Math 261-00
4–9–2004
Review Problems for Test 3
These problems are provided to help you study. The presence of a problem on this handout does not
imply that there will be a similar problem on the test. And the absence of a topic does not imply that it
won’t appear on the test.
Z 2 Z 1+√2x−x2
dy dx.
1. (a) Compute the exact value of
0
(b) Compute the exact value of
Z
4
−4
Z
0
√
16−x2
√
− 16−x2
Z
4
dz dy dx.
√
x2 +y 2
2. Find the volume of the solid lying below the paraboloid z = x2 + y2 and above the region in the x-y plane
bounded by y = x2 and y = x + 2.
Z 1Z 1
1
√
3. (a) Compute
dx dy.
√
3
x +4
y
0
(b) Compute
Z
0
4. Compute
Z
0
2Z
√
2x−x2
2
Z
1
3y2 cos(x4 ) dx dy +
y/2
Z
0
−2
Z
1
3y2 cos(x4 ) dx dy.
−y/2
p
x2 + y2 dy dx.
√
− 2x−x2
1 p 2
5. Find the volume of the region which lies above the cone z = √
x + y2 and below the hemisphere
3
p
z = 1 − x2 − y 2 .
6. Compute
Z
0
1Z
√
1−x2
0
Z
x+y
0
7. Find the area of the surface
p
x2 + y2 dz dy dx.
x = 2u cos v,
y = u2 ,
z = 2u sin v,
0 ≤ u ≤ 1,
0 ≤ v ≤ 2π.
8. Find the center of mass of the region in the first octant cut off by the plane 2x + 2y + z = 4, if the density
is ρ(x, y, z) = 2z + 1.
ZZ
1
(x − 2y) dx dy, where R is the parallelogram bounded by y = x + 1, y = x − 2, y = − x + 1,
9. Compute
2
R
1
and y = − x + 4.
2
Z
(x + y) dx − (x − y) dy, where σ is the path consisting of the segment from (0, 1) to
10. (a) Compute
σ
~
(−1, 0), the segment from (−1, 0) to (0, −1), the segment from (0, −1) to (1, 0), and the segment from (1, 0)
to (0, 1).
Z
F~ · d~s, where ~σ is the curve of intersection of x2 + y2 = 1 and the plane z = 2 + 2x + 3y,
(b) Compute
~
σ
traversed counterclockwise as viewed from above, and F~ = h−2y, 2x, 2i.
11. Let F~ (x, y, z) = hx2 y + z, xz, x + 3yzi. Compute curl F~ and div F~ .
1
12. Compute
Z
σ
~
(y2 + z 3 ) dx + (2xy − 2y) dy + (3xz 2 + 4) dz,
where ~σ (t) is the path which consists of the curve
1
3t
, te2(t−1), t2 (t2 + 1)3
2t + 1
8
for 0 ≤ t ≤ 1, followed by
the segment from (1, 1, 1) to (1, 2, −1).
Z
1 3
2
2
2
13. Compute
dy, where ~σ is the boundary of the square 0 ≤ x ≤ 1,
(x y − xy ) dx + 2x y + x
3
~
σ
0 ≤ y ≤ 1, traversed in the counterclockwise direction.
14. Find the area of the region enclosed by the ellipse
x2
y2
+
= 1.
4
9
Solutions to the Review Problems for Test 3
1. (a) Compute the exact value of
Z
0
√
2 Z 1+ 2x−x2
dy dx.
0
√
Rewrite y = 1 + 2x − x2 :
p
y − 1 = 2x − x2 , (y − 1)2 = 2x − x2 ,
x2 − 2x + (y − 1)2 = 0,
x2 − 2x + 1 + (y − 1)2 = 1,
(x − 1)2 + (y − 1)2 = 1.
√
Thus, y = 1 + 2x − x2 is the top half of the circle of radius 1 centered at (1, 1). The region is bounded
above by this semicircle, below by the x-axis, and on the sides by x = 0 and x = 2:
2
1.5
1
0.5
1
0.5
1.5
2
Since the integrand is 1, the integral represents the area of the region. The area is the sum of the area
of the semicircle (which has radius 1) and the rectangle below it (which is 2 by 1). Thus,
Z
0
√
2 Z 1+ 2x−x2
(b) Compute the exact value of
dy dx =
0
Z
4
−4
Z
√
16−x2
√
− 16−x2
Z
1
π
π · 12 + 2 · 1 = 2 + .
