This

**preview**shows page 1. to view the full**4 pages of the document.**1. Suppose where a is a constant. kjiF yxazzyx ++=),,(

(a) [5 marks] Is there a real number a such that 0])[( >

Ć

FFcurldiv . Justify.

We have

)1,,1( a

yxaz zyx

curl =

ā

ā

ā

ā

ā

ā

=

kji

F and

=

Ć

FF)(curl ),,(11 2zaxyazxay

yxaz

aāāā=

kji

Hence for all values of a and so there is no value of a 011])[( 2<āāā=Ć acurldiv FF

such that . 0])[( >Ć FFcurldiv

(b) [7 marks] In case , evaluate , where C is the portion of the curve of

2ā=aā«ā

C

dxF

intersection of surfaces and in the first octant from o

2

xz =4

22 =+ yx )4,0,2( t )0,2,0(.

In this case kjiF yxzzyx

+

+ā= 2),,(. From and (both equations should

be satisfied) we get a required parametrization of C :

222 2=+ yx

x2

2

xz =

tcos

=

, tsiny2

=

, , tz 2

cos4=

2

0

Ļ

ā¤ā¤ t and we have

ā«ā

C

dxF = =

ā«āāā
ā

2/

0

2)sincos8,cos2,sin2()sin2,cos2,cos8(

Ļ

dtttttttt

= =

ā«ā+

2/

0

222 )cossin16cos4sincos16(

Ļ

dtttttt

=

ĻĻ

Ļ

=ā++āāā++=ā++ā )000

3

16

()

3

16

00(]sin

3

16

2sin2cos

3

16

[2/

0

33 tttt

2

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