19 Nov 2004
[25] 1.
MATH 263
UBC ID:
Page 2 of 5 pages
An antenna at the origin emits a signal whose strength at the point with polar coordinates [r, θ] is
f (r, θ) =
1 + cos(4θ)
,
r
r > 0, −
π
π
<θ< .
4
4
π
π
<θ< .
4
4
(a)
Write the level curve f (r, θ) = 2 in polar function form r = r(θ), −
(b)
Sketch the region in the xy–plane consisting of all points whose polar coordinates obey the
equation r = r(θ) of part (a). Indicate the region where f (r, θ) ≥ 2.
(c)
Find the area of the region described in part (b).
(a)
The level curve has equation 2 = (1 + cos(4θ))/r. Solving for r gives the polar form:
r = r(θ) =
(b)
1 + cos(4θ)
,
2
−
π
π
<θ< .
4
4
The curve r = r(θ) encloses a single lobe along the x-axis. The rightmost point of the lobe is at
(x, y) = (1, 0). One has f (r, θ) ≥ 2 at points on and inside the closed curve just mentioned.
θ=π/4
y
r = 0.5*(1+cos(4θ))
1
f(r,θ)>2
x
θ=−π/4
(c)
Call the region R. Its area is
Z
ZZ
π/4
Z
dA =
R
−π/4
1
=
8
Z
1
2 (1+cos(4θ))
Z
π/4
r dr dθ =
−π/4
0
π/4
1
(1 + cos(4θ)) dθ =
8
−π/4
Z
r2
2
1
r= 2 (1+cos(4θ))
dθ
r=0
π/4
2
def
(1 + 2 cos(4θ) + cos2 (4θ))dθ =
−π/4
There are several ways to find J. One is to let u = 4θ, du = 4 dθ:
Z π
ZZ
1
1
u 1
2
dA =
(1 + 2 cos u + cos u)du =
u + 2 sin u + + sin 2u
32 −π
32
2 4
R
3
3
1 3
π− − π
π
=
=
32 2
2
32
1
J.
8
u=π
u=−π
Or, one could use basic geometry to make three simple observations:
Z π/4
Z π/4
Z π/4
π
π
dθ = ,
2 cos(4θ) dθ = 0,
cos2 (4θ) dθ = .
2
4
−π/4
−π/4
−π/4
Summing these values gives J = 3π/4, so A = J/8 = 3π/32, as before.
Continued on page 3
19 Nov 2004
[25] 2.
MATH 263
UBC ID:
Page 3 of 5 pages
Let R denote the solid defined by the system of inequalities
x ≥ 0, y ≥ 0, z ≥ 0, z ≤ 1 − x2 ,
x + y + z ≤ 2.
(a)
Express the volume of R as an iterated triple integral.
(b)
Compute the volume of R.
(a)
Looking at the figure below from the side (standing far out on the y axis)
z
x+y+z =2
z = 1 − x2
y
x
we see a base region in the xz–plane consisting of 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x2 . The corresponding
triple integral is
Z 1−x2
Z 2−x−z
Z 1
dx
dz
dy.
V =
0
0
(b)
The volume is
Z
Z
1
1−x2
dz (2 − x − z)
dx
V =
0
0
Z
1
=
0
Z
2
dx (2 − x)(1 − x2 ) − 12 (1 − x2 )
1
dx
=
0
0
3
2
− x − x2 + x3 − 12 x4
1
3 1 1 1
− − + −
2 2 3 4 10
49
=
60
=
Continued on page 4
19 Nov 2004
[25] 3.
MATH 263
UBC ID:
Page 4 of 5 pages
Let C be the curve from P = (1, 0, 0) to Q = (0, π/2, π/2) along the intersection of these surfaces:
x = cos(y), y = z.
Choose specific numbers A and B (state your choices clearly!) and then use them to evaluate both
Z
I1 = (yex − Ax2 cos(z)) dx + (ex + By 4 z 2 ) dy + (2y 5 z − x3 sin(z)) dz
Z
and
I2 =
C
C
yex − Ax2 cos(z) + 3 sin2 (y), ex + By 4 z 2 , 2y 5 z − x3 sin(z) • dr.
Hint : You can replace A and B with any values you like. Efficient choices would be best; taking A = 0
and B = 0 is not efficient at all.
Z
Both I1 and I2 are line integrals of vector fields: I1 =
C
Z
F • dr and I2 = I1 +
F(x, y, z) = yex − Ax2 cos(z), ex + By 4 z 2 , 2y 5 z − x3 sin(z) ,
C
G • dr, where
G(x, y, z) = 3 sin2 (y), 0, 0 .