2
2
4
dz dy dx.
√
x2 +y 2
The projection of the region into the x-y plane is
−4 ≤ x ≤ √
4
√
.
− 16 − x2 ≤ y ≤ 16 − x2
2
This is the circle of radius 4 centered at the
p origin.
The bottom of the region is the cone z = x2 + y2 and the top is the plane z = 4.
-4
-2
0
2
4
4
3
2
1
0
-4
-2
0
2
4
Thus, the region is a cone with height h = 4 and radius r = 4.
Since the integrand is 1, the integral represent the volume of the region. A cone of height h and radius
1
r has volume πr 2 h. Therefore,
3
Z
4
−4
Z
√
16−x2
√
− 16−x2
Z
4
√
dz dy dx =
x2 +y 2
1
64π
π · 42 · 4 =
.
3
3
2. Find the volume of the solid lying below the paraboloid z = x2 + y2 and above the region in the x-y plane
bounded by y = x2 and y = x + 2.
The projection into the x-y plane is shown in the first picture:
4
4-1
3
0
2
1
2
1
0
3
20
15
2
10
5
1
0
-1
-0.5
0.5
1
1.5
2
Since x2 and x +2 intersect at x = −1 and at x = 2, the region is described by the following inequalities:
−1 ≤ x ≤ 2,
x2 ≤ y ≤ x + 2.
The top of the solid is z = x2 + y2 . The bottom is the x-y plane z = 0. The second picture shows the
top and the bottom; the solid is the region between them.
The volume is
x+2
Z 2 Z x+2
Z 2
Z 2
1
1
1
(x2 + y2 ) dy dx =
dx =
x2 y + y 3
x3 + 2x2 + (x + 2)3 − x5 − x6 dx =
3
3
3
−1 x2
−1
−1
x2
1 4 2 3
1
1
1
x + x + (x + 2)4 − x6 − x7
4
3
12
6
21
3
2
−1
=
639
≈ 18.25714.
35
3. (a) Compute
1
Z
Z
1
√
y
0
√
1
x3
dx dy.
+4
Interchange the order of integration:
1
0.8
0.6
0≤y≤1
√
y≤x≤1
→
→
0.4
0≤x≤1
0 ≤ y ≤ x2
0.2
0.2
0.4
0.6
1
0.8
Thus,
Z
0
1Z 1
√
y
√
1
x3 + 4
dx dy =
1
Z
0
Z
x2
√
0
1
x3 + 4
2 3
(x + 4)1/2
3
1
dy dx =
Z
1
1
√
x3 + 4
0
=
0
x2
[y] 0
dx =
Z
1
0
√
x2
dx =
x3 + 4
2 √
( 5 − 2) ≈ 0.15738.
3
Here’s the work for the integral:
Z 2
Z
Z
2p 3
x
x2
du
1
du
2√
√
√ · 2 =
√ =
u+c=
x + 4 + c.
dx =
3
u 3x
3
u
3
3
x +4
du
3
2
u = x + 4, du = 3x dx, dx = 2
3x
(b) Compute
Z
0
2
Z
1
2
4
3y cos(x ) dx dy +
y/2
Z
0
−2
Z
1
3y2 cos(x4 ) dx dy.
−y/2
Interchange the order of integration:
2
)
0≤y≤2
y
≤x≤1
2
(
)
−2 ≤ y ≤ 0
y
− ≤x≤1
2
(
1
→
0.2
0.4
0.6
0.8
1
→
0≤x≤1
−2x ≤ y ≤ 2x
-1
-2
Thus,
Z
0
2Z 1
y/2
3y2 cos(x4 ) dx dy +
Z
0
−2
Z
1
3y2 cos(x4 ) dx dy =
Z
0
−y/2
4
1
Z
2x
−2x
3y2 cos(x4 ) dy dx =
Z
1
0
2x
cos(x4 ) y3 −2x dx = 16
Z
1
0
1
x3 cos(x4 ) dx = 4 sin(x4 ) 0 = 4 sin 1 ≈ 3.36588.
