Review for Exam 4.
I Sections 16.1-16.5, 16.7, 16.8.
I 50 minutes.
I 5 problems, similar to homework problems.
I No calculators, no notes, no books, no phones.
I No green book needed.
Review for Exam 4.
I (16.1) Line integrals.
I (16.2) Vector fields, work, circulation, flux (plane).
I (16.3) Conservative fields, potential functions.
I (16.4) The Green Theorem in a plane.
I (16.5) Surface area, surface integrals.
I (16.7) The Stokes Theorem.
I (16.8) The Divergence Theorem.
Conservative fields, potential functions (16.3).Example
Is the field F = 〈y sin(z), x sin(z), xy cos(z)〉 conservative?If “yes”, then find the potential function.
Solution: We need to check the equations
∂yFz = ∂zFy , ∂xFz = ∂zFx , ∂xFy = ∂yFx .
∂yFz = x cos(z) = ∂zFy ,
∂xFz = y cos(z) = ∂zFx ,
∂xFy = sin(z) = ∂yFx .
Therefore, F is a conservative field, that means there exists ascalar field f such that F = ∇f . The equations for f are
∂x f = y sin(z), ∂y f = x sin(z), ∂z f = xy cos(z).
Conservative fields, potential functions (16.3).
Example
Is the field F = 〈y sin(z), x sin(z), xy cos(z)〉 conservative?If “yes”, then find the potential function.
Solution: ∂x f = y sin(z), ∂y f = x sin(z), ∂z f = xy cos(z).Integrating in x the first equation we get
f (x , y , z) = xy sin(z) + g(y , z).
Introduce this expression in the second equation above,
∂y f = x sin(z) + ∂yg = x sin(z) ⇒ ∂yg(y , z) = 0,
so g(y , z) = h(z). That is, f (x , y , z) = xy sin(z) + h(z).Introduce this expression into the last equation above,
∂z f = xy cos(z) + h′(z) = xy cos(z) ⇒ h′(z) = 0 ⇒ h(z) = c .
We conclude that f (x , y , z) = xy sin(z) + c . C
Conservative fields, potential functions (16.3).
Example
Compute I =
∫C
y sin(z) dx + x sin(z) dy + xy cos(z) dz , where C
given by r(t) = 〈cos(2πt), 1 + t5, cos2(2πt)π/2〉 for t ∈ [0, 1].
Solution: We know that the field F = 〈y sin(z), x sin(z), xy cos(z)〉conservative, so there exists f such that F = ∇f , or equivalently
df = y sin(z) dx + x sin(z) dy + xy cos(z) dz .
We have computed f already, f = xy sin(z) + c .Since F is conservative, the integral I is path independent, and
I =
∫ (1,2,π/2)
(1,1,π/2)
[y sin(z) dx + x sin(z) dy + xy cos(z) dz
]I = f (1, 2, π/2)− f (1, 1, π/2) = 2 sin(π/2)− sin(π/2) ⇒ I = 1.
Conservative fields, potential functions (16.3).
Example
Show that the differential form in the integral below is exact,∫C
[3x2 dx +
z2
ydy + 2z ln(y) dz
], y > 0.
Solution: We need to show that the field F =⟨3x2,
z2
y, 2z ln(y)
⟩is conservative. It is, since,
∂yFz =2z
y= ∂zFy , ∂xFz = 0 = ∂zFx , ∂xFy = 0 = ∂yFx .
Therefore, exists a scalar field f such that F = ∇f , or equivalently,
df = 3x2 dx +z2
ydy + 2z ln(y) dz .
Review for Exam 4.
I (16.1) Line integrals.
I (16.2) Vector fields, work, circulation, flux (plane).
I (16.3) Conservative fields, potential functions.
I (16.4) The Green Theorem in a plane.
I (16.5) Surface area, surface integrals.
I (16.7) The Stokes Theorem.
I (16.8) The Divergence Theorem.
The Green Theorem in a plane (16.4).
Example
Use the Green Theorem in the plane to evaluate the line integral
given by
∮C
[(6y + x) dx + (y + 2x) dy
]on the circle C defined by
(x − 1)2 + (y − 3)2 = 4.