Line integrals are easy to evaluate when they represent work done by a conservative vector field. Could F
be conservative? Only when it passes the screening test, i.e., when
∂F3
∂F1
=
,
∂z
∂x
∂F3
∂F2
=
,
∂z
∂y
and
i.e.,
Ax2 sin(z) = −3x2 sin(z),
i.e.,
A = −3,
i.e.,
2By 4 z = 10y 4 z
i.e.,
B = 5.
With these choices, ∇ × F ≡ 0, and it is not hard to see that F ≡ ∇φ for the function
φ(x, y, z) = yex + y 5 z 2 + x3 cos(z).
Consequently
"
#
7
7 Z
Z
π
π
π
π
+
F • dr =
∇φ • dr = φ(Q) − φ(P ) =
+ 0 − [0 + 0 + 1] =
+
− 1.
I1 =
2
2
2
2
C
C
With the same choices for A and B,
Z
I2 = I1 +
A simple parametrization for C is given by
x = cos(t), y = t, z = t, 0 ≤ t ≤ π/2;
Hence
Z
π/2
I2 = I1 + 3
t=0
3 sin2 (y) dx.
C
note dx = − sin(t) dt, dy = dt, dz = dt.
π/2
7 π
π
sin2 (t)(− sin(t) dt) = I1 − cos3 (t) − 3 cos(t)
= I1 − 2 =
+
− 3.
2
2
t=0
[The integral of sin3 (t) is given on the formula sheet. One may also write sin3 (t) = [1 − cos2 (t)] sin(t) and
then substitute u = cos(t).]
Continued on page 5
19 Nov 2004
[25] 4.
MATH 263
UBC ID:
Page 5 of 5 pages
Let S be the piece of the paraboloid z = 10 − x2 − y 2 where 1 ≤ z ≤ 6. Compute
ZZ p
4x2 + 4y 2 + 1 dS.
S
Method 1: Rectangular Coordinates (then switch to polar).
We parametrize S by r(x, y) = hx, y, f (x, y)i, where f (x, y) = 10 − x2 − y 2 . Then we know that
q
p
(fx )2 + (fy )2 + 1 dx dy =
4x2 + 4y 2 + 1 dx dy.
dS =
Also note that if z = 6 then r2 = 4 and if z = 1 then r2 = 9, so we are integrating over an annulus with
inner radius 2 and outer radius 3, which we will denote by R. Hence
ZZ
ZZ p
2
2
4x + 4y + 1 dS =
(4x2 + 4y 2 + 1) dx dy
S
Z
R
2π
Z
3
(4r2 + 1)r dr dθ
=
0
Z
2
3
(4r3 + r) dr
= 2π
2
3
= 2π r + r /2
4
2
2
= 2π[(3 + 3 /2) − (24 + 22 /2)]
4
2
= 2π(81 + 9/2 − 16 − 2) = π(162 + 9 − 32 − 4) = 135π.
Method 2: Cylindrical Coordinates.
We parametrize S in terms of (r, θ) by s(r, θ) = hr cos θ, r sin θ, g(r, θ)i, where g(r, θ) = 10 − r2 . Then we
know that
dS = ((gθ )2 + (rgr )2 + r2 )1/2 dr dθ = (4r4 + r2 )1/2 dr dθ.
Also note that if z = 6 then r2 = 4 and if z = 1 then r2 = 9, so we are integrating over an annulus with
inner radius 2 and outer radius 3, which we will denote by R. Hence
ZZ
ZZ p
4x2 + 4y 2 + 1 dS =
(4r2 + 1)1/2 (4r4 + r2 )1/2 dr dθ
S
Z
R
2π
Z
3
r(4r2 + 1) dr dθ
=
0
Z
2
3
(4r3 + r) dr
= 2π
2
3
= 2π r + r /2
4
2
2
= 2π[(34 + 32 /2) − (24 + 22 /2)] = 135π.
The End
This examination has 5 pages including this cover
The University of British Columbia
Midterm Examination – 19 Nov 2004
Mathematics 263
Multivariable and Vector Calculus
Closed book examination
Time: 50 minutes
Signature
Name
Student Number
Special Instructions:
To receive full credit, all answers must be supported with clear and correct
derivations. No calculators, notes, or other aids are allowed. A formula
sheet is provided with the test.
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examiners.
(b) Speaking or communicating with other candidates.
1
25
2
25
3
25
4
25
Total
100
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