Austin Cholley Zhu MAT 267 ONLINE A Spring 2021
Assignment Section 13.7 due 03/01/2021 at 11:59pm MST
1. (1 point) Evaluate Z ZSp1+x2+y2dS where Sis the he-
licoid: r(u,v) = ucos(v)i+usin(v)j+vk, with 0 ≤u≤5,0≤
v≤2π
Solution:
SOLUTION:
ru×rv=hcosv,sinv,0i × h−usin v,ucosv,1i=
hsinv,−cosv,ui.
Thus |ru×rv|=psin2v+cos2v+u2=√1+u2.
Let D={(u,v)|0≤u≤5,0≤v≤2π}.
Then
RRSp1+x2+y2dS =RRDp1+u2cos2v+u2sin2v|ru×rv|dA
=RRD√1+u2√1+u2dA
=R5
0R2π
0(1+u2)dv du
=2πR5
0(1+u2)du
=2πhu+u3
3i5
0
=280
3π
Answer(s) submitted:
•((280pi )/3)
(correct)
Correct Answers:
•293.215314335047
2. (1 point) Let Sbe the part of the plane 3x+5y+z=3
which lies in the first octant, oriented upward. Find the flux of
the vector field F=4i+2j+3kacross the surface S.
Solution:
SOLUTION
Sis the region in the plane z=3−3x−5yover D=
{(x,y)|0≤x≤1,0≤y≤3
5−3
5x}.
Using xand yas parameters, we have r(x,y) = hx,y,3−3x−5yi.
Then rx×ry=h1,0,−3i×h0,1,−5i=h3,5,1i.
Since the zcomponent is positive, this normal has the correct
orientation.
Then
ZZS
F·dS=ZZD
F(r(x,y)) ·(rx×ry)dA
=ZZDh4,2,3i·h3,5,1idA
=ZZD
25dA
=25Z1
0Z3
5−3
5x
0
dy dx
=25Z1
03
5−3
5xdx
=15
2
Answer(s) submitted:
•48
(incorrect)
Correct Answers:
•7.5
3. (1 point) A fluid has density 3 kg/m3and flows in a veloc-
ity field v=−yi+xj+4zkwhere x,y,and zare measured in
meters and the components of vin meters per second.
Find the rate of flow outward through the sphere x2+y2+z2=1
Solution:
SOLUTION
The rate of flow through the sphere is the flux RRSρv·dS,0≤
θ≤2π,0≤φ≤πand ρ=3 kg/m3
A parametric representation of the sphere is r(φ,θ) =
1sinφcosθi+1 sin φsin θj+1 cos φkand the outward orien-
tation is given by rφ×rθ=1 sin2φcos θi+1 sin2φsin θj+
1sinφcosφk.
We have v(r(φ,θ)) = −1sinφsinθi+1sinφcosθj+4cosφk
Thus the rate of flow through Sis
RRSρv·dS=3R2π
0Rπ
0v(r(θ,φ)·rφ×rθdφdθ
=3R2π
0Rπ
0−1cosφsin2φsinθcosθ+1 cos φsin2φcos θsin θ
+4cos2φsinφdφdθ
=3R2π
0Rπ
04cos2φsinφdφdθ
=24πRπ
0cos2φsinφdφ
Using the substitution u=cosφ, yields
RRSρv·dS=24πR1
−1u2du
=48
3πKg/s
Answer(s) submitted:
•16pi
(correct)
Correct Answers:
•50.26544
4. (1 point) Let Mbe the closed surface that consists of the
hemisphere
M1:x2+y2+z2=1,z≥0,
and its base
M2:x2+y2≤1,z=0.
Let Ebe the electric field defined by E=h20x,20y,20zi. Find
the electric flux across M. Write the integral over the hemi-
sphere using spherical coordinates, and use the outward point-
ing normal.
ZZM1
E·dS=Zb
aZd
c
f(θ,φ)dθdφ,
where
a=,b=,c=,d=,
1
Using tfor θand pfor φ,
f(θ,φ) =
RRM1E·dS=
RRM2E·dS=, so
RRME·dS=.
Solution:
SOLUTION
On M1:
E(r(θ,φ)) = h20sinφcosθ,20sinφsinθ,20cos φi,0≤θ≤
2π,0≤φ≤π/2
rφ×rθ=hsin2φcosθ,sin2φsinθ,sinφcosφiis the outward
pointing normal.
