MATH 255
Applied Honors Calculus III
Winter 2011
Midterm 2 Review Solutions
14.1: #11
This is a circle of radius 1 parallel to the xz plane, oriented in the direction such that at the
point (0, 3, 1), which corresponds to t = 0, the x-coordinate is increasing and the z-coordinate
is decreasing.
14.1: #35
First notice that z ≥ 0 (according to the cone) and thus y ≥ −1 (according to the plane).
According to the plane, we have y = z − 1, and the intersection of the cone and the plane lie on
the surface z 2 = x2 + (z − 1)2 , which simplifies to z = x2 /2 + 1/2. If we parametrize by x = t,
this becomes z = t2 /2 + 1/2 and thus y = t2 /2 − 1/2. Thus the curve is parametrized by
x=t
y=
t2 − 1
,
2
y=
t2 + 1
.
2
14.2: #17
r′ (t) = h30t4 , 12t2, 2i ,
thus
r′ (1) = h30, 12, 2i ,
and
It follows that
|r′ (1)| =
T(1) =
√
√
302 + 122 + 22 = 2 262 .
h30, 12, 2i h15, 6, 1i
r′ (1)
√
=
= √
.
′
|r (1)|
2 262
262
14.2: #25
We first differentiate the parametric equations, giving
x′ (t) = −e−t (cos(t) + sin(t)) ,
y ′(t) = e−t (cos(t) − sin(t)) ,
1
z ′ (t) = −e−t .
The point (1, 0, 1) corresponds to t = 0. Plugging this value into the above equations gives that
the direction vector for the tangent line is
h−1, 1, −1i.
The parametric equations for the tangent line are therefore
x = −t + 1 ,
y = t,
z = −t + 1 .
14.3: #5
r′ (t) = h0, 2t, 3t2i .
The length of this vector is
|r′ (t)| =
√
√
4t2 + 9t4 = t 4 + 9t2 .
The length is then given by
Z 13
Z 1 √
√
1
1
1 2
3/2
3/2
3/2
2
u du =
=
13 − 4
13 − 8 .
L=
t 4 + 9t dt =
18 4
18 3
27
0
14.3: #19
r′ (t) = h3, 4 cos(t), −4 sin(t)i .
r′′ (t) = h0, −4 sin(t), −4 cos(t)i.
q
√
|r′ (t)| = 9 + 16 cos2 (t) + 16 sin2 (t) = 9 + 16 = 5.
i
j
k
r′ (t)×r′′ (t) = 3 4 cos(t) −4 sin(t) = i(−16 cos2 (t)−16 sin2 (t))−j(−12 cos(t))+k(−12 sin(t)).
0 −4 sin(t) −4 cos(t)
q
√
|r′ (t) × r′′ (t)| = (−16)2 + 122 cos2 (t) + (−12)2 sin2 (t) = 256 + 144 = 20.
Thus we have
κ=
|r′ (t) × r′′ (t)|
20
20
4
= 3 =
= .
′
3
|r (t)|
5
125
25
14.3: #39
The given point corresponds to t = 1.
r′ (t) = h2t, 2t2 , 1i .
√
|r′(t)| = 4t2 + 4t4 + 1 = 2t2 + 1.
h2t, 2t2 , 1i
.
T(t) =
2t2 + 1
h2t, 2t2 , 1i h2, 4t, 0i
T′ (t) = −4t 2
+ 2
.
(2t + 1)2
2t + 1
h2, 2, 1i
h2, 2, 1i
T(1) =
=
= h2/3, 2/3, 1/3i .
(2 + 1)
3
−2 4 −4
h2, 2, 1i h2, 4, 0i
+
=h , ,
i.
T′ (1) = −4
2
(2 + 1)
2+1
9 9 9
1√
6
2
|T′ (1)| =
4 + 16 + 16 = = .
9
9
3
T′ (1)
3 −2 4 −4
−1 2 −2
=
h
,
,
i
=
h
, ,
i.
|T′(1)|
2 9 9 9
3 3 3
i
j
k B(1) = T(1) × N(1) = 2/3 2/3 1/3 = h−2/3, 1/3, 2/3i.
