MATH 251.504
Examination 2 Solutions
April 12, 2011
1.
Suppose f (x, y) is a differentiable function such that it and its derivatives take on the
following values:
(x, y) f (x, y) fx (x, y) fy (x, y) fxx (x, y) fyy (x, y) fxy (x, y)
(−2, 2)
9
0
0
−2
−2
−1
3
1
(1, 2)
0
0
−2
−2
2
2
2
(0, 6)
5
0
0
−3
0
1
Which of the following statements is true?
I. f (x, y) has a local minimum at (x, y) = (−2, 2).
II. f (x, y) has a critical point at (x, y) = (1, 2).
III. f (x, y) has a saddle point at (x, y) = (0, 6).
(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) II and III only
2.
Solution: B Consider each statement individually. For I, we see from the table that
(−2, 2) is a critical point since fx (−2, 2) = 0 and fy (−2, 2) = 0, so we try the Second
Derivative Test to determine which kind of critical point it is. We calculate D =
fxx (−2, 2)fyy (−2, 2) − fxy (−2, 2)2 = 4 − 1 = 3 > 0 and observe fxx (−2, 2) = −2 < 0,
so f has a local maximum at (−2, 2). Therefore I is false . Since, fy (1, 2) = 32 6= 0,
we see that f does not have a critical point at (1, 2), so II is false . Finally, for III we
are in a similar situation as I: f has a critical point at (0, 6), but now for the Second
Derivative Test D = fxx (0, 6)fyy (0, 6) − fxy (0, 6)2 = 0 − 1 = −1 < 0. Therefore, f
has a saddle point at (0, 6) and III is true .
√
Consider the point (r, θ, z) = (2 3, π3 , 2) in cylindrical coordinates. What are its
coordinates in spherical coordinates (ρ, θ, φ)?
(A) (16, π3 , π6 )
(B) (16, π2 , π4 )
(C) (4, π3 , π3 )
(D) (4, π3 , π6 )
(E) (4, π3 , π4 )
Solution: C Since ρ2 = z 2 + r2 , we see that ρ2 = 4 + 12 = 16, and so ρ = 4. Also
z = ρ cos φ, so cos φ = 42 = 12 , which implies that φ = π3 . Finally, θ is unchanged
between cylindrical and spherical coordinates, so the answer is (C).
1
Z
3.
3
√
Z
f (x, y) dy dx, we obtain
If we change the order of integration on the integral
0
which integral below?
Z 3 Z y2 +1
f (x, y) dx dy
(A)
1
1
1
y 2 −1
2Z
Z
x+1
f (x, y) dx dy
(B)
0
0
2
Z
y 2 −1
Z
f (x, y) dx dy
(C)
1
Z
0
2
Z
1
(D)
f (x, y) dx dy
y 2 −1
0
Z
2
Z
3
f (x, y) dx dy
(E)
1
y 2 −1
Solution: E We read from the original
integral that the region of integration is
√
given by 0 ≤ x ≤ 3 and 1 ≤ y ≤ x + 1. By sketching this region we find that
by switching the order of x and y the region is also described
by 1 ≤ y ≤ 2 and
√
y 2 − 1 ≤ x ≤ 3 (we obtain x = y 2 − 1 when solving y = x + 1 for x), and so the
answer is (E).
4.
Let D be the disk of radius 2 centered at (x, y) = (2, 0), and let R be the upper half
of D. Suppose that R is given the mass density ρ(x, y) defined by
ρ(x, y) = distance from (x, y) to the origin.
Suppose that m is the mass of R. Which integral below gives the x-coordinate of the
center of mass of R?
Z Z
1 π 4 sin θ 2
r cos θ dr dθ
(A)
m 0 0
Z π Z 4 cos θ
2
1
(B)
r3 cos θ dr dθ
m 0 0
Z Z
1 π 2 3
(C)
r sin θ dr dθ
m 0 0
Z Z √4−x2 p
1 4
x x2 + y 2 dy dx
(D)
m 0 0
Z Z √4−(x−2)2 p
1 4
(E)
x2 + y 2 dy dx
m 0 0
Solution: B The boundary of the disk D is given by the polar equation r = 4 cos θ.
