Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 13.7 due 10/05/2021 at 11:59pm MST
Problem 1. (1 point)
Evaluate Z ZSp1+x2+y2dS where Sis the helicoid: r(u,v) =
ucos(v)i+usin(v)j+vk, with 0 ≤u≤2,0≤v≤4π
Solution:
SOLUTION:
ru×rv=hcosv,sinv,0i×h−usinv,ucos v,1i=hsinv,−cosv,ui.
Thus |ru×rv|=psin2v+cos2v+u2=√1+u2.
Let D={(u,v)|0≤u≤2,0≤v≤4π}.
Then
RRSp1+x2+y2dS =RRDp1+u2cos2v+u2sin2v|ru×rv|dA
=RRD√1+u2√1+u2dA
=R2
0R4π
0(1+u2)dvdu
=4πR2
0(1+u2)du
=4πhu+u3
3i2
0
=56
3π
Answer(s) submitted:
•58.643
(correct)
Correct Answers:
•58.6430628670095
Problem 2. (1 point)
Let Sbe the part of the plane 2x+5y+z=4 which lies in the
first octant, oriented upward. Find the flux of the vector field
F=1i+4j+1kacross the surface S.
Solution:
SOLUTION
Sis the region in the plane z=4−2x−5yover D={(x,y)|0≤
x≤2,0≤y≤4
5−2
5x}.
Using xand yas parameters, we have r(x,y) = hx,y,4−2x−5yi.
Then rx×ry=h1,0,−2i×h0,1,−5i=h2,5,1i.
Since the zcomponent is positive, this normal has the correct ori-
entation.
Then
ZZS
F·dS=ZZD
F(r(x,y)) ·(rx×ry)dA
=ZZDh1,4,1i·h2,5,1idA
=ZZD
23dA
=23Z2
0Z4
5−2
5x
0
dydx
=23Z2
04
5−2
5xdx
=92
5
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•18.4
1
Problem 3. (1 point)
A fluid has density 2 kg/m3and flows in a velocity 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=9
Solution:
SOLUTION
The rate of flow through the sphere is the flux RRSρv·dS,0≤
θ≤2π,0≤φ≤πand ρ=2 kg/m3
A parametric representation of the sphere is r(φ,θ) =
3sinφcosθi+3 sin φsin θj+3 cos φkand the outward orientation
is given by rφ×rθ=9 sin2φcos θi+9 sin2φsin θj+9 sin φcos φk.
We have v(r(φ,θ)) = −3sinφsinθi+3sinφcosθj+12cosφk
Thus the rate of flow through Sis
RRSρv·dS=2R2π
0Rπ
0v(r(θ,φ)·rφ×rθdφdθ
=2R2π
0Rπ
0−27cosφsin2φsinθcosθ+27 cos φsin2φcos θsin θ
+108cos2φsinφdφdθ
=2R2π
0Rπ
0108cos2φsinφdφdθ
=432πRπ
0cos2φsinφdφ
Using the substitution u=cosφ, yields
RRSρv·dS=432πR1
−1u2du
=864
3πKg/s
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•904.77792
Problem 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=h19x,19y,19zi. Find
the electric flux across M. Write the integral over the hemisphere
using spherical coordinates, and use the outward pointing normal.
ZZM1
E·dS=Zb
aZd
c
f(θ,φ)dθdφ,
where
a=,b=,c=,d=,
Using tfor θand pfor φ,
f(θ,φ) =
RRM1E·dS=
RRM2E·dS=, so
RRME·dS=.
Solution:
SOLUTION
On M1:
E(r(θ,φ)) = h19sinφcosθ,19sinφsinθ,19cos φi,0≤θ≤
2π,0≤φ≤π/2
rφ×rθ=hsin2φcosθ,sin2φsinθ,sinφcosφiis the outward point-
ing normal.
Thus
ZZM1
E·dS=Zπ/2
0Z2π
0
E(r(θ,φ)) ·rφ×rθdθdφ
=Zπ/2
0Z2π
019sin3φcos2θ+19sin3φsin2θ+19sin φcos2φdθdφ
=Zπ/2
0Z2π
019sin3φcos2θ+sin2θ+19sinφcos2φdθdφ
=Zπ/2
0Z2π
019sin3φ+19sinφcos2φdθdφ
=Zπ/2
0Z2π
0
19sinφsin2φ+cos2φdθdφ
=Zπ/2
0Z2π
0
19sinφdθdφ
=2π(19)Zπ/2
0
sinφdφ
=38π[−cosφ]π/2
0
=38π
On M2:
r(x,y) = hx,y,0iwith D=(x,y)|x2+y2≤1.