Solution: Recall:
∮C
F · dr =
∫∫S
(∂xFy − ∂yFx
)dx dy .
Here F = 〈(6y + x), (y + 2x)〉. Since ∂xFy = 2 and ∂yFx = 6,Green’s Theorem implies∮
C
[(6y + x) dx + (y + 2x) dy
]=
∮C
F · dr =
∫∫S
(2− 6) dx dy .
Since the area of the disk S = {(x − 1)2 + (y − 3)2 6 4} is π(22),∮C
F · dr = −4
∫∫S
dx dy = −4(4π) ⇒∮
C
F · dr = −16π.
Review for Exam 4.
I (16.1) Line integrals.
I (16.2) Vector fields, work, circulation, flux (plane).
I (16.3) Conservative fields, potential functions.
I (16.4) The Green Theorem in a plane.
I (16.5) Surface area, surface integrals.
I (16.7) The Stokes Theorem.
I (16.8) The Divergence Theorem.
Surface area, surface integrals (16.5).Example
Integrate the function g(x , y , z) = x√
4 + y2 over the surface cutfrom the parabolic cylinder z = 4− y2/4 by the planes x = 0,x = 1 and z = 0.
Solution:
R
z
4
1
4
x
y
S
We must compute: I =
∫∫S
g dσ.
Recall dσ =|∇f ||∇f · k|
dx dy , with k ⊥ R
and in this case f (x , y , z) = y2 + 4z − 16.
∇f = 〈0, 2y , 4〉 ⇒ |∇f | =√
16 + 4y2 = 2√
4 + y2.
Since R = [0, 1]× [−4, 4], its normal vector is k and |∇f · k| = 4.Then, ∫∫
S
g dσ =
∫∫R
(x√
4 + y2) 2
√4 + y2
4dx dy .
Surface area, surface integrals (16.5).
Example
Integrate the function g(x , y , z) = x√
4 + y2 over the surface cutfrom the parabolic cylinder z = 4− y2/4 by the planes x = 0,x = 1 and z = 0.
Solution:
∫∫S
g dσ =
∫∫R
(x√
4 + y2) 2
√4 + y2
4dx dy .
∫∫S
g dσ =1
2
∫∫R
x(4 + y2) dx dy =1
2
∫ 4
−4
∫ 1
0x(4 + y2) dx dy
∫∫S
g dσ =1
2
[∫ 4
−4(4+y2) dy
][∫ 1
0x dx
]=
1
2
(4y+
y3
3
)∣∣∣4−4
(x2
2
)∣∣∣10∫∫
S
g dσ =1
22(42 +
43
3
)1
2= 8
(1 +
4
3
)⇒
∫∫S
g dσ =56
3.
Review for Exam 4.
I (16.1) Line integrals.
I (16.2) Vector fields, work, circulation, flux (plane).
I (16.3) Conservative fields, potential functions.
I (16.4) The Green Theorem in a plane.
I (16.5) Surface area, surface integrals.
I (16.7) The Stokes Theorem.
I (16.8) The Divergence Theorem.
The Stokes Theorem (16.7).
Example
Use Stokes’ Theorem to find the flux of ∇× F outward throughthe surface S , where F = 〈−y , x , x2〉 andS = {x2 + y2 = a2, z ∈ [0, h]} ∪ {x2 + y2 6 a2, z = h}.
Solution: Recall:
∫∫S
(∇× F) · n dσ =
∮C
F · dr.
The surface S is the cylinder walls and its cover at z = h.Therefore, the curve C is the circle x2 + y2 = a2 at z = 0.That circle can be parametrized (counterclockwise) asr(t) = 〈a cos(t), a sin(t)〉 for t ∈ [0, 2π].∫∫
S
(∇× F) · n dσ =
∮C
F · dr =
∫ 2π
0F(t) · r′(t) dt,
where F(t) = 〈−a sin(t), a cos(t), a2 cos2(t)〉 andr′(t) = 〈−a sin(t), a cos(t), 0〉.
The Stokes Theorem (16.7).
