Calculus II: Final Exam Solutions - Math 221, Fall 2017 & 2018

This document presents solutions to final exams for Math 221, covering both Fall 2017 and Fall 2018. The solutions encompass a range of topics, including vector fields, the divergence theorem, Green's theorem, and Stokes' theorem. Specific problems address concepts such as the angle between vectors, the curl of a vector field, path independence, and the application of these theorems in various contexts. The solutions detail the application of theorems, calculations, and provide explanations for each step. The document includes solutions to problems from different exam years, offering a comprehensive guide for students studying calculus and vector analysis. The solutions are designed to aid in understanding and mastering the material covered in the Math 221 course.

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 sint )
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)
¿ f =∇ f = ∂
∂ x i+ ∂
∂ y j+ ∂
∂ z
This does not make sense because we cannot find the dot-product of a vector and scalar value.
Thus nonsense.
d) Grad(div F)
¿ F=∇ F=scalar value function
grad (¿ F )=∇( ∇ . F )
since ∇ is a vector value , ∇ . F is a scalar value
there product is also a vector value
Therefore, grad (div F) is a vector value
e) Div (grad F)
grad F=∇ F
There is no usual multiplication of two vectors
So grad F is nonsense and as a result so is div (grad F)
Question 2
By Greens’ theorem,
∫
c
❑
F . dr=∫
c
❑
Mdx+ Ndy
¿∬
D
❑
( dN
dx − dM
dy ) .dA
where M = y3 −ln x , N= √ y2 +1+3 x
dM
dy =3 y2 ; dN
dy =3
Thus ∫
c
❑
Fdx=∬
D
❑
(3−3 y2) . dA
x= y2=1
y=±1
¿ ∬
−1 y 2
1 1
( 3−3 y2)dx . dy=3∫
−1
1
( 1− y2 ) ∨¿ y2
1 dx ¿
∫
c
❑
F . dr=¿ 3∫
−1
1
( 1− y2 ) (1− y2 ) dy ¿
¿ 3∫
−1
1
( 1+ y4−2 y2 )dy
¿ 3 ( y+ y5
5 − 2
3 y3
)−1
1
¿ 3 (2+ 2
5 − 2
3 (2) )

¿ 3( 16
15 )
¿ 16
5
Question 3
Apply divergence theorem for closed surface as,
∬
s
❑
R . dS=∭
E
❑
(3 x2 +3 y2 +3 z2) dV
¿ 3∭
E
❑
(x2 + y2+ z2 )dV
¿ 3∭
E
❑
P2 dV =3∫
0
2 π
∫
0
π
2
∫
1
2
( P2 )P2 sin∅ dpd ∅ dθ
¿ 3 ( 2 π ) ⌊−cos ∅ ⌋ p5
5 1
2
¿ 3(2 π )(1) 32−1
5
¿ 6 π ( 32
5 )
¿ 186
5 π
Question 4
a) For fields to be path-independent, the curl should be zero
Thus, ∇ × F=0
curl of F=
i j k
δx δy δz
− A ysin x cos x +2 z y e y2
B e y2
+2 z
¿ i ( ( e y2
+2 x ) ( 2 y ) −2 y e y2
) − j ( 0−0 ) + k ¿
atB= ( 0,0,0 ) the field is conservative
→∫− ysinxdx=+ ycos x
∫¿ ¿
→∫ ( e y2
+ 2 z ) dz=z e y2
+¿ z2 ¿
MATH 221
DEC 2016

Question 1
a) v . u
¿ ( 1 ×1 ) + ( 2× 1 ) + (−2 ×0 )
¿ 1+2+0
¿ 3
b) |v|= √((−1¿¿ 2)¿+(22 )+(−22 )) ¿ ¿
¿ √ 17
¿ 4.123
c)
Question 2
a) Velocity , v = dr (t )
dt
¿ 2 t−2
b)
Question 3
a) Gradient
Vf ( x , y , z ) = d x2
dx + dsin( y−z )
dy + dsin( y−z )
dz
¿ ¿
b) The steepest part is found when
Vf ( x , y , z )= ( 0,0,0 )
Question 5
a) ¿ F=∇ F
¿ ¿)
¿ ¿)
b)
b
curl of F=
i j k
δx δy δz
x2 y +2 z 2 y −1
¿ 0−0+0
¿ 0
c) F is a path independent vector because its curl is zero
d) F is not an incompressible vector field because its divergence is not zero

Question 6
a) Parametrization of the cylinder
r ( θ , z ) =( cos θ , sinθ , z ), θ ∈ ( 0 ,2 π ) , z ∈ R