we h
6. Cylindrical and Spherical Coordinates. January 18, 2013
6.1. Cylindrical Coordinates
This is the analogy of polar coordinates in 3D. This is a triple (r, , z), dened as
x = r cos , y = r sin , z = z, r 0, 0 < 2
So the coordinates (r, ) are polar coordi
4. Triple Integrals II. January 14, 2013
4.1. Applications of Triple Integrals
Consider a region E R3 . If the density of a solid body which occupies this region is 1, then its
volume Vol(E) is equal to its mass. This is why
Vol(E) =
1dV.
E
If (x, y, z) i
3. Triple Integrals
3.1. Physical Meaning of Double Integrals
Consider a lamina occupying a region D R2 with density function (x, y). Then its total mass is
(x, y)dA.
D
Indeed, assume for simplicity that D is a rectangle. We can split it into small subrec
2. Double Integrals in Polar Coordinates. January 9, 2013
2.1. Polar Coordinates
Sometimes it is convenient to write a point (x, y) on the plane R2 in polar coordinates (r, ):
x = r cos ,
y = r sin , r 0, 0 < 2.
One of the reasons is that sometimes the cu
5. Change of Variables in Double Integrals. January 16, 2013
5.1. The Main Formula
Assume we have a transformation x = g(u, v), y = h(u, v), which gives a mapping of D on the
xy-plane onto G on the uv-plane. Suppose f : D R is a function. Then
f (x, y)dA
10. Parametric Surfaces. February 4, 2013
10.1. Denition
A curve has one dimension and can be parametrized by x = x(t), y = y(t), z = z(t), a t b. A
surface has two dimensions and can be parametrized by
x = x(u, v), y = y(u, v), z = z(u, v), (u, v) D
wher
11. Surface Integrals. February 6, 2013
11.1 Area of Small Patch
Consider the surface given by parametric equations r = r(u, v), x = x(u, v), y = y(u, v), z =
z(u, v), (u, v) D, where D is some region in R2 . Consider the small patch of this surface corre
7. Change of Variables in Triple Integrals. January 23, 2013
7.1. General Formula
On the Lecture 5, we studied the change of variables for double integrals. A similar formula is valid
for triple integrals. Suppose
x = g1 (u, v, w), y = g2 (u, v, w), z = g
8. The Chain Rule, Directional Derivatives. January 25, 2013
8.1. The Chain Rule I
If z = f (x, y) and x = g(t), y = h(t), then z = f (g(t), h(t) is a function of t. We have:
dz
z dx z dy
=
+
dt
x dt
y dt
Indeed, a change
t in t produces a change
x
dx
t,
13. Curl and Divergence. February 11, 2013
13.1. Denitions
Take a vector eld F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k. Then its divergence is
div F =
P
Q R
+
+
x
y
z
and its curl is
P
R
Q P
R Q
i+
j+
k
y
z
z
x
x
y
You do not really need to me
16. Surface Integrals of Vector Fields I. February 25, 2013
16.1. Motivation and Denition
Assume we have a planar region of area S, with a normal vector n, and a constant vector eld F.
Then the ux or ow of this eld through is dened as
(F n)S.
It has the m
17. Surface Integrals of Vector Fields II. February 27, 2013
17.1. When Surface is a Graph
Suppose is a graph z = g(x, y), (x, y) D of a function. Assume we have a vector eld F =
P i + Qj + Rk. Let us nd an expression for the ux
F dS.
We can parametrize b
MATH 324F
Midterm 1
February 1, 2013
Name
Student ID #
Section
HONOR STATEMENT
I arm that my work upholds the highest standards of honesty and academic integrity at the
University of Washington, and that I have neither given nor received any unauthorized
MATH 324F
Midterm 2
February 22, 2013
Student ID #
Name
Your exam should consist of this cover sheet, followed by 4 problems. Check that you have
a complete exam.
Pace yourself. You have 50 minutes to complete the exam and there are 4 problems. Try
not
Practice Final
Math 324 E, Wi 2013
3/18/2013
Please write your name in the top left corner of the exam.
Answer the questions in the spaces provided on the question sheets. If you
run out of room for an answer, continue on the back of the page. Answer all
MATH 324F
Practice Final Exam
Student ID #
Name
Your exam should consist of this cover sheet, followed by 7 problems. Check that you have
a complete exam.
Unless otherwise indicated, show all your work and justify your answers.
Unless otherwise indicat
Quiz 6 Solution
Math 324F. March 8, 2013
Using Greens Theorem, calculate:
y 2 dx,
C
where C is the triangle with vertices (0, 0), (1, 0), (0, 1), traversed counterclockwise.
Solution. Denote by D the region enclosed by C . This region is given by
D = cfw_
Quiz 5 Solution
Math 324F. March 1, 2013
Find the ux of the eld
F = xi + y j + 2 z k
through the part of the plane z = x + y given by the equations 0 x 1, 0 y 2.
Solution. This is the graph of the function g (x, y ) = x + y for (x, y ) D, where D =
[0, 1]
Quiz 4 Solution
Math 324F. February 15, 2013
For every real value of the parameter , consider the eld
F(r) = |r| r,
r = xi + y j + z k .
(1) Find div F. Simplify as much as possible. For which do we have: div F = 0?
(2) Find curl F. Simplify as much as po
Quiz 3 Solution
Math 324F. February 8, 2013
Calculate
(x z )dS,
where is the part of the plane 2x + 2y + z = 0 which corresponds to 0 x 1, 0 y 2.
Solution. The surface is the graph z = g (x, y ) of the function g (x, y ) = 2x 2y for
(x, y ) D, where D = [
Quiz 2 Solution
Math 324F. January 28, 2013
Calculate
2ydV,
E
where E is the solid region bounded by the xz -coordinate plane and the paraboloid y = 1 x2 z 2 .
Solution. Let D = cfw_x2 + z 2 1 be the unit disc on the xz -plane. Then
E = cfw_(x, z ) D, 0 y
Quiz 1 Solution
Math 324F. January 18, 2013
Calculate
D
x2
1
dA, where D = cfw_1 x2 + y 2 4, y 0, y x y .
+ y2
Solution. Rewrite D in polar coordinates. Since x2 + y 2 = r2 , the inequalities 1 x2 + y 2 4
are written as 1 r2 4 1 r 2. Since y 0, we have: 0