Example
Use Stokes’ Theorem to find the flux of ∇× F outward throughthe surface S , where F = 〈−y , x , x2〉 andS = {x2 + y2 = a2, z ∈ [0, h]} ∪ {x2 + y2 6 a2, z = h}.
Solution: F(t) = 〈−a sin(t), a cos(t), a2 cos2(t)〉 andr′(t) = 〈−a sin(t), a cos(t), 0〉. Hence∫∫
S
(∇× F) · n dσ =
∫ 2π
0F(t) · r′(t) dt,
∫∫S
(∇× F) · n dσ =
∫ 2π
0
(a2 sin2(t) + a2 cos2(t)
)dt =
∫ 2π
0a2 dt.
We conclude that
∫∫S
(∇× F) · n dσ = 2πa2.
Review for Exam 4.
I (16.1) Line integrals.
I (16.2) Vector fields, work, circulation, flux (plane).
I (16.3) Conservative fields, potential functions.
I (16.4) The Green Theorem in a plane.
I (16.5) Surface area, surface integrals.
I (16.7) The Stokes Theorem.
I (16.8) The Divergence Theorem.
The Divergence Theorem (16.8).
Example
Use the Divergence Theorem to find the outward flux of the fieldF = 〈x2,−2xy , 3xz〉 across the boundary of the regionD = {x2 + y2 + z2 6 4, x > 0, y > 0, z > 0}.
Solution: Recall:
∫∫S
F · n dσ =
∫∫∫D
(∇ · F)dv .
∇ · F = ∂xFx + ∂yFy + ∂zFz = 2x − 2x + 3x ⇒ ∇ · F = 3x .∫∫S
F · n dσ =
∫∫∫D
(∇ · F)dv =
∫∫D
3x dx dy dz .
It is convenient to use spherical coordinates:∫∫S
F ·n dσ =
∫ π/2
0
∫ π/2
0
∫ 2
0
[3ρ sin(φ) cos(φ)
]ρ2 sin(φ) dρ dφ dθ.
The Divergence Theorem (16.8).
Example
Use the Divergence Theorem to find the outward flux of the fieldF = 〈x2,−2xy , 3xz〉 across the boundary of the regionD = {x2 + y2 + z2 6 4, x > 0, y > 0, z > 0}.
Solution:∫∫S
F ·n dσ =
∫ π/2
0
∫ π/2
0
∫ 2
0
[3ρ sin(φ) cos(φ)
]ρ2 sin(φ) dρ dφ dθ.
∫∫S
F · n dσ =[∫ π/2
0cos(θ) dθ
][∫ π/2
0sin2(φ) dφ
][∫ 2
03ρ3 dρ
]∫∫
S
F · n dσ =[sin(θ)
∣∣∣π/2
0
][1
2
∫ π/2
0
(1− cos(2φ)
)dφ
][3
4ρ4
∣∣20
]∫∫
S
F · n dσ = (1)1
2
(π
2
)(12) ⇒
∫∫S
F · n dσ = 3π.
Review for the Final Exam.
I Monday, December 13, 10:00am - 12:00 noon. (2 hours.)I Places:
I Sctns 001, 002, 005, 006 in E-100 VMC (Vet. Medical Ctr.),I Sctns 003, 004, in 108 EBH (Ernst Bessey Hall);I Sctns 007, 008, in 339 CSE (Case Halls).
I Chapters 12-16.
I Problems, similar to homework problems.
I No calculators, no notes, no books, no phones.
I No green book needed.
Review for Final Exam.
I Chapter 16, Sections 16.1-16.5, 16.7, 16.8.
I Chapter 15, Sections 15.1-15.4, 15.6.
I Chapter 14, Sections 14.1-14.7.
I Chapter 13, Sections 13.1, 13.3.
I Chapter 12, Sections 12.1-12.6.
Chapter 16, Integration in vector fields.Example
Use the Divergence Theorem to find the flux of F = 〈xy2, x2y , y〉outward through the surface of the region enclosed by the cylinderx2 + y2 = 1 and the planes z = −1, and z = 1.