Thus
ZZM1
E·dS=Zπ/2
0Z2π
0
E(r(θ,φ)) ·rφ×rθdθdφ
=Zπ/2
0Z2π
020sin3φcos2θ+20sin3φsin2θ+20sin φcos2φdθdφ
=Zπ/2
0Z2π
020sin3φcos2θ+sin2θ+20sinφcos2φdθdφ
=Zπ/2
0Z2π
020sin3φ+20sinφcos2φdθdφ
=Zπ/2
0Z2π
0
20sinφsin2φ+cos2φdθdφ
=Zπ/2
0Z2π
0
20sinφdθdφ
=2π(20)Zπ/2
0
sinφdφ
=40π[−cosφ]π/2
0
=40π
On M2:
r(x,y) = hx,y,0iwith D=(x,y)|x2+y2≤1.
Then E(r(x,y)) = h20x,20y,0iand ry×rx=h0,0,−1iso
ZZM2
E·dS=ZZD
E(r(x,y)) ·(ry×rx)dA =0.
Hence ZZM
E·dS=ZZM1
E·dS+ZZM2
E·dS=40π
Answer(s) submitted:
•0
•(pi /2)
•0
•2pi
•(sinˆ(2)(p)cos(t)sinˆ(2)(p)sin(t)sinpcosp)
•
•
•40pi
(score 0.625)
Correct Answers:
•0
•1.5707963267949
•0
•6.28318530717959
•20*(sin(p)**3 + sin(p)*cos(p)**2)
•125.663706143592
•0
•125.663706143592
5. (1 point) Determine whether the flux of the vector field ~
F
through each surface is positive, negative, or zero. In each case,
the orientation of the surface is indicated by the gray normal
vector.
? ? ?
? ?
(Click and drag to rotate)
Solution:
SOLUTION
The vector field is parallel to the surface. Thus the flux is
zero.
The vector field is parallel to the surface. Thus the flux is
zero.
The vector field is transverse to the surface and it has the same
orientation as the surface. Thus the flux is positive
The vector field is transverse to the surface and oriented in
the direction opposite of the orientation of the surface. Thus the
flux is negative.
The vector field is parallel to the surface. Thus the flux is
zero.
Answer(s) submitted:
•Zero
•Zero
•Positive
2
•Zero
•Negative
(score 0.6)
Correct Answers:
•ZERO
•ZERO
•POSITIVE
•NEGATIVE
•ZERO
6. (1 point)
Compute the flux of the vector field ~
F=y
~
i+3~
j−xz
~
kthrough
the surface S, which is the surface y=x2+z2, with x2+z2≤4,
oriented in the positive y-direction.
flux =
Solution:
SOLUTION
Since y=f(x,z) = x2+z2, we have
d~
A= (−fx
~
i+~
j−fz
~
k)dx dz = (−2x
~
i+~
j−2z
~
k)dx dz.
Thus, substituting y=x2+z2into ~
F, we have
ZS
~
F·d~
A=Zx2+z2≤4
((x2+z2)
~
i+3~
j−xz
~
k)·(−2x
~
i+~
j−2z
~
k)dx dz
=Zx2+z2≤4
(−2x3−2xz2+3+2xz2)dxdz.
So
flux =Z2
−2Z√4−z2
−√4−z2(3−2x3)dx dz =Z2
−2Z√4−z2
−√4−z23dx dz −Z2
−2Z√4−z2
−√4−z22x3dx dz.
The first of these is just the 3 times the area of the disk, and the
second is zero by symmetry, so
flux =3(π4) = 12π.
Answer(s) submitted:
•12pi
(correct)
Correct Answers:
•pi*3*4
7. (1 point)
Compute the flux of the vector field ~
F=9x2y2z
~
kthrough the
surface Swhich is the cone px2+y2=z, with 0 ≤z≤R, ori-
ented downward.
