Austin Cholley Zhu MAT 267 ONLINE A Spring 2021
Assignment Midterm 3 final due 03/02/2021 at 08:00pm MST
Problem 1.
2. (1 point) Find the volume of the solid enclosed by the
paraboloids z=9x2+y2and z=32 −9x2+y2.
Solution:
SOLUTION
The two paraboloids intersect when
9x2+y2=32 −9x2+y2or x2+y2=16
9. So
V=ZZx2+y2≤16
932 −9x2+y2−9x2+y2dA
=Z2π
0Z4
3
0
(32 −18r2)r dr dθ
=Z2π
0
dθZ4
3
0
(32r−18r3)dr
= [θ]2π
016r2−9
2r4
4
3
0
=2π16(4
3)2−9
2(4
3)4
=2π128
9=256
9π
Answer(s) submitted:
•89.3609
(correct)
Correct Answers:
•89.3609
Problem 2.
3. (1 point) (a) Calculate the flux of the vector field
~
F(x,y,z) = 4
~
i−7
~
kthrough a sphere of radius 2 centered at
the origin, oriented outward.
Flux =
(b) Calculate the flux of the vector field ~
F(x,y,z) =~
i−8~
j+3
~
k
through a cube of side length 2 with sides parallel to the axes,
oriented outward.
Flux =
Answer(s) submitted:
•
•
(incorrect)
Correct Answers:
•0
•0
Problem 3.
7. (1 point) Evaluate the line integral RCF·drwhere
F=h−5sinx,−4cos y,10xziand Cis the path given by r(t) =
(2t3,−t2,−2t)for 0 ≤t≤1
RCF·dr=
Solution:
The relevant vectors are:
F(r(t)) = −5sin(2t3),−4cos(−t2),−40t4
r0(t) = 6t2,−2t,−2
F(r(t))·r0(t) = −5sin(2t3),−4cos(−t2),−40t4·6t2,−2t,−2
=−30t2sin(2t3) + 8tcos(−t2) + 80t4
The line integral is then:
ZC
F·dr=Z1
0
F(r(t))·r0(t)dt =Z1
0−30t2sin(2t3)+8tcos(−t2)+80t4dt
=h5cos(2t3)−4sin(−t2) + 16t5i1
0
=5cos(2)−4sin(−1) + 11
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•12.2851
Problem 4.
12. (1 point) Use cylindrical coordinates to evaluate the triple
integral RRREpx2+y2dV , where Eis the solid bounded by the
circular paraboloid z=1−16x2+y2and the xy-plane.
Solution:
SOLUTION
The paraboloid z=1−16x2+y2=1−16r2intersects the
xy-plane in the circle x2+y2=1
16 or r2=1
16 ⇒r=1
4, so in
cylindrical coordinates, Eis given by
E=(r,θ,z)|0≤θ≤2π,0≤r≤1
4,0≤z≤1−16r2.
Thus
ZZZEpx2+y2dV =Z2π
0Z1
4
0Z1−16r2
0
√r2r dz dr dθ
=Z2π
0
dθZ1
4
0Z1−16r2
0
r2dzdr
=2πZ1
4
0
r2(1−16r2)dr
=2πZ1
4
0
(r2−16r4)dr
=2πr3
3−16r5
5
1
4
0
=1
240 π
Answer(s) submitted:
•-18.01179
1
(incorrect)
Correct Answers:
•0.01309
Problem 5.
10. (1 point) A)
Consider the vector field F(x,y,z) = h−5yz,0,xyi.
Find the divergence and curl of F.
div(F) = ∇·F=.
curl(F) = ∇×F=h, , i.
B)
Consider the vector field F(x,y,z) = h−2x2,−4(x+y)2,−(x+
y+z)2i.
Find the divergence and curl of F.
div(F) = ∇·F=.
curl(F) = ∇×F=h, , i.