The region R of integration is then given by 0 ≤ θ ≤ π2 (since we are using only
2
the
p top half of the disk) and 0 ≤ r ≤ 4 cos θ. Also, in polar coordinates ρ(x, y) =
x2 + y 2 = r on R. Therefore,
ZZ
Z π Z 4 cos θ
2
1
1
x=
(r cos θ) · r · r dr dθ,
xρ(x, y) dA =
m R
m 0 0
and so the answer is (B). [For the answers in rectangular coordinates, the integrand in
(D) is correct but the limits of integration do not describe R; the limits of integration
in (E) are right for R, but the integrand does not give the x-coordinate of the center
of mass.]
5.
Let E be the region in the first octant bounded below by the plane z = 5 and above
by the paraboloid z = 9 − x2 − y 2 . Which integral below expresses the volume of E?
Z 3 Z √9−x2 Z 9−x2 −y2
dz dy dx
(A)
5
0
0
Z Z √
Z
2
2
9−y 2
3
(B)
9−x −y
dz dx dy
√
− 9−y 2 0
√
2 Z
4−x2 Z 9−x2 −y 2
−3
Z
(C)
−2
Z
2
√
− 4−x2
dz dy dx
0
Z √4−y2 Z
9−x2 −y 2
(D)
dz dx dy
0
0
5
(E) none of the above
Solution: D By setting 5 = 9−x2 −y 2 , we see that the curve of intersection projects
onto the circle x2 + y 2 = 4 in the xy-plane. In the first quadrant (which sits below
the
p
first octant) this projects down to the quarter disk 0 ≤ y ≤ 2 and 0 ≤ x ≤ 4 − y 2 ,
which then gives the bounds for x- and y-coordinates of points in E. Finally the
z-coordinate of a point in E is bounded by 5 ≤ z ≤ 9 − x2 − y 2 , so the answer is (D).
6.
Let f (x, y) = xy + 1. What are the maximum and minimum values attained by
f (x, y) on the ellipse x2 + 2y 2 = 16? At what points (x, y) do they occur? You must
use the method of Lagrange multipliers.
Solution: We let g(x, y) = x2 + 2y 2 − 16, and so
multipliers
fx = λgx ⇒
y
fy = λgy ⇒
x
g = 0
⇒ x2 + 2y 2
we set up equations for Lagrange
= λ(2x)
= λ(4y)
= 0
There are several ways to solve these equations; here is one. First of all, we note that
x 6= 0 and y 6= 0 in any solution: if x = 0, then the first equation implies that y = 0,
but (x, y) = (0, 0) is not a solution of the third equation (similarly for y). Therefore
we are free to divide by x and y to obtain
λ=
y
2x
and λ =
3
x
.
4y
Therefore
y
x
=
2x
4y
⇒
4y 2 = 2x2
⇒
x2 = 2y 2 .
We substitute this into the third equation and obtain
2y 2 + 2y 2 − 16 = 4y 2 − 16 = 0.
Solving for y we see that y 2 = 4, and so y = ±2. We substitute each of y = 2 and
y = −2√into the equation x2 = 2y 2 , which in both cases yields x2 = 8 and thus
x = ±2 2. We see then there are four points (x, y) we need to consider:
√
√
√
√
(2 2, 2), (−2 2, 2), (2 2, −2), (−2 2, −2).
We substitute each of these into f (x, y):
√
√
f (2 2, 2) = 4 2 + 1,
√
√
f (−2 2, 2) = −4 2 + 1,
√
√
f (2 2, −2) = −4 2 + 1,
√
√
f (−2 2, −2) = 4 2 + 1.
Therefore on the ellipse x2 + 2y 2 = 16, f (x, y) = xy + 1 attains a maximum value of
√
√
√
4 2 + 1 at the points (2 2, 2) and (−2 2, −2) , and it attains a minimum value of
√
√
√
−4 2 + 1 at the points (−2 2, 2) and (2 2, −2) .