Solution: Recall:
∫∫S
F · n dσ =
∫∫∫D
(∇ · F) dv . We start with
∇ · F = ∂x(xy2) + ∂y (x2y) + ∂z(y) ⇒ ∇ · F = y2 + x2.
The integration region is D = {x2 + y2 6 1, z ∈ [−1, 1]}. So,
I =
∫∫∫D
(∇ · F) dv =
∫∫∫D
(x2 + y2) dx dy dz .
We use cylindrical coordinates,
I =
∫ 2π
0
∫ 1
0
∫ 1
−1r2 dz r dr dθ = 2π
[∫ 1
0r3 dr
](2) = 4π
( r4
4
∣∣∣10
).
We conclude that
∫∫S
F · n dσ = π. C
Chapter 16, Integration in vector fields.
Example
Use Stokes’ Theorem to find the work done by the forceF = 〈2xz , xy , yz〉 along the path C given by the intersection of theplane x + y + z = 1 with the first octant, counterclockwise whenviewed from above.
Solution:
C
z
x + y + z = 11
1
x
1 y
x + y = 1
S
R
Recall:
∫C
F · dr =
∫∫S
(∇× F) · n dσ.
The surface S is the level surface f = 0 of
f = x + y + z − 1
therefore, ∇f = 〈1, 1, 1〉, |∇f | =√
3 and|∇f · k| = 1.
n =∇f
|∇f |=
1√3〈1, 1, 1〉, dσ =
|∇f ||∇f · k|
dx dy =√
3 dx dy .
Chapter 16, Integration in vector fields.
Example
Use Stokes’ Theorem to find the work done by the forceF = 〈2xz , xy , yz〉 along the path C given by the intersection of theplane x + y + z = 1 with the first octant, counterclockwise whenviewed from above.
Solution: n =1√3〈1, 1, 1〉 and dσ =
√3 dx dy .
We now compute the curl of F,
∇× F =
∣∣∣∣∣∣i j k
∂x ∂y ∂z
2xz xy yz
∣∣∣∣∣∣ = 〈(z − 0),−(0− 2x), (y − 0)〉
so ∇× F = 〈z , 2x , y〉. Therefore,∫∫S
(∇× F) · n dσ =
∫∫R
(〈z , 2x , y〉 · 1√
3〈1, 1, 1〉
)√3 dx dy
Chapter 16, Integration in vector fields.
Example
Use Stokes’ Theorem to find the work done by the forceF = 〈2xz , xy , yz〉 along the path C given by the intersection of theplane x + y + z = 1 with the first octant, counterclockwise whenviewed from above.
Solution:
I =
∫∫S
(∇× F) · n dσ =
∫∫R
(〈z , 2x , y〉 · 1√
3〈1, 1, 1〉
)√3 dx dy .
I =
∫∫R
(z + 2x + y) dx dy , z = 1− x − y ,
I =
∫ 1
0
∫ 1−x
0(1+x) dy dx =
∫ 1
0(1+x)(1−x) dx =
∫ 1
0(1−x2) dx .
I = x∣∣∣10− x3
3
∣∣∣10
= 1− 1
3=
2
3⇒
∫C
F · dr =2
3.
Chapter 16, Integration in vector fields.
Example
Find the area of the cone S given by z =√
x2 + y2 for z ∈ [0, 1].Also find the flux of the field F = 〈x , y , 0〉 outward through S .
Solution:
1
x
y
z
S
R
1
1
Recall: A(S) =
∫∫S
dσ. The surface S is the
level surface f = 0 of the functionf = x2 + y2 − z2. Also recall that
dσ =|∇f ||∇f · k|
dx dy .
Since ∇f = 2〈x , y ,−z〉, we get that
|∇f | = 2√
x2 + y2 + z2, z2 = x2 + y2 ⇒ |∇f | = 2√
2 z .
Also |∇f · k| = 2z , therefore, dσ =√
2 dx dy , and then we obtain
A(S) =
∫∫R
√2 dx dy =
∫ 2π
0
∫ 1
0
√2r dr dθ = 2π
√2r2
2
∣∣∣10
=√
2 π.
Chapter 16, Integration in vector fields.
