Mathematics 205
HWK 20 Solutions
Section 16.3 p755
Problem 5, §16.3, p755. Sketch (or describe) the region of integration for the integral
Z
1
0
Z
1
−1
Z
√
1−z 2
f (x, y, z) dy dz dx.
0
Solution. For the inner integral both z and x are fixed. Imagine shooting an arrow in the positive
y-direction (i.e. parallel to the y-axis, in the direction of increasing y). The limits of integration
tell us that the arrow
enters the region when y = 0 (so when it hits the xz-plane) and leaves the
√
region when y = 1 − z 2 , so when it hits the right half of the circular cylinder y 2 + z 2 = 1, which
is a cylinder that encloses the x-axis, having circles of radius 1 as cross sections perpendicular to
the x-axis. Then the limits on z and x tell us that y-arrows are needed for every fixed pair of
values x and z satisfying −1 ≤ z ≤ 1 and 0 ≤ x ≤ 1. So the region in the xz-plane that is needed
for the outer two integrals is a rectangle. The conditions on z don’t actually limit our half a solid
cylinder any further than it’s already been limited. The conditions on x tell us to chop off the part
of the half-cylinder that lies behind the yz-plane (where x < 0) and to chop off the part of the
half-cylinder that lines in front of the line x = 1. Thus we’re left with something that looks like
half of a cylindrical log. The log lies to the right of the √
xz-plane, between the planes x = 0 and
x = 1, and to the left of the semi-cylindrical surface y = 1 − z 2 .
Here’s an effort at a sketch. Note: I accidentally did a sketch of the top half of the region, and
am not having very good luck with attempts to sketch the full region on screen, so this sketch
shows only the top half. This is a quarter of a log. The full region is half of a log. There should
be another half below the xy-plane. (See the next page for a computer-generated sketch.)
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Math 205 HWK 20 Solns continued
§16.3 p755
Here’s a computer-generated sketch (with√the axes rotated to afford a better view) showing the
four surfaces (x = 0, x = 1, y = 0, y = 1 − z 2 ) that bound the half-log.
Problem 15, §16.3, p755. Find the volume of the pyramid (or tetrahedron) with base in the
plane z = −6 and sides formed by the three planes y = 0 and y − x = 4 and 2x + y + z = 4.
Solution. We need to work out what this tetrahedron is. First think about the planes z = −6,
y = 0, and y − x = 4. The first is parallel to the xy-plane, 6 units down. The second, perpendicular
to the base plane, is just the xz-plane. The third, also perpendicular to the base plane, is a plane
that could be created by sketching the line y − x = 4 in the xy-plane and then translating it up
and down along the z-axis. Let’s see what we have so far tells us about the base of the pyramid.
Draw an xy-plane and label it z = −6. Draw in the two lines y = 0 and y − x = 4. The third
side of the base will come from where the fourth and final plane 2x + y + z = 4 meets the plane
z = −6. On this intersection we will have 2x + y = 10, so sketch that line in as well. See that the
base of the pyramid is a triangular region T , say, in the plane z = −6.
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Math 205 HWK 20 Solns continued
§16.3 p755
Two sides of the tetrahedron (the side where y = 0 and the side where y − x = 4) rise straight
up. The third side (where 2x + y + z = 4) rises on a slant until it hits the other two sides, which
happens at the point (−4, 0, 12). Here’s a sketch showing the four planes that create the four faces
of the tetrahedron.
Call
the solid region whose volume we wish to find W , say. We need to express the volume integral
R
1
W dW as a triple iterated integral. If we shoot a z-arrow upward through W , it will enter the
region W where z = −6 and it will leave the region W where 2x+y +z = 4, or in other words where
R R z=4−2x−y
z = 4 − 2x − y. Using a little shorthand, we can write the volume of W as T z=−6
1 dz dA.
Now we need to set up limits of integration for x and y, so that (x, y) will cover the base region
T . It will work best to fix y and shoot an x-arrow. Then the x-arrow enters T where x = y − 4
10 − y
and leaves T where x =
. The values for y must then vary from y = 0 to y = 6. Putting all
2
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Math 205 HWK 20 Solns continued
§16.3 p755
these pieces together, we have
volume of W =
Z
6
0
=
Z
10−y
2
y−4
6
0
Z
Z
Z
10−y
2
y−4
6
Z
4−2x−y
dz dx dy
−6
(10 − 2x − y) dx dy
£
¤x= 10−y
2
(10 − y)x − x2 x=y−4
dy
0
¶
Z 6µ
1
1
2
2
2
=
(10 − y) − (10 − y) − (10 − y)(y − 4) + (y − 4)
dy
2
4
0
¶
Z 6µ
1
2
2
=
(10 − y) − (10 − y)(y − 4) + (y − 4)
dy
4
0
¶
Z 6µ
1 2
2
2
=
25 − 5y + y − (14y − y − 40) + (y − 8y + 16) dy
4
0
¶
Z 6µ
9 2
=
y − 27y + 81 dy
4
0
·
¸6
3 3 27 2
=
y − y + 81y
4
2
0
=
= 162
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Math 205 HWK 20 Solns continued
§16.3 p755
Problem 18, §16.3, p755. Find the average value of the sum of the squares of three numbers
x, y, z, where each number is between 0 and 2.