7.
Let R be the first quadrant sector (pie wedge) of the circle x2 + y 2 = 4 between x = 0
and p
y = x. Let E be the solid in 3-space that is above R and below the surface
z = 9 + x2 + y 2 .
(a) Express the volume of E as an iterated double integral in rectangular coordinates.
(You need not evaluate the integral).
(b) Convert the integral from part (a) into an iterated double integral in polar coordinates
and evaluate it.
Solution: (a) The region R is the sector inside the circle x2 + y 2 = 4 between the
angles θ = π4 and θ = π2 . (The line x = 0 is the y-axis.) The line y = x intersects the
√ √
circle x2 + y 2 = 4 at the point (x, y) = ( 2, 2), and by sketching R we see that it
is bounded by the inequalities
√
√
R : 0 ≤ x ≤ 2, x ≤ y ≤ 4 − x2 .
The double integral expressing the volume of E is then
√
ZZ p
Z
2
2
9 + x + y dA =
R
0
4
2
√
Z
x
4−x2
p
9 + x2 + y 2 dy dx.
(b) Now converting this integral to polar coordinates, we first see that R is bounded
by the inequalities
π
π
R : ≤ θ ≤ , 0 ≤ r ≤ 2.
4
2
Thus, since x2 + y 2 = r2 and dy dx = r dr dθ,
Z √2 Z √4−x2 p
Z πZ 2 √
2
2
2
r 9 + r2 dr dθ.
9 + x + y dy dx =
0
π
4
x
0
We now evaluate this integral, by first using the substitution u = 9 + r2 , du = 2r dr:
Z π Z 13
Z πZ 2 √
2
2
1√
r 9 + r2 dr dθ =
u du dθ
π
π
2
0
9
4
4
13
Z π
1 2 2 3/2
=
u
dθ
2 π4 3
9
Z π
1 2
133/2 − 93/2 dθ
=
3 π4
π
1 3/2
=
13 − 27 θ π2
4
3
√
π
=
(13 13 − 27).
12
8.
Let F be the solid in 3-space that is above the xy-plane and between the sphere
x2 + y 2 + z 2 = 18 and the cone z 2 = x2 + y 2 . Suppose also that F has mass density
d(x, y) = x2 + y 2 .
(a) Express the volume of F as an iterated triple integral in spherical coordinates.
(You need not evaluate the integral.)
(b) Express the mass of F as an iterated triple integral in cylindrical coordinates.
(You need not evaluate the integral.)
Solution: The region E is the snow cone shaped region between the cone and the
sphere. (Some students read the problem to mean that E was the solid below the
cone and inside the sphere; but assuming they solved the problem correctly for this
region they were not penalized.)
(a) We calculate the bounds on E in spherical coordinates. First 0 ≤ θ ≤ 2π, since
we are rotating completely around the z-axis. The cone z 2 = x2 + y 2 determines the
range of φ. Since z = ρ cos φ and r = ρ sin φ, we see that on this cone, where z 2 = r2 ,
we have tan φ = 1, which means the cone is described by the equation
π
φ= .
4
Thus the φ range of E is 0 ≤ φ ≤ π4 . Finally, ρ ranges from 0 to the radius of the
√
sphere, 18. Thus
ZZZ
Z 2π Z π Z √18
4
Vol(E) =
dV =
ρ2 sin φ dρ dφ dθ.
E
0
5
0
0
(b) To convert the limits of integration on E into cylindrical coordinates, we first see
that E is contained in the cylinder x2 + y 2 = 9 (obtained by substituting z 2 = x2 + y 2
into x2 +y 2 +z 2 = 18). Thus E sits over the disk of radius 3 centered at the origin and
is bounded above by the sphere and below by the cone. In cylindrical coordinates,
the sphere is given by z 2 + r2 = 18 and the cone is given by z = r. The mass density
is given by d = x2 + y 2 = r2 . All together then, the mass of E is
ZZZ
Z
mass(E) =
2π
Z
3
√
Z
d dV =
E
0
6
0
r
18−r2
r3 dz dr dθ.