2
Then E(r(x,y)) = h19x,19y,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=38π
Answer(s) submitted:
•0
•1.5707
•0
•6.28318
•19(sin(p)ˆ(3)+sin(p)*cos(p)ˆ(2))
•38pi
•0
•38pi
(correct)
Correct Answers:
•0
•1.5707963267949
•0
•6.28318530717959
•19*(sin(p)**3 + sin(p)*cos(p)**2)
•119.380520836412
•0
•119.380520836412
Problem 5. (1 point)
Determine whether the flux of the vector field ~
Fthrough each sur-
face 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 transverse to the surface and it has the same
orientation as the surface. Thus the flux is positive
The vector field is parallel to the surface. Thus the flux is zero.
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.
The vector field is parallel to the surface. Thus the flux is zero.
Answer(s) submitted:
•Positive
•Zero
•Negative
•Zero
•Zero
(correct)
Correct Answers:
•POSITIVE
•ZERO
•NEGATIVE
3
•ZERO
•ZERO
Problem 6. (1 point)
Compute the flux of the vector field ~
F=2y
~
i+2~
j−2xz
~
kthrough
the surface S, which is the surface y=x2+z2, with x2+z2≤9,
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≤9
(2(x2+z2)
~
i+2~
j−2xz
~
k)·(−2x
~
i+~
j−2z
~
k)dx dz
=Zx2+z2≤9
(−4x3−4xz2+2+4xz2)dxdz.
So
flux =Z3
−3Z√9−z2
−√9−z2(2−4x3)dx dz =Z3
−3Z√9−z2
−√9−z22dx dz−Z3
−3Z√9−z2
−√9−z24x3dx dz.
The first of these is just the 2 times the area of the disk, and the
second is zero by symmetry, so
flux =2(π9) = 18π.
Answer(s) submitted:
•18pi
(correct)
Correct Answers:
•pi*2*9
Problem 7. (1 point)
Compute the flux of the vector field ~
F=2x2y2z
~
kthrough the sur-
face Swhich is the cone px2+y2=z, with 0 ≤z≤R, oriented
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 param-
eterized 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= (rcosθ
~
i+rsinθ~
j−r
~
k)dr dθ.
Hence
ZS
~
F·d~
A=Z2π
0ZR
0
2(r5cos2θsin2θ
~
k)·(rcosθ
~
i+rsinθ~
j−r
~
k)dr dθ
=Z2π
0ZR
0−2r6cos2θsin2θdr dθ
=−2R7
7Z2π
0
sin2θcos2θdθ
=−2R7
7Z2π
0
sin2θ(1−sin2θ)dθ
=−2R7
7Z2π
0
(sin2θ−sin4θ)dθ
=−2R7
7π
4=−1
14 πR7.
The cone is not differentiable at the point (0,0). However the flux
integral, which is improper, converges.
Answer(s) submitted:
•rcos(theta)
4
•rsin(theta)
•r
•0
•R
•0
•2pi
•(rcos(theta)i+r*sin(theta)j-rk)*dr*dtheta
•-2pi R.5
(score 0.888889)
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
•-2*pi*Rˆ7/28
Problem 8. (1 point)
Compute the flux of ~
F=x
~
i+y~
j+z
~
kthrough the curved surface of
the cylinder x2+y2=9 bounded below by the plane x+y+z=3,
above by the plane x+y+z=9, 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=3cost
~
i+3sint~
j+s
~
k,
where 0 ≤t≤2πand 3 −3cost−3sint≤s≤9−3 cost−3sint.
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
−3sint3cost0
0 0 1
ds dt = (3xcost+3ysint)ds dt.
Plugging in for xand y, this is
~
F·d~
A=9(cos2t+sin2t)ds dt =9ds dt.
Hence,
ZS
~
F·d~
A=Z2π
0Z9−3cost−3 sint
3−3cost−3 sint
9ds dt =Z2π
0
54dt =108π.
Answer(s) submitted:
•108pi
(correct)
Correct Answers:
•108*pi
Problem 9. (1 point)
Calculate RRSf(x,y,z)dS For
y=4−z2, 0 ≤x,z≤9; 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 [0,9]×[0,9]in the xz-
plane. By (1)we get:
ZZS
f(x,y,z)dS =Z9
0Z9
0
zp1+4z2dz dx =
Z9
0
1dxZ9
0
zp1+4z2dz=9Z9
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 =9Z9
0
zp1+4z2dz =
9Z325
1
u1/2
8du =18
3· 3253/2−1
8!≈4393.52
Answer(s) submitted:
•4393.515
(correct)
Correct Answers:
•4393.52
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