Example
Find the area of the cone S given by z =√
x2 + y2 for z ∈ [0, 1].Also find the flux of the field F = 〈x , y , 0〉 outward through S .
Solution: We now compute the outward flux I =
∫∫S
F · n dσ.
Since
n =∇f
|∇f |=
1√2 z〈x , y ,−z〉.
I =
∫∫R
1√2 z
(x2 + y2)√
2 dx dy =
∫∫R
√x2 + y2 dx dy .
Using polar coordinates, we obtain
I =
∫ 2π
0
∫ 1
0r r dr dθ = 2π
r3
3
∣∣∣10
⇒ I =2π
3.
Review for Final Exam.
I Chapter 16, Sections 16.1-16.5, 16.7, 16.8.
I Chapter 15, Sections 15.1-15.4, 15.6.
I Chapter 14, Sections 14.1-14.7.
I Chapter 13, Sections 13.1, 13.3.
I Chapter 12, Sections 12.1-12.6.
Chapter 15, Multiple integrals.
Example
Find the volume of the region bounded by the paraboloidz = 1− x2 − y2 and the plane z = 0.
Solution:
y
D
Rx
z
1
1
1
So, D = {x2 + y2 6 1, 0 6 z 6 1− x2 − y2},and R = {x2 + y2 6 1, z = 0}. We know that
V (D) =
∫∫∫D
dv =
∫∫R
∫ 1−x2−y2
0dz dx dy .
Using cylindrical coordinates (r , θ, z), we get
V (D) =
∫ 2π
0
∫ 1
0
∫ 1−r2
0dz r dr dθ = 2π
∫ 1
0(1− r2) r dr .
Substituting u = 1− r2, so du = −2r dr , we obtain
V (D) = 2π
∫ 0
1u
(−du)
2= π
∫ 1
0u du = π
u2
2
∣∣∣10
⇒ V (D) =π
2.
Chapter 15, Multiple integrals.
Example
Set up the integrals needed to compute the average of the functionf (x , y , z) = z sin(x) on the bounded region D in the first octantbounded by the plane z = 4− 2x − y . Do not evaluate theintegrals.
Solution: Recall: f =1
V (D)
∫∫∫D
f dv .
D
z
x
yR
2x + y + z = 4
2x + y = 42
4
4
Since V (D) =
∫ 2
0
∫ 4−2x
0
∫ 4−2x−y
0dz dy dx ,
we conclude that
f =
∫ 2
0
∫ 4−2x
0
∫ 4−2x−y
0z sin(x) dz dy dx∫ 2
0
∫ 4−2x
0
∫ 4−2x−y
0dz dy dx
.
Chapter 15, Multiple integrals.Example
Reverse the order of integration and evaluate the double integral
I =
∫ 4
0
∫ 2
y/2ex2
dx dy .
Solution: We see that y ∈ [0, 4] and x ∈ [0, y/2], that is,
y = 2x
y
x2
4Therefore, reversing the integration ordermeans
I =
∫ 2
0
∫ 2x
0ex2
dy dx .
This integral is simple to compute,
I =
∫ 2
0ex2
x dx , u = x2, du = 2x dx ,
I =
∫ 4
0eu du ⇒ I = e4 − 1.
Review for the Final Exam.
I Monday, December 13, 10:00am - 12:00 noon. (2 hours.)I Places:
I Sctns 001, 002, 005, 006 in E-100 VMC (Vet. Medical Ctr.),I Sctns 003, 004, in 108 EBH (Ernst Bessey Hall);I Sctns 007, 008, in 339 CSE (Case Halls).
I Chapters 12-16.
I ∼ 12 Problems, similar to homework problems.
I No calculators, no notes, no books, no phones.
Plan for today: Practice final exam: April 30, 2001.
Remark on Chapter 16.
Remark: The normal form of Green’s Theorem is atwo-dimensional restriction of the Divergence Theorem.
I The Divergence Theorem:
∫∫S
F · n dσ =
∫∫∫D
(∇ · F) dv .
I Normal form of Green’s Thrm:
∮C
F · n ds =
∫∫S
(∇ · F) dA.
