Grading scheme: All double and triple integral problems (#2, #3, #5 and #6, but exclude #1 and #8) will be
graded according to the following (4:4:2) grading scheme.
• Diagram (4 points) – one or more diagram(s) should be drawn to show the region of integration. Some
sample strips should be included to explain how the lower and upper limits of the integral are determined.
• Set-up of an integral (4 points) – the lower and upper limits, and the order of integration should be
consistent with the diagrams drawn. Also, the order of integration must be chosen so that it is possible to
evaluate the integral without the help of machine.
• Evaluation (2 points) – for evaluating the integral and the correctness of the answer.
If the order of integration was badly chosen so that the computation cannot be carried out completely, points
will be deducted from both the set-up of an integral and the evaluation components.
1. Answer the following short questions. Each part is independent from the others.
(a) The volume of a piece of cake (yes, another one) can be computed by the following triple integral in
cylindrical coordinates:
Z π Z π/6 Z π/3
rdrdθdz.
0
0
/2
0
Sketch the region of integration (i.e. the cake) in the (x, y, z)-space. Clearly indicate on your diagram
the height, the radius and the angle of the cake. You do not need to evaluate the integral.
Solution: The cake:
(b) Suppose f (x, y) is a differentiable function defined on the whole (x, y)-plane. Given (a, b) is a point such
that ∇f (a, b) = 0. What can you say about the point (a, b)? Check all that apply:
√
(a, b) is a critical point of the function f (x, y).
√ ∂f
∂f
(a, b) = ∂y (a, b) = 0.
∂x
√
The tangent plane at the point (x, y) = (a, b) is horizontal.
The function f (x, y) attains either a local maximum or a local minimum at (a, b).
/2
(c) The Lagrange’s Multiplier method can be used to find critical points of a given function f (x, y) on a level
set constraint g(x, y) = 0 by solving the vector equation ∇f (x, y) = λ∇g(x, y). With the help of a diagram,
explain briefly why the solutions of this vector equation are the critical points on the constraint.
/2
Solution:
(d) What is the most suitable coordinate system of the three-dimensional space for evaluating the integral
below? Please justify your choice. You do not need to evaluate the integral.
Z ∞ Z ∞ Z ∞ −(x2 +y 2 +z2 )
e
dxdydz.
p
−∞ −∞ −∞ x2 + y 2 + z2
/2
Solution: Spherical coordinates (ρ, φ, θ): under this coordinates, the term x2 + y 2 + z2 becomes ρ2
and the integral is rewritten as:
Z
2π Z π Z ∞ −ρ2
e
0
0
0
ρ
· ρ2 sin φdρdφdθ
The square-root is gone so it is much easier to integrate.
(e) Let u(x, y) be a function with domain all of R2 and assume that all second partial derivatives of u exist
and are continuous. Suppose u satisfies:
∂2 u ∂2 u
+
>0
∂x2 ∂y 2
at all points. Explain why u does not have any local maximum points.
2 >0
Solution: In order for (x0 , y0 ) to be a local maximum point, we need uxx < 0 and uxx uyy − uxy
2
at (x0 , y0 ). If (x0 , y0 ) were such a point, then at this point uxx uyy > uxy ≥ 0, and so uxx and uyy are
either both > 0, or both < 0. However, given that uxx + uyy > 0 at all points, uxx and uyy cannot be
both < 0, and so uxx and uyy must be both > 0. Therefore, it is impossible to have uxx < 0 and so u
cannot have any local maximum points.
Alternatively, you may argue this way: since uxx + uyy > 0 everywhere, at any critical point at least
one of the second partials uxx and uyy must be strictly positive. This means that at a critical point
(x0 , y0 ) either the second derivative of the x = x0 slice or the second derivative of the y = y0 slice is
positive (possibly both are), so the value of u increases as you move away from u in at least one of
these planes. The function cannot attain a local maximum anywhere.
Page 2
/2
√
2. Let R be the region given by x2 + y 2 ≤ 4 and y ≥ −x 3 in the (x, y)-plane. Evaluate:
"
2xydA.
R
Solution:
It is a circular region,
it’s best to use polar coordinates. If rectangular coordinates are used, the integral
√
2
would involve 4 − x which makes the computation difficult.
"
2π
3
Z 2Z
2xydA =
− π3
0
R
2π
3
Z 2Z
=
−π
3
0
2
Z
=
0
Z
2
=
2 · r cos θ · r sin θ · rdθdr
2r 3 cos θ sin θdθdr
θ= 2π
3
r sin θ dr
−π
3
2
θ=
(since
d
sin2 θ = 2 sin θ cos θ)
dθ
3
r 3 · 0dr
(since sin(2π/3) = sin(π/3) = − sin(−π/3))
0
=0
Remark: it is absolutely fine to use the order rdrdθ for the computation, provided the upper/lower
limits are switched accordingly.
Page 3
/ 10
3. A tetrahedron T has four sides bounded the planes x = 0, y = 0, z = 0 and x + 2y + 3z = 6. Evaluate the
following triple integral:
$
1
dV .
