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HINTS TO ASSIGNMENT #5
1.
Just evaluate both sides and see that they are equal (mixed second partials equal!). For
notations see formula 5.33 in your text.
2.
In all parts it is useful to apply identity 5.28 of your text and simple result
r
r
r
=
∇
.
(a)
r
r
r
r
r
r
r
r
r
12
)
3
(
3
]
)
(
)
(
[
3
3
)
3
(
2
3
=
+
=
⋅
∇
+
⋅
∇
=
⋅
∇
=
⋅
∇
=
∇
⋅
r
r
r
r
∇
⇒
12
=
k
(b)
3
2
2
)
2
(
1
2
)
1
(
2
)
1
3
1
(
)
(
r
r
r
r
r
r
r
r
r
r
r
r
r
−
=
−
=
∇
=
⋅
−
∇
=
⋅
∇
∇
2
−
=
⇒
k
(c)
k
k
k
k
r
k
r
r
k
r
r
−
=
⋅
−
=
⋅
+
3
3
1
)
(
1
r
r
r
∇
so
3
=
k
(d)
)
2
(
)
1
(
)
1
(
)
(
3
2
2
2
2
2
r
r
r
r
r
r
r
−
⋅
∇
=
∇
⋅
∇
=
∇
=
⋅
∇
∇
4
5
4
2
]
4
3
[
)
2
(
r
r
r
r
=
⋅
−
−
=
r
r
,
k
.
2
=
3.
The area is
, here
π
80
)]
(
01
.
0
4
[
2
2
=
−
+
∫
ds
y
x
C
2
0
,
sin
10
,
cos
10
:
≤
≤
=
=
t
t
y
t
x
C
.
For both sides the area is
160
. Hence one needs
2
m
6
.
1
liters of paint.
4.
(a)
3
2
6
2
sin
4
2
2
)]
2
)(sin
2
cos
2
)(
2
sin
[(
4
0
3
4
0
−
=
−
=
−
=
∫
t
dt
t
t
t
I
(b)
t
dt
t
s
t
2
2
2
2
)
(
0
=
=
∫
, that is
2
2
s
=
t
. For
4
,
0
=
t
we get
2
2
,
0
=
s
so the
arc length parametrization of
C
is
2
2
0
),
2
sin
,
2
cos
2
,
2
sin
(
)
(
≤
≤
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This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto Toronto.
 Fall '09
 RomauldStanczak
 Calculus

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