Remark: The tangential form of Green’s Theorem is a particularcase of the Stokes Theorem when C , S are flat (on z = 0 plane).
I The Stokes Theorem:
∮C
F · dr =
∫∫S
(∇× F) · n dσ.
I Tang. form of Green’s Thrm:
∮C
F · dr =
∫∫S
(∇× F) · k dA.
Practice final exam: April 30, 2001. Prbl. 1.
Example
Given A = (1, 2, 3), B = (6, 5, 4) and C = (8, 9, 7), find thefollowing:
I−→AB and
−→AC .
Solution:−→AB = 〈(6− 1), (5− 2), (4− 3)〉, hence
−→AB = 〈5, 3, 1〉. In the same way
−→AC = 〈7, 7, 4〉.
I−→AB +
−→AC = 〈12, 10, 5〉.
I−→AB ·
−→AC = 35 + 21 + 4.
I−→AB ×
−→AC =
∣∣∣∣∣∣i j k5 3 17 7 4
∣∣∣∣∣∣ = 〈(12− 7),−(20− 7), (35− 21)〉,
hence−→AB ×
−→AC = 〈5,−13, 14〉.
Practice final exam: April 30, 2001. Prbl. 2.
Example
Find the parametric equation of the line through the point(1, 0,−1) and perpendicular to the plane 2x − 3y + 5x = 35. Thenfind the intersection of the line and the plane.
Solution: The normal vector to the plane 〈2,−3, 5〉 is the tangentvector to the line. Therefore,
r(t) = 〈1, 0,−1〉+ t 〈2,−3, 5〉,so the parametric equations of the line are
x(t) = 1 + 2t, y(t) = −3t, z(t) = −1 + 5t.
The intersection point has a t solution of
2(1+2t)−3(−3t)+5(−1+5t) = 35 ⇒ 2+4t+9t−5+25t = 35
38t = 38 ⇒ t = 1 ⇒ r(1) = 〈3,−3, 4〉.
Practice final exam: April 30, 2001. Prbl. 3.
Example
The velocity of a particle is given by v(t) = 〈t2, (t3 + 1)〉, and theparticle is at 〈2, 1〉 for t = 0.
I Where is the particle at t = 2?
Solution: r(t) =⟨( t3
3+ rx
),( t4
4+ t + ry
)⟩. Since
r(0) = 〈2, 1〉, we get that r(t) =⟨( t3
3+ 2
),( t4
4+ t + 1
)⟩.
Hence r(2) = 〈8/3 + 2, 7〉.I Find an expression for the particle arc length for t ∈ [0, 2].
Solution: s(t) =
∫ t
0
√τ4 + (τ3 + 1)2 dτ .
I Find the particle acceleration.
Solution: a(t) = 〈2t, 3t2〉.
Practice final exam: April 30, 2001. Prbl. 4.Example
I Draw a rough sketch of the surface z = 2x2 + 3y2 + 5.
Solution: This is a paraboloid along the vertical direction,opens up, with vertex at z = 5 on the z-axis, and the x-radiusis a bit longer than the y -radius.
I Find the equation of the tangent plane to the surface at thepoint (1, 1, 10).
Solution: Introduce f (x , y) = 2x2 + 3y2 + 5, then
L(1,1)(x , y) = ∂x f (1, 1) (x − 1) + ∂y f (1, 1) (y − 1) + f (1, 1).
Since f (1, 1) = 10, and ∂x f = 4x , ∂y f = 6y , then
z = L(1,1)(x , y) = 4(x − 1) + 6(y − 1) + 10.
Practice final exam: April 30, 2001. Prbl. 5.
Example
Let w = f (x , y) and x = s2 + t2, y = st2. If ∂x f = x − y and∂y f = y − x , find ∂sw and ∂tw in terms of s and t.
Solution:
∂sw = ∂x f ∂sx+∂y f ∂sy = (x−y)2s+(y−x)t2 = (x−y)(2s−t2).
Therefore, ∂sw = (s2 + t2 − st2)(2s − t2).