Solution:
SOLUTION
A)
∇·F=∂
∂x(−5yz) + ∂
∂y(0) + ∂
∂z(xy) = 0+0+0=0
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
−5yz 0xy
=∂
∂y(xy)−∂
∂z(0)i−∂
∂x(xy)−∂
∂z(−5yz)j
+∂
∂x(0)−∂
∂y(−5yz)k
= (x−0)i−(y+5y)j+ (0+5z)k
=hx,−6y,5zi
B)
∇·F=∂
∂x(−2x2) + ∂
∂y(−4(x+y)2) + ∂
∂z(−(x+y+z)2)
=−4x−8(x+y)−2(x+y+z)
=−14x−10y−2z
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
−2x2−4(x+y)2−(x+y+z)2
=∂
∂y(−(x+y+z)2)−∂
∂z(−4(x+y)2)i
−∂
∂x(−(x+y+z)2)−∂
∂z(−2x2)j
+∂
∂x(−4(x+y)2)−∂
∂y(−2x2)k
= (−2(x+y+z)−0)i−(−2(x+y+z)−0)j+ (−8(x+y)−0)k
=h−2(x+y+z),+2(x+y+z),−8(x+y)i
Answer(s) submitted:
•0
•1x
•-6y
•5z
•-4x+(2)(-4)(x+y)+(-1)(2)(x+y+z)
•-1(2)(x+y+z)
•2(x+y+z)
•-4(2)(x+y)
(correct)
Correct Answers:
•0
•(1 - 0)*x
•(-5 - 1)*y
•(0 - -5)*z
•2*-2*x + 2*-4*(x+y) + 2*-1*(x+y+z)
•2*-1*(x+y+z)
•-2*-1*(x+y+z)
•2*-4*(x+y)
Problem 6.
4. (1 point)
Evaluate I=RC(sinx+4y)dx +(5x+y)dy for the nonclosed
path ABCD in the figure.
A=
(0,0),B= (4,4),C= (4,8),D= (0,12)
I=
Solution:
Solution:
A=
(0,0),B= (4,4),C= (4,8),D= (0,12)
Let F=hsinx+4y,5x+yi, hence P=sinx+4yand Q=
5x+y.
2
We denote by C1the closed path determined by Cand the
segment DA. Then by Green’s Theorem,
ZC1
Pdx +Q dy =ZZD∂Q
∂x−∂P
∂ydA =
ZZD
(5−4)dA =1ZZD
dA =1Area(D) (1)
The area of Dis the area of the trapezoid ABCD, that is,
Area(D) = BC +ADh
2=(4+12)·4
2=32.
A= (0,0),B= (4,4),C= (4,8),D= (0,12)
Combining with (1)we get
ZC1
Pdx +Q dy =1·32 =32
Using properties of line integrals, we have
ZC
Pdx +Q dy +ZDA
Pdx +Q dy =32 (2)
We compute the line integral over DA, using the parametrization
DA :x=0,y=t,tvaries from 12 to 0.
We get
ZDA
Pdx +Q dy =Z0
12
F(0,t)·d
dt h0,tidt =
Z0
12 hsin0 +4t,5·0+ti·h0,1idt =
Z0
12 h4t,ti·h0,1idt =Z0
12
t dt =t2
2
0
t=12
=−72
We substitute in (2)and solve for the required integral:
ZC
Pdx +Q dy −72 =32
or
ZC
Pdx +Q dy =104.
Answer(s) submitted:
•-72
(incorrect)
Correct Answers:
•104
Problem 7.
8. (1 point) Find the volume of the ellipsoid x2+y2+5z2=
36.
Solution:
SOLUTION
The ellipsoid x2+y2+5z2=36 intersects the xy-plane in the
circle x2+y2=36.
Solving for zyields z=1
√5p36 −x2−y2.
By symmetry,
V=2ZZx2+y2≤36
1
√5p36 −x2−y2dA
=2
√5Z2π
0Z6
0p36 −r2r drdθ
=2
√5Z2π
0
dθZ6
0p36 −r2r dr
=4π
√5Z6
0p36 −r2r dr
[Substituting u=36 −r2,du =−2rdr]:
=−2π
√5Z0
36
√udu
=2π
√52
3u3/236
0
=4π
3√5(36)3/2
=4π
3√5·63
Answer(s) submitted:
•404.62932
(correct)
Correct Answers:
•404.629328507946
3
Problem 8.
6. (1 point) Let F=7xyi +2y2jbe a vector field in the plane,
and Cthe path y=5x2joining (0,0) to (1,5) in the plane.