(a) Parameterize the cone using cylindrical coordinates
(write θas theta ).
x(r,θ) =
y(r,θ) =
z(r,θ) =
with ≤r≤
and ≤θ≤
(b) With this parameterization, what is d~
A?
d~
A=
(c) Find the flux of ~
Fthrough S.
flux =
Solution:
SOLUTION
Using cylindrical coordinates, we see that the surface Sis
parameterized by
~r(r,θ) = rcos θ
~
i+rsinθ~
j+r
~
k,
with 0 ≤r≤Rand 0 ≤θ≤2π. We have
∂~r
∂r×∂~r
∂θ =
~
i~
j~
k
cosθsinθ1
−rsinθrcosθ0
=−rcosθ
~
i−rsinθ~
j+r
~
k.
Since the vector ∂~r/∂r×∂~r/∂θ points upward, in the direction
opposite to the specified orientation, we use
d~
A=−(∂~r/∂r×∂~r/∂θ)dr dθ= (rcos θ
~
i+rsinθ~
j−r
~
k)dr dθ.
Hence
ZS
~
F·d~
A=Z2π
0ZR
0
9(r5cos2θsin2θ
~
k)·(rcosθ
~
i+rsinθ~
j−r
~
k)dr dθ
=Z2π
0ZR
0−9r6cos2θsin2θdr dθ
=−9R7
7Z2π
0
sin2θcos2θdθ
=−9R7
7Z2π
0
sin2θ(1−sin2θ)dθ
=−R7
7Z2π
0
(sin2θ−sin4θ)dθ
=−9R7
7π
4=−9
28 πR7.
The cone is not differentiable at the point (0,0). However the
flux integral, which is improper, converges.
Answer(s) submitted:
•9
•9
•9
•0
•9
•0
•9
•
•
(score 0.222222222222222)
Correct Answers:
•r*cos(theta)
•r*sin(theta)
•r
•0
•R
•0
•2*pi
•(r*cos(theta)i+r*sin(theta)j-rk)*dr*dtheta
•-9*pi*Rˆ7/28
3
8. (1 point)
Compute the flux of ~
F=x
~
i+y~
j+z
~
kthrough just the curved
surface of the cylinder x2+y2=4 bounded below by the plane
x+y+z=1, above by the plane x+y+z=4, and oriented
away from the z-axis.
flux =
Solution:
SOLUTION
The curved surface of the circular cylinder is parameterized
by
~r=x
~
i+y~
j+z
~
k=2cost
~
i+2sint~
j+s
~
k,
where 0 ≤t≤2πand 1−2cost−2 sint≤s≤4−2cost−2 sint.
The vector ∂~r/∂t×∂~r/∂spoints away from the z-axis, so
d~
A= (∂~r/∂t×∂~r/∂s)dsdt and
~
F·d~
A=
x y z
−2sint2 cost0
0 0 1
ds dt = (2xcost+2ysint)ds dt.
Plugging in for xand y, this is
~
F·d~
A=4(cos2t+sin2t)ds dt =4ds dt.
Hence,
ZS
~
F·d~
A=Z2π
0Z4−2cost−2 sint
1−2cost−2 sint
4ds dt =Z2π
0
12dt =24π.
Answer(s) submitted:
•24pi
(correct)
Correct Answers:
•24*pi
9. (1 point)
Calculate RRSf(x,y,z)dS For
y=4−z2, 0 ≤x,z≤5; f(x,y,z) = z
RRSf(x,y,z)dS =
Solution:
Solution: We use the formula for the surface integral over a
graph y=g(x,z):
ZZS
f(x,y,z)dS =
ZZD
f(x,g(x,z),z)q1+g2
x+g2
zdx dz (1)
Since y=g(x,z) = 4−z2, we have gx=0, gz=−2z, hence:
q1+g2
x+g2
z=p1+4z2
f(x,g(x,z),z) = z
The domain of integration is the square [0,5]×[0,5]in the xz-
plane. By (1)we get:
ZZS
f(x,y,z)dS =Z5
0Z5
0
zp1+4z2dz dx =
Z5
0
1dxZ5
0
zp1+4z2dz=5Z5
0
zp1+4z2dz
We use the substitution u=1+4z2,du =8z dz to compute the
integral. This gives:
ZZS
f(x,y,z)dS =5Z5
0
zp1+4z2dz =
5Z101
1
u1/2
8du =10
3· 1013/2−1
8!≈422.516
Answer(s) submitted:
•422.5155
(correct)
Correct Answers:
•422.516
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