Here’s the work for the integral:
Z
Z
Z
1
1
1
du
cos u du = sin u + c = sin(x4 ) + c.
x3 cos(x4 ) dx = x3 cos u · 3 =
4x
4
4
4
du
u = x4 , du = 4x3 dx, dx = 3
4x
4. Compute
Z
0
2Z
√
2x−x2
√
− 2x−x2
p
x2 + y2 dy dx.
√
Note that y = ± 2x − x2 may be rewritten as follows:
y2 = 2x − x2 ,
x2 − 2x + y2 = 0,
x2 − 2x + 1 + y2 = 1,
(x − 1)2 + y2 = 1.
This is a circle of radius 1 centered at (1, 0).
I’ll convert to polar:
1
0.5
0 ≤ x ≤ 2√
√
− 2x − x2 ≤ y ≤ 2x − x2
→
0.5
1
1.5
2
→
(
π )
π
− ≤θ≤
2
2
0 ≤ r ≤ 2 cos θ
-0.5
-1
To get the polar equation for the circle, start with x2 − 2x + y2 = 0. Then
x2 + y2 = 2x,
r 2 = 2r cos θ,
r = 2 cos θ.
π
π
Note that the whole circle is traced out once as θ goes from − to (not, for example, from 0 to 2π).
2
2
p
√
The integrand is x2 + y2 = r 2 = r. So
Z
0
2
Z
√
2x−x2
√
− 2x−x2
p
8
3
x2
+
y2
dy dx =
Z
π/2
−π/2
Z
2 cos θ
2
r dr dθ =
0
Z
π/2
−π/2
1 3
r
3
2 cos θ
dθ =
0
π/2
1
32
8
3
sin θ − (sin θ)
=
.
(cos θ) dθ =
3
3
9
−π/2
−π/2
Z
π/2
Here’s the work for the integral:
Z
Z
Z
(cos θ)3 dθ = (cos θ)2 (cos θ) dθ =
3
1 − (sin θ)2 (cos θ) dθ =
u = sin θ,
du = cos θ dθ,
5
Z
dθ =
(1 − u2 ) cos θ ·
du
cos θ
du
=
cos θ
Z
(1 − u2 ) du =
1
1
u − u3 + c = sin θ − (sin θ)3 + c.
3
3
1 p 2
5. Find the volume of the region which lies above the cone z = √
x + y2 and below the hemisphere
3
p
z = 1 − x2 − y 2 .
-0.5
0
0.5
1
0.75
0.5
0.25
0
-0.5
0
0.5
I’ll do the integral in spherical coordinates. It’s pretty clear that the ranges for θ and ρ are 0 ≤ θ ≤ 2π
and 0 ≤ ρ ≤ 1. What is the range for φ? I need to figure out the angle between the side of the cone and the
z-axis.
1
To do this, take a random point on the cone: For instance, if x = 1 and y = 0, then z = √ . Here’s
3
the picture:
1
3
1/ 3
1
2/ 3
p/3
2
p
√
1
x2 + y2 = 12 + 02 = 1) and vertical side √ (the
3
value of z).
√ I found the hypotenuse using Pythagoras. Then I scaled the triangle up by multiplying all the
sides by 3 so I could see the ratios better. In the second triangle, I can clearly see that the cone angle is
π
.
3
π
Therefore, the range on φ is 0 ≤ φ ≤ .
3
The volume is
I drew a triangle with horizontal side 1 (since r =
ZZZ
dx dy dz =
R
Z
2π
0
Z
π/3
0
Z
1
ρ2 sin φ dρ dφ dθ = 2π
0
Z
π/3
sin φ
0
1 3
ρ
3
1
dφ =
0
2π
3
Z
π/3
sin φ dφ =
0
2π
π
π/3
[− cos φ]0 = ≈ 1.04720.
3
3
6. Compute
Z
0
1Z
0
√
1−x2
Z
0
x+y
p
x2 + y2 dz dy dx.
I’ll convert to cylindrical coordinates. The ranges 0 ≤ x ≤ 1 and 0 ≤ y ≤
6
√
1 − x2 describe the interior
of the circle of radius 1 centered at the origin which lies in the first quadrant:
1
0.8
0.6
0.4
0.2
0.2
In polar coordinates, it is
0.4
0.6
1
0.8
π
0≤θ≤
2.
0≤r≤1
The limits on z become
0 ≤ z√
≤ r cos θ + r sin θ.
p
The integrand is x2 + y2 = r 2 = r.
