MATH 209, Lab 8
Richard M. Slevinsky∗
Part I
Problems
Z
1 Z 2x Z y
2 x y z dz dy dx;
1. Evaluate
0
x
0
We find:
Z
1 Z 2x Z y
Z
2 x y z dz dy dx =
0
x
1 Z 2x 0
0
Z
x
1 Z 2x
=
0
Z
x y z2
y
0
dy dx,
x y 3 dy dx,
x
1
=
0
x y4
4
2x
dx,
x
1
x5 (24 − 1)
dx,
4
0
6 4
1
24 − 1
x (2 − 1)
15
5
=
=
=
= .
4·6
24
24
8
0
Z
=
ZZZ
n
o
p
2 x dV where E = (x, y, z) ∈ R3 : 0 ≤ x ≤ 4 − y 2 , 0 ≤ y ≤ 2, 0 ≤ z ≤ y ;
2. Evaluate
E
We find:
√
ZZZ
Z
2Z
4−y 2
Z
y
2 x dV =
2 x dz dx dy,
E
0
Z
0
√
2Z
0
4−y 2
=
0
0
Z
2Z
√
[2 x z]y0 dx dy,
4−y 2
=
2 x y dx dy,
0
Z
=
0
2
0
Z
=
x2 y
√4−y2
0
dy,
2
y(4 − y 2 ) dy,
0
(4 − y 2 )2
= −
4
2
= 4.
0
3. Find the volume of the tetrahedron enclosed by the coordinate planes and the plane 2 x + y + z = 4.
The plane is z = 4 − 2 x − y and is therefore above the xy-plane in the first octant. Projecting the plane
∗
Contact: [email protected]
1
onto the xy-plane (z = 0 plane), we find 2 x + y = 4, or y = 4 − 2 x. When y = 0, x = 2. Therefore,
the region of the tetrahedron is E = (x, y, z) ∈ R3 : 0 ≤ z ≤ 4 − 2 x − y, 0 ≤ y ≤ 4 − 2 x, 0 ≤ x ≤ 2 ,
and we find:
ZZZ
Z 2 Z 4−2 x Z 4−2 x−y
V =
dV =
dz dy dx,
E
0
0
0
Z 2 Z 4−2 x
(4 − 2 x − y) dy dx,
=
0
0
2
Z
=
0
y2
(4 − 2 x)y −
2
4−2 x
dx,
0
2
(4 − 2 x)2
dx,
2
0
2
(4 − 2 x)3
43
16
= −
=
= .
2·2·3 0 4·3
3
Z
=
Exercises
3
1. Le B be the rectangular box B = (x, y, z) ∈ R : 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c . Evaluate
ZZZ
(x y 2 +
B
z 3 ) dV ;
ZZZ
(3 + 2 x y) dV , where E is the region lying above the xy-plane and below the sphere
2. Evaluate
E
x2 + y 2 + z 2 = 4;
3. Find the volume of the region E = (x, y, z) ∈ R3 : 0 ≤ x ≤ 5, 0 ≤ y ≤ 3 x, y ≤ z ≤ x + 2 .
Part II
Problems
1. Use cylindrical coordinates to find the volume of the region lying in the first octant, bounded above
by the paraboloid z = 4 − x2 − y 2 and lying within the cylinder x2 + y 2 = 2 x;
In cylindrical coordinates, the two functions are z = 4 − r2 and r2 = 2 r cos θ or r = 2 cos θ. Therefore,
the region in the problem is E = (r, θ, z) : 0 ≤ z ≤ 4 − r2 , 0 ≤ r ≤ 2 cos θ, 0 ≤ θ ≤ π/2 , and we find:
ZZZ
Z
V =
π/2 Z 2 cos θ
Z
dV =
E
4−r2
r dz dr dθ,
0
Z
0
π/2 Z 2 cos θ
=
0
0
r(4 − r2 ) dr dθ,
0
2 cos θ
(4 − r2 )2
dθ,
=
−
2·2
0
0
Z π/2 2
4
(4 − 4 cos2 θ)2
−
dθ,
=
4
4
0
Z π/2 2
4
42 sin4 θ
=
−
dθ,
4
4
0
Z π/2
=4
(1 − sin4 θ) dθ.
Z
π/2 0
2
To integrate the sin4 θ term, we expand using the angle doubling identity:
2 !
Z π/2
Z π/2
1
−
cos
2θ
1−
dθ,
(1 − sin4 θ) dθ = 4
4
2
0
0
Z π/2 Z π/2 1 − cos 2θ + cos2 2θ
1 − cos 2θ + (1 + cos 4θ)/2
1−
=4
1−
dθ = 4
dθ,
4
4
0
0
Z π/2 Z π/2
cos 4θ
5/2 + cos 2θ −
(4 − 1 + cos 2θ − (1 + cos 4θ)/2) dθ =
=
dθ,
2
0
0
sin 2θ sin 4θ π/2 5π
= 5θ/2 +
=
−
.
2
8
4
0
2. Find the volume of the region that lies inside the cone φ = α and the sphere ρ = a;
The region can be characterized as E = {(ρ, θ, φ) : 0 ≤ ρ ≤ a, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ α}. Therefore, we
find:
Z α Z 2π Z a
ZZZ
ρ2 sin φ dρ dθ dφ,
dV =
V =
0
0
0
E
Z α Z 2π 3 a
Z
ρ
a3 α
=
sin φ dφ,
sin φ dθ dφ = 2π
3 0
3 0
0
0
= 2π
ZZZ
z
3. Compute
a3
2πa3 (1 − cos α)
[− cos φ]α0 =
.
3
3
p
x2 + y 2 + z 2 dV where E is the region in problem 2.
E
The p
region in problem 2 is E = {(ρ, θ, φ) : 0 ≤ ρ ≤ a, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ α}. Since z = ρ cos φ and
ρ = x2 + y 2 + z 2 , we find:
ZZZ p
Z α Z 2π Z a
2
2
2
z x + y + z dV =
ρ cos φ ρ ρ2 sin φ dρ dθ dφ,
E
0
0
0
Z α Z 2π Z a
ρ4 cos φ sin φ dρ dθ dφ,
=
0
0
0
Z
a5 α
= 2π
cos φ sin φ dφ.
5 0
Since sin 2φ = 2 sin φ cos φ:
ZZZ p
Z
a5 α sin 2φ
2
2
2
z x + y + z dV = 2π
dφ,
5 0
2
E
a5 − cos 2φ α πa5 (1 − cos 2α)
= 2π
=
.
5
4
10
0
Exercises
ZZZ
x2 dV where E is the interior of the unit sphere centered on the origin;
1. Compute
E
2. Consider the plane y = z and the paraboloid z = x2 + y 2 . Convert these equations to cylindrical
coordinates. Solve them simultaneously to determine the r and θ values where these two surfaces
intersect. Find the volume of the region that lies between these two surfaces;
3. Find the volume of the region that lies between the paraboloids z = 10−x2 −y 2 and z = 2 x2 + y 2 − 1 .
3