−1/3 2/3 −2/3
N(1) =
14.4: #13
v(t) = r′ (t) = et h− sin(t), cos(t), 1i + et hcos(t), sin(t), ti
= et hcos(t) − sin(t), cos(t) + sin(t), 1 + ti.
a(t) = r′′ (t) = et hcos(t) − sin(t), cos(t) + sin(t), 1 + ti + et h− sin(t) − cos(t), − sin(t) + cos(t), 1i
= et h−2 sin(t), 2 cos(t), 2 + ti.
p
|v(t)| = e2t (cos(t) − sin(t))2 + (cos(t) + sin(t))2 + (1 + t)2 = et (t + 1).
14.4: #15
v(t) =
Z
a(t)dt = hc1 , c2 , t + c3 i .
To have v(0) = h1, −1, 0i, we need to take c1 = 1, c2 = −1, c3 = 0. Thus
v(t) = h1, −1, ti ,
and
t2
+ c6 i .
2
To have r(0) = h0, 0, 0i, we need to take c4 = c5 = c6 = 0. Thus
r(t) =
Z
v(t)dt = ht + c4 , −t + c5 ,
r(t) = ht, −t,
t2
i.
2
14.4: #31
r′ (t) = h3 − 3t2 , 6ti .
p
|r′ (t)| = (3 − 3t2 )2 + 36t2 = 3 + 3t2 .
1
1
T(t) =
h3 − 3t2 , 6ti =
h1 − t2 , 2ti .
2
3 + 3t
1 + t2
2(1 − t2 )
4t
,
i.
T′ (t) = h−
(1 + t2 )2 (1 + t2 )2
p
2
2(1 + t2 )
1
2 + 4(1 − t2 )2 =
=
.
16t
|T′ (t)| =
2
2
2
2
(1 + t )
(1 + t )
1 + t2
N(t) =
The tangential component is
T′ (t)
2t 1 − t2
=
h−
,
i.
|T′ (t)|
1 + t2 1 + t2
a(t) = r′′ (t) = h−6t, 6i .
compT(t) a(t) = a(t) · T(t) =
1
(−6t(1 − t2 ) + 6(2t)) = 6t .
2
1+t
The normal component is
compN(t) a(t) = a(t) · N(t) =
1
((−6t)(−2t) + 6(1 − t2 )) = 6 .
1 + t2
Thus
a(t) = 6tT(t) + 6N(t) .
Chapter 14 T-F quiz
√
(1) True, if we reparametrize by t = [3] u then this is the familiar vector equation of a
line.
(2) True, the x-coordinate of r′ (t) is always 1, thus r′ (t) is never zero.
(3) False, r′ (0) = 0, thus the curve is not smooth at t = 0.
(4) True. It just is true.
(5) False, in general, we need to use the product rule.
(6) False. This is really a composition of functions, and therefore should use the chain rule.
(7) False, the derivative should be taken with respect to arc length in order for κ to be well
defined.
(8) False, the cross product is taken in the wrong order, which will change the sign.
(9) True, although we didn’t talk about this in class.
(10) False, the tangent vectors will be parallel, but not necessarily of the same magnitude.
Chapter 14 Review #11
r′ (t) = ht2 , t, 1i .
√
|r′ (t)| = t4 + t2 + 1 .
1
r′ (t)
=√
ht2 , t, 1i .
′
4
2
|r (t)|
t +t +1
1
ht(t2 + 2), 1 − t4 , −t(2t2 + 1)i
T′ (t) = 4
2
3/2
(t + t + 1)
p
p
(t2 + t + 1)(t2 − t + 1)(t4 + 4t2 + 1)
1
2 (t2 + 2)2 + (1 − t4 )2 + t2 (2t2 + 1)2 =
.
|T′ (t)| = 4
t
(t + t2 + 1)3/2
(t4 + t2 + 1)3/2
T(t) =
T′ (t)
1
ht(t2 + 2), 1 − t4 , −t(2t2 + 1)i .
=p
′
2
2
|T (t)|
(t + t + 1)(t − t + 1)(t4 + 4t2 + 1)
p
p
(t2 + t + 1)(t2 − t + 1)(t4 + 4t2 + 1)
(t2 + t + 1)(t2 − t + 1)(t4 + 4t2 + 1)
|T′ (t)|
√
=
.
κ(t) = ′
=
|r (t)|
(t4 + t2 + 1)2
(t4 + t2 + 1)3/2 t4 + t2 + 1
N(t) =
Chapter 14 Review #13
If y = f (x), then the curvature is given by
|f ′′(x)|
κ(x) =
.