Therefore,
Z
0
Z
π/2
0
1
Z
0
1
Z
√
1−x2
0
Z
x+y
0
Z
p
2
2
x + y dz dy dx =
0
r 2 [z]0r cos θ+r sin θ dr dθ =
1
4
Z
π/2
Z
π/2
(cos θ + sin θ)
0
(cos θ + sin θ) dθ =
0
0
Z
0
π/2
Z
1
1
Z
r cos θ+r sin θ
r 2 dz dr dθ =
0
r 3 dr dθ =
Z
π/2
(cos θ + sin θ)
0
1 4
r
4
1
dθ =
0
1
1
π/2
[sin θ − cos θ]0 = .
4
2
7. Find the area of the surface
x = 2u cos v,
y = u2 ,
z = 2u sin v,
0 ≤ u ≤ 1,
0 ≤ v ≤ 2π.
T~u = h2 cos v, 2u, 2 sin vi, T~v = h−2u sin v, 0, 2u cos vi,
ı̂
̂
k̂
T~u × T~v = 2 cos v
2u 2 sin v = h4u2 cos v, −4u(sin v)2 − 4u(cos v)2 , 4u2 sin vi =
−2u sin v 0 2u cos v h4u2 cos v, −4u, 4u2 sin vi,
p
p
p
kT~u × T~v k = 16u4 (cos v)2 + 16u2 + 16u4 (sin v)2 = 16u4 + 16u2 = 4u u2 + 1.
The area is
Z
0
2π
Z
0
1
Z
p
2
4u u + 1 du dv = 8π
0
1
1
p
1 2
8π 3/2
3/2
2
u u + 1 du = 8π (u + 1)
=
(2 − 1) ≈ 15.31780.
3
3
0
Here’s the work for the integral:
Z
Z p
Z
√
dw
1
1 √
1
2
w dw = w 3/2 + c = (u2 + 1)3/2 + c.
u u + 1 du = u w ·
=
2u
2
3
3
7
w = u2 + 1,
dw = 2u du,
du =
dw
2u
8. Find the center of mass of the region in the first octant cut off by the plane 2x + 2y + z = 4, if the density
is ρ(x, y, z) = 2z + 1.
z
4
y
2
y=2-x
2
2
x
2
y
x
The region is shown in the first picture. The top is the plane z = 4 − 2x − 2y, and the bottom is z = 0.
Thus, the range for z is 0 ≤ z ≤ 4 − 2x − 2y.
The projection into the x-y plane is shown in the second picture. It is
0≤x≤2
.
0≤y ≤2−x
The mass is
Z
2
0
Z
0
2
Z
0
Z
2−x
0
2−x
Z
4−2x−2y
(2z + 1) dz dy dx =
0
(4 − 2x − 2y) + (4 − 2x − 2y) dy dx =
0
2
2
0
2
Z
Z
1
1
(4 − 2x)3 + (4 − 2x)2
6
4
Z
0
2
Z
0
2−x z2 + z
4−2x−2y
0
dy dx =
1
1
− (4 − 2x − 2y)3 − (4 − 2x − 2y)2
6
4
2
1
1
dx = − (4 − 2x)4 − (4 − 2x)3 = 8.
48
24
0
The region is symmetric and x and y, and so is the density function. Therefore, x = y.
I’m going to omit the ugly details of the integrations for the moments.
The x-moment is
Z 2 Z 2−x Z 4−2x−2y
52
.
x(2z + 1) dz dy dx =
15
0
0
0
52
13
13
15
=
. Likewise, y =
.
Hence, x =
8
30
30
The z-moment is
Z 2 Z 2−x Z 4−2x−2y
0
0
z(2z + 1) dz dy dx =
0
56
7
Hence, z = 5 = .
8
5
8
56
.
5
2−x
0
dx =
9. Compute
1
(x − 2y) dx dy, where R is the parallelogram bounded by y = x + 1, y = x − 2, y = − x + 1,
2
R
ZZ
1
and y = − x + 4.
2
y
(2,3)
y = -(1/2)x + 4
y=x+1
(4,2)
(0,1)
y=x-2
y = -(1/2)x + 1
x
(2,0)
I graphed the lines and found the intersection points. Next, I’ll construct a transformation from the
square 0 ≤ u ≤ 1, 0 ≤ v ≤ 1, onto the parallelogram. I’ll use (0, 1) as my reference point.
