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

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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

## Document Summary

2( (i) show that f is conservative on. Hence f is conservative iff k z z. 3r , and of course which actually is the case. 0 (ii) find the potential function for f i. e. a scalar function f such that. 2 y f f f z y x zyxf implies. We may write for we get the point. F: [10 marks] verify the divergence theorem in case zyx y. 2 region in the first octant bounded by the plane. 2 x z i x j z and v is the. The boundary s of v consists of four faces smooth). 0zs dsnf dsnf dsnf dsnf zx zx ds plane. If you do not use the formula for the volume, you have to evaluate the iterated integral dv)001( hb. 0s: (a) [2 marks] let point x of s relative to. A point such that (b) [7 marks] evaluate. C t = is the interior point of s relative to.