6
−
2y
− 3z
T
Solution:
There are six choice of order of integration. However, if we integrate the function first by dy or dz, a
complicated ln-term would appear and it makes further computations even more difficult. We choose
dV = dxdydz, since in this way the ‘annoying’ denominator will be cancelled out after integrating by x:
$
T
1
dV =
6 − 2y − 3z
1 (6−3z)
2
Z 2Z
0
0
=
0
0
1 (6−3z)
2
Z 2Z
=
0
6−2y−3z
0
1 (6−3z)
2
Z 2Z
Z
1
dxdydz
6 − 2y − 3z
1
· (6 − 2y − 3z)dydz
6 − 2y − 3z
dydz
0
= area of the triangular side on the yz-plane
1
= · 3 · 2 = 3.
2
Page 4
/ 10
4. Let f (x, y) = sin x sin y.
(a) Find all critical points of f within the rectangle − π4 < x <
3π
4 ,
− 3π
4 <y<
3π
4 .
/4
Solution: We calculate
∇f = hcos x sin y, sin x cos yi.
Evidently the gradient is defined on all of R2 , so the only critical points occur where it vanishes,
i.e. ∇f = h0, 0i.
Set:
(cos x = 0
cos x sin y = 0
and
sin x cos y = 0
or
and
(sin x = 0
and
(x = 0
(x = π/2
sin y = 0)
or
y = 0)
or
or
cos y = 0)
y = π/2
or
y = −π/2).
Using a little bit logic, we can get three possible solutions:
π π
π π
(x, y) = (0, 0),
,
,
,− .
2 2
2 2
(b) For each of the critical points you found above, indicate if it is a local maximum, a local minimum, a
saddle point, or none of these.
Solution: We have second derivatives
fxx = − sin x sin y
fyx = cos x cos y
so
fxy = cos x cos y
fyy = − sin x sin y,
2
fxx fyy − fxy
= sin2 x sin2 y − cos2 x cos2 y.
Point
(0, 0)
π π
,
2 2 π
π
,
2 −2
fxx
0
-1
fyy
0
-1
fxy
1
0
2
fxx fyy − fxy
-1
1
This point is a:
saddle
local maximum
1
1
0
1
local minimum
Page 5
/6
5. Let R be the triangle with vertices (1, 1), (2, 3), and (3, 1) in the (x, y)-plane. Define
f (x, y) = xe
Find the average value of f on R.
y 2 −6y
.
!
[Reminder: average value is given by
f dA
R
,
A(R)
where A(R) is the area of R.]
Solution:
We choose dA = dxdy for two reasons:
2
1. It is extremely difficult (if not impossible) to integrate ey −6y by y. Hopefully, if one integrates the
function by x first, there may be some extra factor that makes the computation possible. You will
see in the steps below that it really happens to have a factor 6 − 2y so that the integral can be done
by a u-substitution.
2. A less important reason is that doing dydx involves two separate integrals but doing dxdy involves
only one.
"
Z 3Z
f dA =
y+1
2
1
R
Z 3
=
1
Z
=
Z
=
Z
7−y
2
xey
1 2 y 2 −6y
x e
2
3 y 2 −6y
e
=
dxdy
x= 7−y
2
x=
y+1
2
48 − 16y
dy
4
2
(6 − 2y)ey
2 −6y
dy
1
= −ey
= −e
2 −6y
−9
y=3
+e
dy
!
49 − 14y + y 2 1 + 2y + y 2
−
dy
4
4
2
1
2
3 y −6y
e
1
3
2 −6y
y=1
−5
The triangle has area 2, so the average value is
e−5 −e−9
.
2
Page 6
/ 10
6. Let R be the region in the first quadrant bounded by the hyperbolas 6xy = π and 3xy = π, and also the
straight-lines y = x and y = 2x.
Evaluate the following double integral:
"
y
· sec2 (xy) dxdy.
Rx
[Hint: Use a suitable change of variables so that the region R is transformed into a rectangular region.]
Solution:
Make a change of variables so that the region R is transformed into a rectangle:
y
x
v = xy.
u=
Another hint for using this change of variables is that u sec2 v is much easier to integrate when compared
to the given function.
The region is now transformed into a rectangle bounded by:
u = 1,
u = 2,
v = π/6,
Page 7
v = π/3.
/ 10
The Jacobian determinant is:
∂(u, v) ux uy =
∂(x, y) vx vy y
− 2 1x = x
y
x =−
Therefore,
y
2y
1
·x− ·y = − .
x
x
x2
2y
∂(u, v) dxdy =
dudv = dxdy = 2udxdy.
∂(x, y)
x
1
Therefore, we have dxdy = 2u
dudv. After this change of variables, we can evaluate the integral as
follows:
Z π/3 Z 2
"
y
1
· sec2 (xy) dxdy =
u · sec2 v ·
dudv
x
2u
π/6
1
R
Z π/3 Z 2
1
=
sec2 vdudv
π/6
1 2
Z2
Z
1 π/3
=
du ·
sec2 vdv
2
1
π/6
!
