Final Exam Question Paper Assignment

FINAL EXAM FALL 2017
Question 1
a) ∫ Fdr=∫|F |.∨dr∨¿ cos θ ¿
We see that the angle between F and dr is always acute
Thus θ →(0 , π
2 )
→ cos θ>0
Therefore, since all elements are positive, this is always positive
b) The angle between ∇ l∧dr is always acute and hence all elements being
positive the result is always positive.
c) Curl F represents the vector field perpendicular to the plane of F. Thus the
angle between the two is π
2 hence the cosine is zero. This makes the result to
be also zero.
d) S: x2+ y2+ z2=a2
Thus this depends on all the other terms
Question 2
a) C.
b) A
c) A
d) C
e) A
f) C
g) A
Question 3
∫
C
❑
F . dr=∫∫ dθ
dx − ∂ ρ
∂ y
Using Greens’ theorem
¿∫∫(3 x)dA
¿ ∫
− π
4
π
4
∫
0
2
(3 rcosθ)rdrdθ
¿ 3(sin θ)−π
4
π
4 r3
3 0
2
¿ δ ¿

¿ δ ( 1
√2 + 1
√ 2 )
¿ δ √ 2
Question 5
Part (a)
curl of F=
i j k
δx δy δz
− ysin x cos x +2 z y e y2
e y2
+2 z
¿ i ( ( e y2
+2 x ) ( 2 y ) −2 y e y2
) − j ( 0−0 ) + k ¿
at ( 0,0,0 ) the field is conservative
→∫− ysinxdx=+ ycos x
∫¿ ¿
→∫ ( e y2
+ 2 z ) dz=z e y2
+ ¿ z2 ¿
Thus the potential function is given by;
f ( x , y , z=F1=z e y2
+ ycos x + z2)
Part b
r ( t )=¿
r ( 0 ) = ( 0,0,0 )
r ( 1 )= ( π , 1,1+ 0 )=(1,1,0)
1 ( e12
) + ( 1 ) cos π +1 ¿−(0+ 0+0)
¿ e−1+1
¿ e
Question 7
By Stoke’s theorem,
∫∫
s
❑
curl F . dA=∮
c
❑
Fdr

Where C is a circle curved around the y-axis at y=1
r ( t ) =( 2cost t , 1,2 sin t )
Since the radius of the circle is 2,
r' ( t ) =¿
F ( r ( t ) )=(−2 sin t e' , 4 sin2 t eacos2 t , 2 cos t e ' )
F ( r ( t ) ) .r ' ( t )=4 sin2 te+ 0+4 e cos2 te
¿ 4 e ( sin2+ cos2 t ) =4 e ( 1 )=4 e
∮ F . dr =∫
0
2 π
Fr ( t ) . r' ( t ) . dt
¿∫
0
2 π
4 e . dt
¿ 8 πe
FINAL EXAM FALL 2018
Question 1
a) F × curl F
Let F=(F1 , F2 , F3)
curl F=¿
i j k
∂
∂ x
∂
∂ y
∂
∂ z
F1 F2 F3
∨¿
F × curl F=
i j k
F1 F2 F3
∂ F3
∂ y − ∂ F2
∂ x
∂ F1
∂ z − ∂ F3
∂ x
∂ F1
∂ y − ∂ F2
∂ x
Thus F is a vector valued
b) Div(grad f)
grad f =∇ f
Thus,
¿ ( grad F ) =∇. ∇ f
¿ ∇2 f
¿ ∂2 f
∂ x2 + ∂2 f
∂ y2 + ∂2 f
∂ z2
So, div(grad f) is scalar valued.
c) Div(div f)