The vector from (0, 1) to (2, 0) is h2, −1i. The vector from (0, 1) to (2, 3) is h2, 2i. The transformation
is
x
2 2
u
0
=
+
.
y
−1 2
v
1
If I multiply out and combine terms on the right, then equate corresponding components, I get
x = 2u + 2v,
y = −u + 2v + 1,
0 ≤ u ≤ 1,
0 ≤ v ≤ 1.
The Jacobian is
det 2 −1 = 6.
2 2 The integrand is
x − 2y = (2u + 2v) − 2(−u + 2v + 1) = 4u − 2v − 2.
Hence,
ZZ
R
(x − 2y) dx dy =
Z
0
1
Z
0
1
(4u − 2v − 2)(6) du dv = 6
1
−12 v2
2
10. (a) Compute
Z
σ
~
1
0
Z
0
1
2
1
2u − 2uv − 2u 0 dv = 6
Z
1
(−2v) dv =
0
= −6.
(x + y) dx − (x − y) dy, where σ is the path consisting of the segment from (0, 1) to
(−1, 0), the segment from (−1, 0) to (0, −1), the segment from (0, −1) to (1, 0), and the segment from (1, 0)
9
to (0, 1).
(0,1)
A: y = x + 1
D: y = 1 - x
(-1,0)
(1,0)
B: y = -x - 1
C: y = x - 1
(0,-1)
I’ll break the integral up into four segments, as shown in the picture.
Segment A is y = x + 1. x + y = 2x + 1, x − y = −1, dy = dx, and x goes from 0 to −1. The integral is
Z
−1
0
is
−1
Z
((2x + 1) dx − (−1) dx) =
0
−1
(2x + 2) dx = x2 + 2x 0 = −1.
Segment B is y = −x − 1. x + y = −1, x − y = 2x + 1, dy = −dx, and x goes from −1 to 0. The integral
Z 0
Z 0
0
(−dx − (2x + 1) dx) =
(−2x − 2) dx = −x2 − 2x −1 = −1.
−1
−1
Segment C is y = x − 1. x + y = 2x − 1, x − y = 1, dy = dx, and x goes from 0 to 1. The integral is
Z
1
((2x − 1) dx − dx) =
0
Z
1
0
1
(2x − 2) dx = x2 − 2x 0 = −1.
Segment D is y = 1 − x. x + y = 1, x − y = 2x − 1, dy = −dx, and x goes from 1 to 0. The integral is
Z
Therefore,
Z
σ
~
(b) Compute
Z
~
σ
0
1
(dx − (2x − 1)dx) =
Z
1
0
0
(2 − 2x) dx = 2x − x2 1 = −1.
(x + y) dx − (x − y) dy = −1 + (−1) + (−1) + (−1) = −4.
F~ · d~s, where ~σ is the curve of intersection of x2 + y2 = 1 and the plane z = 2 + 2x + 3y,
traversed counterclockwise as viewed from above, and F~ = h−2y, 2x, 2i.
-2
-1
0
1
2
10
5
0
-5
-2
-1
0
1
2
10
The projection of the curve into the x-y plane is the circle x2 + y2 = 1, which may be parametrized
by x = cos t, y = sin t, 0 ≤ t ≤ 2π. Note that this parameter range traverses the circle counterclockwise as
viewed from above.
Plugging these expressions into z = 2 + 2x + 3y, I get z = 2 + 2 cos t + 3 sin t. Hence, the curve of
intersection is
~σ (t) = hcos t, sin t, 2 + 2 cos t + 3 sin ti.
Therefore,
~σ ′ (t) = h− sin t, cos t, −2 sin t + 3 cos ti.
The integrand is
F~ (~σ (t)) · ~σ ′ (t) = h−2 sin t, 2 cos t, 2i · h− sin t, cos t, −2 sin t + 3 cos ti =
2(sin t)2 + 2(cos t)2 − 4 sin t + 6 cos t = 2 − 4 sin t + 6 cos t.
The integral is
Z
σ
~
F~ · d~s =
Z
2π
0
2π
(2 − 4 sin t + 6 cos t) dt = [2t + 4 cos t + 6 sin t]0 = 4π ≈ 12.56637.