∂tw = ∂x f ∂tx+∂y f ∂ty = (x−y)2t+(y−x)2st = (x−y)(2t−2st).
Therefore, ∂tw = (s2 + t2 − st2)2t(1− s).
Practice final exam: April 30, 2001. Prbl. 6.
Example
Find all critical points of the function f (x , y) = 2x2 + 8xy + y4
and determine whether they re local maximum, minimum of saddlepoints.
Solution:
∇f = 〈(4x + 8y), (8x + 4y3)〉 = 〈0, 0〉 ⇒
{x + 2y = 0,
2x + y3 = 0.
−4y + y3 = 0 ⇒
y = 0 ⇒ x = 0 ⇒ P0 = (0, 0)
y = ±2 ⇒ x = ∓4 ⇒
{P1 = (4,−2)
P2 = (−4, 2)
Since fxx = 4, fyy = 12y2, and fxy = 8, we conclude thatD = 3(16)y2 − 4(16).
Practice final exam: April 30, 2001. Prbl. 6.
Example
Find all critical points of the function f (x , y) = 2x2 + 8xy + y4
and determine whether they re local maximum, minimum of saddlepoints.
Solution:P0 = (0, 0), P1 = (4,−2), P2 = (−4, 2), D = 3(16)y2 − 4(16).
D(0, 0) = −4(16) < 0 ⇒ P0 = (0, 0) saddle point.
D(4,−2) = 12(16)− 4(16) > 0, fxx = 4 ⇒ P1 = (4,−2) min.
D(−4, 2) = 12(16)− 4(16) > 0, fxx = 4 ⇒ P1 = (−4, 2) min.
Practice final exam: April 30, 2001. Prbl. 7.
Example
Evaluate the integral I =
∫ 1
0
∫ √x
xy dy dx by reversing the order
of integration.
Solution: The integration region is the set in the square[0, 1]× [0, 1] in between the curves y = x and y =
√x . Therefore,
I =
∫ 1
0
∫ y
y2
y dx dy =
∫ 1
0y(y − y2) dy =
∫ 1
0(y2 − y3) dy
I =(y3
3− y4
4
)∣∣∣10
=1
3− 1
4⇒ I =
1
12.
Practice final exam: April 30, 2001. Prbl. 8.
Example
Find the work done by the force F = 〈yz , xz ,−xy〉 on a particlemoving along the path r(t) = 〈t3, t2, t〉 for t ∈ [0, 2].
Solution:
W =
∫C
F · dr =
∫ 2
0F(t) · r′(t) dt,
where F(t) = 〈t3, t4,−t5〉 and r′(t) = 〈3t2, 2t, 1〉. Hence
W =
∫ 2
0(3t5 + 2t5 − t5) dt =
∫ 2
04t5 dt =
4
6t6
∣∣∣20
=2
326.
Therefore, W = 27/3.
Practice final exam: April 30, 2001. Prbl. 9.
Example
Show that the force fieldF = 〈(y cos(z)− yzex), (x cos(z)− zex), (−xy sin(z)− yex)〉 isconservative. Then find its potential function. Then evaluate
I =
∫C
F · dr for r(t) = 〈t, t2, πt3〉.
Solution: The field F is conservative, since
∂xFy = cos(z)− zex = ∂yFx ,
∂xFz = −xy sin(z)− yex = ∂zFx ,
∂yFz = −x sin(z)− ex = ∂zFy .
The potential function is a scalar function f solution of
∂x f = y cos(z)− yzex , ∂y f = x cos(z)− zex , ∂z f = −xy sin(z)− yex .
Practice final exam: April 30, 2001. Prbl. 9.
Example
Show that the force fieldF = 〈(y cos(z)− yzex), (x cos(z)− zex), (−xy sin(z)− yex)〉 isconservative. Then find its potential function. Then evaluate
I =
∫C
F · dr for r(t) = 〈t, t2, πt3〉.
Solution: Recall:
∂x f = y cos(z)− yzex , ∂y f = x cos(z)− zex , ∂z f = −xy sin(z)− yex .
