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Ma 227 Exam III Solutions 12 / 7 / 09

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1

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1a [15 pts.] Evaluate the line integral

C 

F  dr 

where F  zi y2 j xk  and C is the curve given by r t  t  1i et  j t 2k , 0 ≤ t  ≤ 2.

Solution:

C 

F  dr  a

bF t  r ′t dt 

0

2t 2i e2t  j t  1k  i et  j 2tk  dt 

0

2t 2 e3t  2t 2 2t dt 

0

23t 2 e3t  2t dt  t 3

e3t 

3 t 2

0

2 12

e6

3− 1

3

353

e6

3

1b [15 pts.] Let f  f  x, y, z Show that

∇ ∇ f  0

Assume that the cross partials of f are equal.

Solution:

∇ f  f  xi f  y j f  zk 

∇ ∇ f 

i  j k 

∂∂ x

∂∂ y

∂∂ z

∂ f ∂ x

∂ f ∂ y

∂ f ∂ z

i  j k 

∂∂ x

∂∂ y

∂∂ z

∂ f ∂ x

∂ f ∂ y

∂ f ∂ z

i  j

∂∂ x

∂∂ y

∂ f ∂ x

∂ f ∂ y

f  yzi f  zx j f  xyk  − f  yxk  − f  zyi− f  xz j 0

2 [20 pts.] he vector function

F  2 xy z i x2  j x k 

is conservative. Evaluate C 

F  dr  where C is any path from 1,−1, 2 to 2,2,2.

Solution: Since F  is conservative, then there exists a  x, y, z such that ∇ F . Once we find , then

C  F 

dr  2,2,2 − 1,−1, 2

We have that

 x 2 xy z,  y x2, and  z x

Integrating the first equation with respect to x while holding y and z constant yields

x2 y xz g y, z

Hence

2

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 y x2 g y x2

So g y 0 and g h z.

x2 y xz h z

 z x h′ z x

Hence h′ z 0 and h z c, a constant and

x2 y xz c

C 

F  dr  2,2,2 − 1,−1, 2 8 4 1 − 2 11

3a 15 pts. Evaluate

C 

sin xdx x2 y3dy

directly without using Green’s Theorem, where C is the triangle shown below.

0.0 0.5 1.0 1.5 2.00.0

0.5

1.0

1.5

2.0

x

y

Solution: The triangle consists of the segment y 0, 0 ≤ x ≤ 2, followed by the segment x 2, 0 ≤ y ≤ 2, followed by the line y x, where x goes from 2 to 0. Thus

C 

sin xdx x2 y3dy 0

2sin xdx

0

24 y3dy

2

0sin xdx x5dx

−cos2 1 24 − 1 cos2 − 26

6 24 − 25

3 1 − 2

324

163

: 163

3

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3b 15 pts. Evaluate the line integral in 3a by using Green’s Theorem.

Solution:

C 

Pdx Qdy  R

Q x − P y dA

Here Q x2 y3 and P sin x. Therefore

C 

sin xdx x2 y3dy  R

2 xy3 − 0 dA

0

2

 y

22 xy3dxdy

0

2

0

 x2 xy3dydx

163

4 a [5 pts.] Let S be the surface x2 y2 4, 0 ≤ z ≤ 4. Sketch S and give a parametrization of S incylindrical coordinates.

Solution: S is the shell of the cylinder of radius 2 parallel to the z axis between z 0 and z 4. Weparametrize S as

r , z 2cosi 2sin j zk  0 ≤ ≤ 2, 0 ≤ z ≤ 4

2

4 b [15 pts.] Give an expression for

S

 x y zdS

in cylindrical coordinates where S is the surface in part 4a. Do not evaluate your expression.

Solution:

r  −2sini 2cos j

r  z k 

4

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r  r  z

i  j k 

−2sin 2cos 0

0 0 1

i  j k 

−2sin 2cos 0

0 0 1

i  j

−2sin 2cos

0 0

2cosi 2sin j

|r  r  z | 1

S

 f  x, y, zds G

 f u, v| r u r v|dudv,

Thus

S

 x y zdS 0

2

0

42cos 2sin z 1dzd 

5