(1 + f ′ (x)2 )3/2
In this case, f ′ (x) = 4x3 and f ′′ (x) = 12x2 . Thus, at the point x = 1, we have
κ=
12
12
= 3/2 .
3/2
(1 + 16)
17
Chapter 14 Review #17
v(t) = r′ (t) = hln t + 1, 1, −e−t i .
p
v(t) = |v(t)| = (ln t + 1)2 + 1 + e−2t .
1
a(t) = r′′ (t) = h , 0, e−t i .
t
15.1 #37
15.2 #13
We will approach along an arbitrary parabola y = λx2 . On this line, we have
2λ2 x4
λ2
f (x, λx2 ) = 4
=
.
x + λ2 x4
1 + λ2
This expression depends of λ, thus the limit is different as we approach along different parabolas.
Thus the limit does not exist.
15.3 #17
2y
(x + y)2
2x
fy (x, y) = −
(x + y)2
fx (x, y) =
15.3 #49
−2y
∂2z
=
2
∂x
(x + y)3
∂2z
2x
=
2
∂y
(x + y)3
∂2z
x−y
∂2z
=
=
∂x∂y
∂y∂x
(x + y)3
15.3 #63
∂3w
4
=
,
∂x∂y∂z
(y + 2z)3
∂3w
= 0.
∂x2 ∂y
15.4 #5
∂z = −2 sin(0) = 0 .
=
−y
sin(x
−
y)
∂x (2,2)
(2,2)
= cos(0) + 2 sin(0) = 1 .
= (cos(x − y) + y sin(x − y))
∂z ∂y (2,2)
Thus, equation for the tangent plane is
(2,2)
(z − 2) = (y − 2)
or
y = z.
15.4 #17
fx (x, y) = p
thus
−x
20 − x2 − 7y 2
fy (x, y) = p
−7y
20 − x2 − 7y 2
,
−7
−2
, fy (2, 1) =
, f (2, 1) = 3
3
3
and the linear approximation at the point (2, 1) is
2
7
L(2,1) (x, y) = 3 − (x − 2) − (y − 1) .
3
3
Plugging in x = 1.95 and y = 1.08 gives and the linear approximation at the point (2, 1) is
2
7
L(2,1) (1.95, 1.08) = 3 − (−0.05) − (0.08) ≈ 2.8467 .
3
3
fx (2, 1) =
15.5 #5
dw
t
x y/z
xy y/z
1−t
y/z
2 (1−t)/(1+2t)
= e (2t) + e (−1) − 2 e (2) = 2t e
−
.
dt
z
z
1 + 2t (1 + 2t)2
15.5 #11
s
∂z
r
r
.
= e cos θ(t) − e sin θ 2
∂s
s + t2
∂z
t
r
r
.
= e cos θ(s) − e sin θ 2
∂t
s + t2
15.5 #25
∂u
= 2x(r cos(θ)) + z(r sin(θ)) + y(1) .
∂p
∂u = 2(6)(3 cos(0)) + 5(3 sin(0)) + 0 = 36 .
∂p P
∂u
= 2x(p cos(θ)) + z(p sin(θ)) + y(1) .
∂r
∂u = 2(6)(2 cos(0)) + 5(2 sin(0)) + 0 = 24 .
∂p P
∂u
= 2x(−pr sin(θ)) + z(pr cos(θ)) + y(0) .
∂θ
15.5 #35
Thus
∂u = 2(6)(−(2)(3) sin(0)) + 5((2)(3) cos(0)) + 0 = 30 .
∂θ P
dT dt t=3
dx
dy
dT
= Tx (x, y) + Ty (x, y) .
dt
dt
dt
dy
1
dx
1
.
= ,
= √
dt
3
dt
2 1+t
1
1
+3
=4 √
= 1 + 1 = 2 deg/sec. .
3
2 1+3
15.6 #11
We first need to find the unit vector in the direction of v. It is given by
h4, −3i
h4, −3i
v
=√
.
=
2
2
|v|
5
4 +3
We now compute partial derivatives. The dot product of the gradient and the unit vector in
the direction of v is the directional derivative we are looking for.
√
x
fx (x, y) = 2 y , fy (x, y) = √ .
y
√
8 y
3
3x
4
− √ .
Dv f (x, y) = fx (x, y) − fy (x, y) =
5
5
5
5 y
√
9
23
16
8 4 3(3)
+ √ =
−
=
.