11. Let F~ (x, y, z) = hx2 y + z, xz, x + 3yzi. Compute curl F~ and div F~ .
curl F~ = ı̂
∂
∂x
x2 y + z
̂
∂
∂y
xz
k̂
d
= h3z − x, 1 − 1, z − x2 i = h3z − x, 0, z − x2 i.
dz x + 3yz
div F~ = 2xy + 0 + 3y = 2xy + 3y.
12. Compute
Z
σ
~
(y2 + z 3 ) dx + (2xy − 2y) dy + (3xz 2 + 4) dz,
where ~σ (t) is the path which consists of the curve
the segment from (1, 1, 1) to (1, 2, −1).
3t
1
, te2(t−1), t2 (t2 + 1)3
2t + 1
8
for 0 ≤ t ≤ 1, followed by
It would be very tedious to compute the line integral directly, and it should lead you to ask yourself
whether there might not be an easier way. Well,
ı̂
̂
k̂
∂
∂
∂
2
3
2
curlhy + z , 2xy − 2y, 3xz + 4i = = h0, 3z 2 − 3z 2 , 2y − 2yi = h0, 0, 0i.
∂x
∂y
∂z
2
y + z 3 2xy − 2y 3xz 2 + 4
The field is conservative. I’ll find a potential function f. I want
∂f
= y2 + z 3 ,
∂x
∂f
= 2xy − 2y,
∂y
∂f
= 3xz 2 + 4.
∂z
Integrate the first equation with respect to x:
Z
f(x, y, z) = (y2 + z 3 ) dx = xy2 + xz 3 + C(y, z).
11
C(y, z) is an arbitrary constant depending on y and z. Differentiate with respect to y and set the result
∂f
equal to
= 2xy − 2y:
∂y
∂f
∂C
=
= 2xy − 2y.
2xy +
∂y
∂y
∂C
Cancelling 2xy’s, I get
= −2y, so
∂y
Z
C = −2y dy = −y2 + D(z).
D(z) is an arbitrary constant depending on z. Then
f(x, y, z) = xy2 + xz 3 − y2 + D(z).
∂f
= 3xz 2 + 4:
Differentiate with respect to z and set the result equal to
∂z
∂f
3xz 2 + D′ (z) =
= 3xz 2 + 4.
∂z
Cancelling 3xz 2 ’s, I get D′ (z) = 4, so D(z) = 4z + E. Now E is a numerical arbitrary constant, and
since I need some potential function, I can take E = 0. Then
f(x, y, z) = xy2 + xz 3 − y2 + 4z.
3t
2(t−1) 1 2 2
3
Now ~σ starts at (0, 0, 0) (as you see by plugging t = 0 into
, te
, t (t + 1) ), and it ends
2t + 1
8
at (1, 2, −1). By path independence,
Z
(y2 + z 3 ) dx + (2xy − 2y) dy + (3xz 2 + 4) dz = f(1, 2, −1) − f(0, 0, 0) = −5.
~
σ
1
2x2 y + x3 dy, where ~σ is the boundary of the square 0 ≤ x ≤ 1,
3
~
σ
0 ≤ y ≤ 1, traversed in the counterclockwise direction.
13. Compute
Z
(x2 y − xy2 ) dx +
By Grren’s Theorem,
Z 1Z 1
Z 1Z 1
1 3
2
2
2
(x y − xy ) dx + 2x y + x dy =
(4xy + x2 ) − (x2 − 2xy) dx dy =
6xy dx dy =
3
~
σ
0
0
0
0
1
Z 1
Z 1
2 1
3
3 2
y
= .
3x y 0 dy =
3y dy =
2
2
0
0
0
Z
14. Find the area of the region enclosed by the ellipse
x2
y2
+
= 1.
4
9
Parametrize the ellipse by
~σ (t) = h2 cos t, 3 sin ti, 0 ≤ t ≤ 2π.
This curve traverses the ellipse counterclockwise. By Green’s Theorem, the area is
Z 2π
Z 2π
Z 2π
Z
Z 2π
dy
2
dt =
(2 cos t)(3 cos t) dy = 6
(cos t) dt = 3
(1 + cos 2t) dt =
x dy =
x·
dt
0
0
0
σ
~
0
2π
1
3 t + sin 2t
= 6π.
2
0
Soon you will have forgotten the world, and the world will have forgotten you. - Marcus Aurelius
c 2006 by Bruce Ikenaga
12