MATH 248
SHORT SOLUTIONS TO ASSIGNMENT 6
1. A proper parametrization of is x(t) = cos t, y (t) = 2 cos sin t, z (t) = 2 sin sin t,
0 t 2 . The value of the integral is 8 (sin cos ).
2. A proper parametrization of is
x(t) =
aa
+ cos t,
22
a
sin t,
2
y (t)
MATH 248
MIDTERM EXAM II
There are 5 problems on the exam worth the total of 30 points. Question
5 is on the oposite side.
Your solutions should be written in a clear, complete and logical way you
must convince the marker that your solutions are correct.
MATH 248
PRACTICE ASSIGNMENT
This assignment will be neither collected nor graded. The problems on the
assignment concern the Stokes and the divergence theorem. You should solve
these problems as a part of your preparation for the nal exam.
Problems
1. Pr
MATH 248
MIDTERM EXAM I
There are 6 problems on the exam worth the total of 30 points. Questions
5 and 6 are on the oposite side.
Your solutions should be written in a clear, complete and logical way you
must convince the marker that your solutions are co
MATH 248
SHORT SOLUTIONS TO ASSIGNMENT 7
1. The parametrization of S is x(u, v ) = u, y (u, v ) = v , z (u, v ) = a u v , where
u, v satisfy u2 + v 2 a2 . Tu Tv = 3 and the area is
3dudv =
3a2 .
u2 + v 2 a 2
2. Let D = cfw_(u, v ) : 0 u
x(u, v ) = u cos
MATH 248
SOLUTION TO ASSIGNMENT 1
1. (a)
Df (x, y, z ) =
2x 1 1
,
2 1 2z
2 2
vw
2uvw2 2uv 2 w
w2 cos v 2w sin v
Dg (u, v, w) = 0
2uev
u2 ev
0
(b)
h(u, v, w) = f (uv 2 w2 , w2 sin v, u2 ev )
= (u2 v 4 w4 + w2 sin v + u2 ev , 2uv 2 w2 + w2 sin v + u4 e2v )
MATH 248
SOLUTION TO ASSIGNMENT 4
1. (a) The change of variables x = r cos , y = r sin yields that
r
drd,
I (p, a) =
2 + r 2 )p
D (p
where D = cfw_(r, ) : 0 r a, 0 2 . Hence,
a
2
I (p, a) =
0
(p2
0
a
(p2
rdr
+ r2 )p
(p2
du
.
+ u)p
= 2
0
a2
=
0
r
ddr
+ r2
MATH 248
SOLUTION TO ASSIGNMENT 2
1. (a) We have shown in class that
g
f
f
= cos
+ sin ,
r
x
y
g
f
f
= r sin
+ r cos .
x
y
(1)
Hence
g
r
2
+
1
r2
g
2
=
cos
f
f
+ sin
x
y
= cos
f
x
+ sin2
f
x
2
=
f
x
2
+
2
2
+ sin
2
+ sin
f
y
f
y
2
+ cos2
f
f
+ co
MATH 248
SOLUTION TO ASSIGNMENT 3
2. Draw a picture. The region of integration is the union of two elementary regions
D1 and D2 described by
D1 = cfw_(x, y ) : /2 x 0, (x + /2) y x + /2,
D2 = cfw_(x, y ) : 0 x /2, x /2 y x + /2.
Hence,
0
x+/2
cos x cos yd
MATH 248
SHORT SOLUTIONS TO THE MIDTERM EXAM II
1.
f (x, y ) =
sin x sin y
,
x
y
.
The critical points are (n, m ), where n, m are positive integers and n < m. If n and m
are both even or both odd, then the critical point is a saddle point. If n is even a
MATH 248
SHORT SOLUTIONS TO ASSIGNMENT 5
1. The solid D can be described as
(x, y, z ) : x2 + y 2 4,
D=
x2 + y 2
z
4
5 x2 y 2 .
Hence
x2 + y 2
4
5 r2
r2
4
rdr =
5
dxdydz =
D
y2
x2
dxdy
x2 + y 2 4
2
= 2
0
2
(5 5 4).
3
2. The solid D can be described as
D