1
1
1 √
v=π/3
= (2 − 1) · tan v|v=π/6 =
3− √ .
2
2
3
√
You may rationalize the final answer as:
3
3 ,
but it is optional.
Page 8
7. Consider the level surface given by the equation:
x2 + y 2 + z2 + 2xz = 16.
It is a right elliptic cylinder in the (x, y, z)-space: the cylinder’s central
axis passes through the origin but it may not be one of the coordinate
axes. The uniform cross-sections sliced by planes orthogonal to the central axis are congruent ellipses centered at the central axis.
(a) Find the point(s) on the cylinder which is/are closest to the origin.
/ 10
p
Solution: To minimize the distance-from-the-origin function x2 + y 2 + z2 , we instead minimize
f (x, y, z) = x2 +y 2 +z2 to ease our calculations. The constraint is given by: g(x, y, z) = x2 +y 2 +z2 +2xz =
16, and so we need to solve the system
2x = 2λx + 2λz
2y = 2λy
2z = 2λz + 2λx
x2 + y 2 + z2 + 2xz = 16.
The right-hand sides of the first and third equations are equal, we must have x = z for all possible
candidate points.
Now the system is simplified to:
x = 2λx
y = λy
4x2 + y 2 = 16.
If the second equation (y = λy) , 0, then divide the first by the second equation of the new system,
we have: yx = 2x
y , which is only true when x = 0.
Substitute x = 0 into 4x2 + y 2 = 16, we get y = ±4. Two possible candidates are: (0, 4, 0) and
(0, −4, 0) . Note that x = z.
For another case, if y = λy = 0, then y = 0, and we have 4x2 = 16 by the third equation, and so
x = ±2. We get two other possible candidates: (2, 0, 2) and (−2, 0, −2) .
Evaluate f at these four candidate points:
f (−2, 0, −2) = 8
f (2, 0, 2) = 8
f (0, −4, 0) = 16
f (0, 4, 0) = 16.
Therefore, the points closest to the origin are (−2, 0, −2) and (2, 0, 2).
(b) Show that the cylinder’s central axis is parallel to the vector h1, 0, −1i.
Solution: The four solutions obtained above from the equation solving in part (a) are the points
on the cylinder at which the level sets of f and g are tangent to each other. From the given geometric information (aided by some intuition), these four points are the vertex points on the ellipse obtained by intersecting the cylinder by a plane through the origin and perpendicular to
the central axis. Therefore, the central axis is perpendicular to both h2, 0, 2i and h0, 4, 0i. Since
h2, 0, 2i × h0, 4, 0i = h−8, 0, 8i = −8h1, 0, −1i, the central axis is parallel to the vector h1, 0, −1i.
Page 9
/2
8. The diagram below shows a spherical shell S with thickness r2 − r1 . The outer sphere has radius r2 and the
inner spherical cavity has radius r1 . Both spheres are centered at the origin. Suppose this shell has uniform
density δ and mass m.
Show that its moment of inertia about the z-axis is:
!
2m r25 − r15
.
Iz =
5 r23 − r13
#
[Reminder: moment of inertia is defined to be S δD 2 dV where D(x, y, z)
denotes the perpendicular distance between the point (x, y, z) and the zaxis.]
Grading scheme: set-up of integral (4 points); evaluation and completion of the problem (4 points)
Solution: We use spherical coordinates: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Then dV =
ρ2 sin θdρdφdθ.
$
Iz =
δ(x2 + y 2 )dV
Z
S
2π Z π Z r2
δ(ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ) · ρ2 sin φdρdφdθ
=
0
0
r1
Z 2π Z π Z r2
δ · ρ2 sin2 φ · (cos2 θ + sin2 θ) · ρ2 sin φdρdφdθ
=
0
0
r1
Z 2π Z π Z r2
δ · ρ4 sin3 φdρdφdθ
=
0
Z
0
2π
=δ
r1
Z
π
dθ ·
0
sin3 φdφ ·
0
Z
r2
ρ4 dρ
r1
4 1
(see below for the calculation of
= δ · 2π · · (r25 − r15 )
3 5
m
4 1
= · 2π · · (r25 − r15 )
V
3 5
m
4 1
= 4
· 2π · · (r25 − r15 )
3
3
3
5
3 π(r2 − r1 )
!
2m r25 − r15
=
.
5 r23 − r13
Rπ
Below we supply the detail of the calculation of 0 sin3 φdφ:
Z
π
3
Z
π
sin φdφ =
0
(1 − cos2 φ) sin φdφ
0
Z
φ=π
=−
(1 − cos2 φ)d(cos φ)
φ=0
φ=π
cos3 φ = − cos φ +
3 φ=0
1
1
4
= 1−
− −1 +
= .
3
3
3
Page 10
Z
π
0
sin3 φdφ.)
/8
This page is intentionally blank. You may detach it from the exam paper and use it as a scratch paper.