The x-integral of the first equation impliesf = xy cos(z)− yzex + g(y , z). Introduce f into the secondequation above,
x cos(z)− zex + ∂yg = x cos(z)− zex ⇒ ∂yg(y , z) = 0,
so we conclude g(y , z) = h(z), hence f = xy cos(z)− yzex + h(z).
Practice final exam: April 30, 2001. Prbl. 9.
Example
Show that the force fieldF = 〈(y cos(z)− yzex), (x cos(z)− zex), (−xy sin(z)− yex)〉 isconservative. Then find its potential function. Then evaluate
I =
∫C
F · dr for r(t) = 〈t, t2, πt3〉.
Solution: Recall: f = xy cos(z)− yzex + h(z).
Introduce f into the equation ∂z f = −xy sin(z)− yex , that is,
−xy sin(z)− ex + h′(z) = −xy sin(z)− yex ⇒ h′(z) = 0.
So, h(z) = c , a constant, hence f = xy cos(z)− yzex + c .
Finally
∫C
F · dr =
∫ (1,1,π)
(0,0,0)df = f (1, 1, π)− f (0, 0, 0).
So we conclude that
∫C
F · dr = −(1 + πe).
Practice final exam: April 30, 2001. Prbl. 10.
Example
Use the Green Theorem to evaluate the integral
∫C
Fx dx + Fy dy
where Fx = y + ex and Fy = 2x2 + cos(y) and C is the trianglewith vertices (0, 0), (0, 2) and (1, 1) traversed counterclockwise.
Solution: Denoting F = 〈Fx ,Fy 〉, Green’s Theorem says∫C
F · dr =
∫∫S
(∇× F) · k dA =
∫∫S
(∂xFy − ∂yFx) dA.
∫C
F · dr =
∫∫S
(4x − 1) dx dy =
∫ 1
0
∫ 2−y
y(4x − 1) dx dy .
A straightforward calculation gives
∫C
F · dr = 3.
Practice final exam: April 30, 2001. Prbl. 11.
Example
Find the surface area of the portion of the paraboloidz = 4− x2 − y2 that lies above the plane z = 0. Use polarcoordinates to evaluate the integral.
Solution:
A(S) =
∫∫S
dσ, dσ =|∇f ||∇f · k|
dx dy
where f = x2 + y2 + z − 4. Therefore,
∇f = 〈2x , 2y , 1〉 ⇒ |∇f | =√
1 + 4x2 + 4y2, ∇f · k = 1.
A(S) =
∫ 2π
0
∫ 2
0
√1 + 4r2 r dr dθ, u = 1 + 4r2, du = 8r dr .
The finally obtain A(S) = (π/6)(173/2 − 1).
Practice final exam: April 30, 2001. Prbl. 12.Example
Use the Stokes Theorem to evaluate I =
∫∫S
[∇× (y i)
]· n dσ
where S is the hemisphere x2 + y2 + z2 = 1, with z > 0.
Solution: F = 〈y , 0, 0〉. The border of the hemisphere is given bythe circle x2 + y2 = 1, with z = 0. This circle can be parametrizedfor t ∈ [0, 2π] as
r(t) = 〈cos(t), sin(t), 0〉 ⇒ r′(t) = 〈− sin(t), cos(t), 0〉,
and we also have F(t) = 〈sin(t), 0, 0〉. Therefore,∫∫S
(∇× F
)· n dσ =
∮ 2π
0F(t) · r′(t) dt = −
∫ 2π
0sin2(t) dt
∫∫S
(∇× F
)· n dσ = −1
2
∫ 2π
0
[1− cos(2t)
]dt.
Practice final exam: April 30, 2001. Prbl. 12.
Example
Use the Stokes Theorem to evaluate I =
∫∫S
[∇× (y i)
]· n dσ
where S is the hemisphere x2 + y2 + z2 = 1, with z > 0.
Solution:
∫∫S
(∇× F
)· n dσ = −1
2
∫ 2π
0
[1− cos(2t)
]dt.
Recall that ∫ 2π
0cos(2t) dt =
1
2
(sin(2t)
∣∣∣2π
0
)= 0.
Therefore, we obtain∫∫S
(∇× F
)· n dσ = −π.
C