A. Evaluate RCF·dr
B. Does the integral in part (A) depend on the path joining
(0,0) to (1,5)? (y/n)
Answer(s) submitted:
•297.9166
•y
(score 0.5)
Correct Answers:
•92.0833333333333
•Y
Problem 9.
9. (1 point) Evaluate the integral.
Z3
0Z√9−x2
−√9−x2Z√9−x2−z2
−√9−x2−z2
1
(x2+y2+z2)1/2dy dz dx =
Solution:
SOLUTION
z=−p9−x2−y2and z=p9−x2−y2represent, respec-
tively, the lower and upper part of the sphere x2+y2+z2=9.
In spherical coordinates, the sphere has equation ρ=3.
y=−√9−x2and y=√9−x2represent, respectively, the lower
and upper part of the circle x2+y2=9, which is the intersection
of the sphere ρ=3 with the xy-plane.
Since 0 ≤x≤3, the region of integration consists of the hemi-
sphere ρ=3 with x>0.
Thus, in spherical coordintes, the integral becomes
Zπ/2
−π/2Zπ
0Z3
0
1
(ρ2)1/2ρ2sinφdρdφdθ=Zπ/2
−π/2Zπ
0Z3
0
ρsinφdρdφdθ
= [θ]π/2
−π/2ρ2
23
0
[−cosφ]π
0
=π32
2(2) = 9π
Answer(s) submitted:
•9pi
(correct)
Correct Answers:
•9*pi
Problem 10.
11. (1 point) Consider the surface with parametric equations
r(s,t) = hst,s+t,s−ti.
A) Find the equation of the tangent plane at (2,3,1).
.
B) Find the surface area under the restriction s2+t2≤1
Solution:
SOLUTION
A)
rt(s,t) = hs,1,−1iand rs(s,t) = ht,1,1i. Since the point
(2,3,1)corresponds to s=2,t=1, a normal vector to the
surface at (2,3,1)is given by
rs(2,1)×rt(2,1) = h1,1,1i×h2,1,−1i=h−2,3,−1i.
An equation of the tangent plane is
−2(x−2) + 3(y−3)−(z−1) = 0
B)
rs×rt=h−2,s+t,t−siand
|rs×rt|=p4+ (s+t)2+ (t−s)2=√4+2s2+2t2.
The surface area under the given restrictions is then
ZZs2+t2≤1p4+2s2+2t2dA =Z2π
0Z1
0
rp4+2r2dr
=2π1
6(4+2r2)3/21
0
=π
3(6√6−8)
=π2√6−8
3
Answer(s) submitted:
•-2(x-2)+3(y-3)-(z-1)=0
•7.01301
(correct)
Correct Answers:
•-2*(x-2)+3*(y-3)-(z-1)=0
•7.01301755236959
Problem 11.
1. (1 point)
Each vector field shown shown represents the force
on a particle at different points in the plane as a result
of another particle at the origin. Match each vector field
with its description.
? 1. An attractive force whose magnitude decreases as dis-
tance increases.
? 2. A repulsive force whose magnitude increases as dis-
tance increases.
? 3. An attractive force whose magnitude increases as dis-
tance increases.
? 4. A repulsive force whose magnitude decreases as dis-
tance increases.
4
A B
C D
(Click on a graph to enlarge it.)
Answer(s) submitted:
•B
•C
•D
•A
(score 0.5)
Correct Answers:
•B
•C
•A
•D
Problem 12.
5. (1 point) Evaluate the triple integral
ZZZE
zdV where Eis the solid bounded by the cylinder
y2+z2=144 and the planes x=0,y=3xand z=0 in the first
octant.
Solution:
SOLUTION
A picture of the solid Eis given below.
ZZZE
zdV =Z4
0Z12
3xZ√144−y2
0
zdz dy dx =Z4
0Z12
3x
1
2144 −y2dy dx
=1
2Z4
0144y−y3
3y=12
y=3x
dx =1
2Z4
01728 −576 −432x+9x3dx
=1
21152x−432 x2
2+9x4
44
0
=1
2(4608 −3456 +576) = 864
Answer(s) submitted:
•-486
(incorrect)
Correct Answers:
•864
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