Solution. Let f (x, y, z) = x2 + y 2 + z 2 , and let R denote the brick in 3-space that is given by
the conditions 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2. Then the number we want is the
R average value
of f over R. It is easy to compute that the volume of R is 8. We need to find R f dV . We can
use an iterated integral, in any one of the 6 possible orders of integration. For instance,
Z
R
f (x, y, z) dV =
Z
2
0
Z
2
0
Z
(x2 + y 2 + z 2 ) dx dy dz
0
¸x=2
x3
+ xy 2 + xz 2
dy dz
3
0
0
x=0
¶
Z 2Z 2µ
8
2
2
=
+ 2y + 2z
dy dz
3
0
0
¸y=2
Z 2·
8y 2y 3
2
=
+
+ 2yz
dz
3
3
0
y=0
¶
Z 2µ
32
=
+ 4z 2 dz
3
0
·
¸z=2
32z
4z 3
=
+
3
3 z=0
=
Z
2
Z
2
·
2
= 32
Now divide by the volume of R to find that the specified average value is
R
f dV
32
=
= 4.
volume of R
8
R
I don’t know about you, but I found this result a little surprising – surprising, that is, that it came
out to be a whole number. I was expecting some weird fraction.
Note that we could regard this result as saying that if the density at a point (x, y, z) is equal to
the square of its distance from the origin, then the average density for this particular brick is 4.
p
Problem 19, §16.3, p755. Let W bepthe solid cone bounded by z = x2 + y 2 and z = 2.
R
Decide (without calculating) whether W x2 + y 2 dV is positive, negative, or zero.
Solution. Since the integrand is definitely positive throughout W (except at the very bottom
tip point of the cone), the integral will be positive.
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Math 205 HWK 20 Solns continued
§16.3 p755
p
Problem 21, §16.3, p755. Let WR be the solid cone bounded by z = x2 + y 2 and z = 2.
Decide (without calculating) whether W x dV is positive, negative, or zero.
Solution. This cone has its tip at the origin, and it expands outward and upward until z = 2.
The cross sections parallel to the xy-plane are all disks with centers on the z-axis. On the half of
the cone that lies in front of the yz-plane, the integrand, x, will be positive. On the other half,
which lies behind the yz-plane, the integrand will be negative. Moreover each point in the front
half has a matching point in the back half where the integrand takes exactly the opposite value.
Therefore the integral over the front half and the integral over the back half will cancel each other
out. The total integral will be zero. A shorter description of the key facts would be: the integrand
is an odd function of the variable x and the region is symmetric with respect to the plane x = 0.
This symmetry makes the total integral zero.
p
Problem 30, §16.3, p755. Let W be the solid half-cone
bounded
by
z
=
x2 + y 2 , z = 2,
R
and the yz-plane with x ≥ 0. Decide whether the integral W x dV is positive, negative, or zero.
Solution. Every point of the region of integration satisfies x ≥ 0, and x is the function we are
integrating. Thus the integrand is ≥ 0 (and it only equals zero on the leftmost face of the region),
so the integral is positive.
Problem 37, §16.3, p755. The figure in the text shows part of a spherical ball of radius 5 cm,
namely the part that goes from the bottom of the sphere up 2 cm (so the part that lies between a
plane just tangent to the bottom of the sphere and a parallel plane 2 cm higher). Write an iterated
triple integral that represents the volume of this region.
R
Solution.
Call the solid region W . Then we want to write W 1 · dV as an iterated triple
integral. First we need to introduce a coordinate system. It will probably be simplest to turn W
upside down first. Then a coordinate system that puts the top of the sphere at the point (0, 0, 5)
so the center is at the origin. Then the sphere has the equation x2 + y 2 + z 2 = 25, and the solid
W is that portion of the ball x2 + y 2 + z 2 ≤ 25 that lies above the plane z = 3. (See sketch on the
next page.)
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Math 205 HWK 20 Solns continued
§16.3 p755
It will be easiest to integrate first with respect to z. Fixing (x, y) and shooting a z-arrow upward,
we see
p that the arrow enters the solid where z = 3 and leaves it where z hits the sphere, so where
z = 25 − x2 − y 2 . Then the point (x, y) must vary over the disk in the (x, y)-plane that is the
shadow of the disk formed by the intersection of the sphere x2 + y 2 + z 2 = 25 with the plane z = 3.
This intersection circle has equation x2 + y 2 + 9 = 25 or x2 + y 2 = 16. In other words, we need
(x, y) to vary over the disk in the xy-plane that lies inside a circle of radius 4 centered at the origin.
This gives us either
volume of W =
Z
Z √25−x2 −y2
dz dy dx
Z √16−y2 Z √25−x2 −y2
dz dx dy.
4
−4
or
volume of W =
Z
4
√
16−x2
Z
√
− 16−x2
√
−4
−
3
16−y 2
3
Better yet, take advantage of the symmetry and use, say,
volume of W = 4
Z
4
0
Z
√
0
16−x2
Z √25−x2 −y2
dz dy dx
3
Of course this integral is still messy to calculate using rectangular coordinates. You might like to
think about how you could find this volume more easily.
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