Dv f (3, 4) =
5
5
10
10
5 4
15.6 #21
y2
2y
, fy (x, y) =
.
2
x
x
42 2(4)
i = h−4, 4i .
∇f (2, 4) = h− 2 ,
2
2
This is the direction of the maximum rate of change. The rate of change is
√
√
|∇f (2, 4)| = 16 + 16 = 4 2 .
fx (x, y) = −
15.6 #39
Let f (x, y, z) = x2 + 2y 2 + 3z 2 . Then the surface in question is the level surface f (x, y, z) = 21.
The gradient of f is normal to the surface, and is
∇f (x, y, z) = h2x, 4y, 6zi .
The normal line is in the direction of the gradient, thus the vector equation for the normal line
is
r(t) = h4, −1, 1i + th2(4), 4(−1), 6(1)i = h4, −1, 1i + th8, −4, 6i .
The gradient is the normal vector to the tangent plane, thus the tangent plane is given by
2(4)(x − 4) + 4(−1)(y + 1) + 6(1)(z − 1) = 0 ,
or 8(x − 4) − 4(y + 1) + 6(z − 1) = 0 .
15.6 #47
∇f (x, y) = h2x, 8yi .
∇f (2, 1) = h4, 8i .
8
A line in the direction h4, 8i has slope 4 = 2. The normal line the the curve has this slope, thus
the tangent line has slope − 12 . The equation of the tangent line is
1
y − 1 = − (x − 2).
2
15.7 #1
For part (a) we have
D = 4(2) − 12 = 7 > 0 ,
fxx (1, 1) = 4 > 0 ,
thus (1, 1) is a local minimum.
For part (b) we have
thus (1, 1) is a saddle point.
D = 4(2) − 32 = −1 < 0 ,
15.7 #13
∇f (x, y) = hex cos y, −ex sin(y)i .
Since the exponential function is never zero, the gradient can only vanish if both cos y and sin y
are zero. This never happens, thus there are no critical points, thus no local extrema.
15.7 #33
∇f (x, y) = h6x2 , 4y 3i .
Thus the only critical point is at the origin. On the boundary, we can make the substitution
y 2 = 1−x2 , and, as a function of one variable, we have f (x) = 2x3 +(1−x2 )2 = x4 +2x3 −2x2 +1.
The derivative is f ′ (x) = 4x3 + 6x2 − 4x = 2x(x + 2)(2x − 1). Therefore, our boundary critical
points are
r !
1
3
(0, ±1),
.
,±
2
4
If we evaluate the function f at each of these points, we see that the maximum is attained at
f (1, 0) = 2
and the minimum is attained at
f (−1, 0) = −2.
15.7 #39
We want to minimize the function f (x, y, z) = x2 + y 2 + z 2 . To reduce this to two variables,
2
notice we are constrained to the surface z 2 = xy + 1. Solving this equation for x gives x = z y−1 .
Substituting this expression into f gives the function of two variables
2
2
z −1
+ y2 + z2 .
f (y, z) =
y
The gradient of this function is
2
2(z − 1)
1 −2(z 2 − 1)2
2
2
4
2
3
+
2y,
2z
+
1
=
−2(z
−
1)
+
2y
,
2z
2y(z
−
1)
+
y
∇f (y, z) =
y3
y2
y3
The first component vanishes when y 2 = ±(z 2 − 1). Subbing this into the second component,
we see that we must have either z = 0 or ±2y 3 + y 3 = 0 which implies that y = 0. If z = 0,
then the requirement that the first component vanishes gives y = ±1. If y = 0, notice that
the original constraint implies that z = ±1. We therefore have 4 critical points: (±1, 0) and
(0, ±1).
The minimum distance corresponds to the points (0, 0, ±1).
15.8 #9
Let g(x, y, z) = x2 + 2y 2 + 3z 2 . The gradients we need are
∇f (x, y, z) = hyz, xz, xyi ,
The Lagrange multiplier equation is
∇g(x, y, z) = h2x, 4y, 6zi .
hyz, xz, xyi = λh2x, 4y, 6zi ,
and so we have to solve the system of equations
yz = 2λx ,
xz = 4λy ,
xy = 6λz ,
x2 + 2y 2 + 3z 2 = 6 .
Combining the first two equations gives
y
x
=
or 2y 2 = x2 .
x
2y
Similarly, we can get
3z 2 = 2y 2 , 3z 2 = x2 .
