# MAT237Y1 Study Guide - Piecewise, Simply Connected Space, Iterated Integral

18 views4 pages
Published on 27 Sep 2011
School
UTSG
Department
Mathematics
Course
MAT237Y1
Professor
1. (a) [6 marks] Let kjiF 222 3)2(),,( zexyyzyx y++=
(i) Show that F is conservative on 3
R
.
The vector field F is of class on
1
C3
R
, and of course 3
R
is simply connected.
Hence F is conservative iff 0F
=
curl which actually is the case
0
kji
F=+=
=)22,00,00(
32 222
yy
zexyy zyx
curl
y
(ii) Find the potential function for F i.e. a scalar function f such that .
f=F
We should have =
),,( z
f
y
f
x
f)3,2,( 222 zexyy y
and
2
y
x
f=
implies , substituting to ),(),,( 2zyhxyzyxf += y
exy
y
f2
2=
we get
y
exy
y
h
xy 2
22 =
+ which implies )(
2
1
),( 2zkezyh y+= so
)(
2
1
),,( 22 zkexyzyxf y+= . Substituting this to 2
3z
z
f=
we get czzk +=
3
)(.
Hence czexyzyxf y++= 322 2
1
),,(
1.(b) [6 marks] Evaluate where and
C
dxG kjiG 222 3)2()(),,( zexyyyzyx y+++=
C is a helix parametrized by
π
=
ttttt 0,),sin,(cos)(g.
HINT: Use the first part and write HG
+
=
f
We may write =),,( zyxG)0,0,()
2
1
(322 yzexy y++. For 0
=
t we get the point and
for
)0,0,1(
π
=t we get the point ),0,1(
π
. Hence
C
dxG = ++
C
y
C
dydzexy xx )0,0,()
2
1
(322 = ++
C
y
C
dxydzexy x)
2
1
(322 =
= ++
π
π
0
),0,1( )0,0,1(
322 )sin)((sin]
2
1
[dtttzexy y = 2
)]0
2
1
0()
2
1
0[( 3
π
π
++ =
= 2
3
π
π
2
Unlock document

This preview shows page 1 of the document.
Unlock all 4 pages and 3 million more documents.