MATH 52 MIDTERM II SOLUTIONS MAY 13, 2009
THIS IS A CLOSED BOOK, CLOSED NOTES EXAM. NO CALCULATORS OR OTHER
ELECTRONIC DEVICES ARE PERMITTED.
YOU DO NOT NEED TO EVALUATE ANY INTEGRALS IN ANY PROBLEM. THERE
ARE 4 PROBLEMS, EACH WORTH 10 POINTS.
THE TERMS D
MATH 52 MIDTERM II NOVEMBER 4, 2009
THIS IS A CLOSED BOOK, CLOSED NOTES EXAM. NO CALCULATORS OR OTHER
ELECTRONIC DEVICES ARE PERMITTED.
INTEGRAL EVALUATIONS ARE SPECIFICALLY REQUIRED IN BOTH PARTS OF
PROBLEMS 1 AND 3. YOU CAN USE SHORTCUTS, FOR EXAMPLE IF
Math 52 - Autumn 2006 - Midterm Exam II
Problem 1. Let R be a region in R2 with area A, and coordinates of the centroid (a, b). Let
L denote the length of the boundary R of the region R and (m, n) denote the coordinates
of the boundarys centroid. Use (som
Math 52 - Autumn 2010 - Midterm Exam II
Name:
Student ID:
Signature:
Instructions:
Print your name and student ID number, select your section number and TAs name,
and write your signature to indicate that you accept the Honor Code.
There are 4 problems
Math 52 - Autumn 2013 - Midterm Exam I
Name:
Student ID:
Signature:
Instructions: Print your name and student ID number, write your signature to indicate
that you accept the honor code. During the test, you may not use notes, books, calculators.
Read each
Math 52 - Winter 2011 - Midterm Exam II
Problem 1. (12 pts.)
compute
Given a parametrized path (t) = (2t, 1 t, t2 ) for 0 t 1
a)
x ds
Solution: Let X(t) = (2t, 1 t, t2 ), then X (t) = (2, 1, 2t), and |X (t)| =
5 + 4t2 .
The integral is thus:
1
1
2 dt =
SOLUTION MATH 52 SECOND MIDTERM
SPRING 2010
1.(10) Let C be the space curve given by x(t) = (t, t, t2 ), 0 t 1.
a. Evaluate the integral
x ds;
C
b. for F = i + z j y k, evaluate he integral
F ds;
C
c. evaluate the integral
x dx + z dy.
C
Solution. a. x =
Math 52
Second Midterm Solutions 2009
1. (8 points)
Decide whether or not the following vector elds are gradient vector elds. If so, nd a
potential function for the vector eld. If not, explain how you know that no potential function
exists.
a) F (x, y, z)
Problem 1. (10 pts.) Show that there is no function f : R3 R for which
2x 5z
3x
f =
3
y x
Solution: Assume there such an f existed. Then, on one hand, by the properties of the
gradient
f = 0.
On the other hand
3y 2
2x 5z
j
i
k
= det
3x
x
y
z = 4 = 0.