Plugging these into the last equation gives
3x2 = 6 ,
p
√
thus x = ± 2. It follows that y = ±1 and z = ± 2/3. Checking each of these values, we see
that the maximum value is
p
√
√
f ( 2, 1, 2/3) = 2/ 3,
and the minimum value is
p
√
√
f (− 2, −1, − 2/3) = −2/ 3.
15.8 #19
The gradient of f is
∇f (x, y) = −e−xy hy, xi ,
thus f has a critical value at the origin. To find the boundary critical values, we can use the
method of Lagrange multipliers with g(x, y) = x2 + 4y 2 . The gradient of g is
∇g(x, y) = h2x, 8yi ,
and so the Lagrange multiplier equations are
−e−xy y = 2λx ,
−e−xy x = 8λy ,
x2 + 4y 2 = 1 .
Combining the first two equations, we get that
x2 = 4y 2.
Plugging this into the last equation gives 2x2 = 1, thus x = ± √12 and y = ± 2√1 2 . Checking our
5 critical points, we see that the maximum occurs at f (± √12 , ∓ 2√1 2 ) = e1/4 , and the minimum
occurs at f (± √12 , ± 2√1 2 ) = e−1/4 .
Ch. 15 Review T-F Quiz
(3) This is false in general, as the derivatives are taken in the opposite order. However, if the
mild assumptions of Clairaut’s Theorem are satisfied, then it is true.
(5) This is false in general. To show that a limit exists, you need to use an ǫ and δ approach.
(7) True.
(9) False, the gradient is a vector, not a scalar.
(10) True, by the second derivative test.
Ch. 15 Review #13
1
fx (x, y) = p
,
2x + y 2
Ch. 15 Review #19
fx (x, y) = 12x2 − y 2 ,
fy (x, y) = −2xy ,
fy (x, y) = p
y
2x + y 2
fxx (x, y) = 24x ,
fxy (x, y) = fyx (x, y) = −2y .
.
fyy (x, y) = −2x ,
Ch. 15 Review #25 Let
g(x, y, z) = 3x2 − y 2 + 2x − z
so that the given surface is the level surface g(x, y, z) = 0. Then the gradient of g is normal to
the surface.
∇g(x, y, z) = h6x + 2, −2y, −1i , ∇g(1, −2, 1) = h8, 4, −1i .
This vector is normal to the surface at the specified point, thus we can use it as the normal
vector to the tangent plane and the direction vector for the normal line. The tangent plane is
given by
8(x − 1) + 4(y + 2) − 1(z − 1) = 0
and the vector equation for the normal line is
r(t) = h1, −2, 1i + th8, 4, −1i.
Ch. 15 Review #44 (a) In the direction of the gradient. (b) In the direction opposite the
gradient. (c) In the direction normal to the gradient (direction of the level curve). (d) In the
direction at an angle cos(1/2) from the gradient.
Ch. 15 Review #45 A vector in the direction of the point (4, 1) is h3, −4i. The unit vector
in this direction is v = h3/5, −4/5i. The gradient of f is ∇f (x, y) = hx−1/2 , −2yi, and at the
specified point it is ∇f (1, 5) = h1, −10i. The directional derivative is
Dv f (1, 5) = h1, −10i · h3/5, −4/5i = 3/5 + 8 = 43/5.
Ch. 15 Review #47 The gradient is ∇f (x, y) = h2xy, x2 + 2√1 y i. At the specified point,
it is h4, 9/2i. This vector
of maximal change. The rate of change in this
p gives the direction
√
direction is |h4, 5i| = 16 + 81/4 = 145/2.
Ch. 15 Review #53
∇f (x, y) = h3y − 2xy − y 2, 3x − x2 − 2xyi .
The critical points are the solutions to the system of equations
y(3 − 2x − y) = 0 ,
x(3 − x − 2y) = 0 .
The solutions are (0, 0), (0, 3), (3, 0) and (1, 1). The second derivatives are
Thus
fxx (x, y) = −2y ,
fyy (x, y) = −2x ,
fxy (x, y) = 3 − 2x − 2y .
fxx (0, 0)fyy (0, 0) − fxy (0, 0)2 = −9 < 0 ,
fxx (0, 3)fyy (0, 3) − fxy (0, 3)2 = −9 < 0 ,
fxx (3, 0)fyy (3, 0) − fxy (3, 0)2 = −9 < 0 ,
and (0, 0), (3, 0) and (0, 3) are all saddle points.
fxx (1, 1)fyy (, 1) − fxy (1, 1)2 = 4 − 1 > 0 ,
fxx (1, 1) = −2 < 0 ,
thus (1, 1) gives a local maximum. The value at this local maximum is f (1, 1) = 1.
