MATH 211 FIRST MIDTERM EXAM SOLUTIONS
(1) Change the order of integration:
9 x2
3
dx
0
0
xey
dy
=
9y
9
9 y
dy
dx
0
0
9
=
0
ey x2
dy
9y 2
9
=
xey
9y
x= 9 y
x=0
y
dy
0
e
2
1
= (e9 1) .
2
(2) Two independent tangent vectors to the helicoid are given by:
u =
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MATH 211 SECOND MIDTERM EXAM SOLUTIONS
(1) Let (t) = (x(t), y (t), f (t). Since F has only a k component:
1
dt f (t) k (t)
F=
0
C
1
dt f (t)
=
0
d
(f )(t)
dt
11
d
dt (f )2 (t)
20
dt
1
= (f (q ) f (p) .
2
=
(2) We have:
(f g ) =
f
g+f
g=
f
g,
since the cu
MATH 211 MAKEUP FINAL EXAM
(1) (a) State the Generalized (i.e. dierential forms version) of Stokes Theorem.
(b) Write down the relationship between vector elds and 1-forms on R3 .
(c) Write down the relationship between vector elds and 2-forms on R3 .
(d)
MATH 211 FINAL EXAM SOLUTIONS
(1) The result follows from two facts:
d
f (r) = f (r) r ,
d =1
d
f (r) = 1 f (r)
d =1
=1
= f (r) .
(2) Multiply by f and use the integration by parts formula, and the fact that f 0 on
:
f f + f 2 = 0 ,
f2 = 0 ,
f f +
f
2
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