Ch. 15 Review #55
∇f (x, y) = h4y 2 − 2xy 2 − y 3 , 8xy − 2x2 y − 3xy 2 i .
The critical points are the solutions to the equations
y 2(4 − 2x − y) = 0 ,
xy(8 − 2x − 3y) = 0 .
The solution to this system which lies on the interior of the triangle is (1, 2). On the horizontal
and vertical boundaries, the functions is identically zero. On the diagonal boundary, we have
y = 6 − x. Subbing into the formula for f , on the boundary
The derivative is
f (x, 6 − x) = (6 − x)2 (4x − x2 − x(6 − x)) = −2x(6 − x)2 .
d
f (x, 6 − x) = (6 − x)(4x − 2(6 − x)) = 6(6 − x)(x − 2) ,
dx
which is zero at x = 6 and x = 2. Of these, only x = 2 corresponds to a point on the interior
of the diagonal boundary segment. On this segment, if x = 2 the y = 4. Our critical points
are thus (1, 2) on the interior of the region, and (2, 4) on the boundary. In addition, we should
note that we could possibly have a max or min value of zero which is attained on the horizontal
and vertical boundaries. Checking our critical points, we have f (1, 2) = 16 − 4 − 8 = 4, and
f (2, 4) = −64. The maximum value is thus 4 and the minimum value is −64.
Ch. 15 Review #59 Let g(x, y) = x2 + y 2, so that the constraint is a level curve of g. The
gradients are
∇f (x, y) = h2xy, x2i ∇g(x, y) = h2x, 2yi ,
so the Lagrange multiplier equations are
2xy = 2λx ,
x2 = 2λy ,
x2 + y 2 = 1 .
√
Combining the first two equations gives, if x √
and y are not zero, 2y 2 = x2 , √so x = ± 2 y.
Using the third equation, we see that if x = ± 2 y, then y = ± √13 and x = ± √23 . Also, x = 0,
y = ±1, λ = 0 is a solution.
So we have the six critical points
(0, ±1),
√
2 1
(± √ , √ ),
3 3
√
1
2
(± √ , − √ ).
3
3
They take the values
√
√
2 1
2
2
1
2
f (0, ±1) = 0, f ((± √ , √ ) = √ , f ((± √ , − √ ) = − √ ,
3 3
3 3
3
3
3 3
2
2
thus the maximum value is 3√3 and the minimum value is − 3√3 .
16.6 #6 The part of the paraboliod lying above the plane lies above the disk x2 + y 2 = 4.
Denote this disk D(2) (the disk of radius 2). This is the region over which we will integrate.
The surface area integration formula thus gives
s 2
ZZ
ZZ
2
p
∂z
∂z
A=
+
+ 1 dA =
(−2x)2 + (−2y)2 + 1 dA.
∂x
∂y
D(2)
D(2)
This integral is easiest to evaluate in polar coordinates, giving
2
Z Z
Z
Z 2π Z 2 √
1 2π 2 √ 2
1 2π 2 2
3/2 2
(4r + 1) dθ
8r 4r + 1 drdθ =
r 4r + 1 drdθ =
8 0
8 0 3
0
0
0
r=0
√
Z 2π
√
1
1
(17
17 − 1)π
=
(173/2 − 13/2 )dθ = (17 17 − 1)(2π) =
.
12 0
12
6
Ch. 16.6 T-F Quiz
(1) True, the function is continuous on the entire region of integration, thus Fubini’s theorem
applies.
(2) False, the limits of integration on the right hand side don’t make sense.
(3) True, if we can separate variables in the integrand and are integrating over a rectangle,
the the double integral is a product of two integrals.
(4) True, the integrand is an odd function of y, and the region of integration is symmetric
with respect to the x-axis.
(5) True. This integral gives half of the volume of a sphere of radius 2, which is (1/2)(4π/3)(23) =
16π/3.
(6) True, the integrand is always less than 3 on the region of integration, and the area of
the region of integration is 3. Thus the integral is less than (3)(3) = 9.
(7) False, the integral is missing a factor of r in the integrand.