MATH 253 WORKSHEET 32
SPHERICAL COORDINATES
(1) Express the following surfaces in spherical coordinates.
(a) The sphere of radius
2
about the origin.
Solution: ρ ≤ 2
(b) The double cone
Solution:
z 2 = x2 + y 2 .
ρ2 cos2 φ = r2 = ρ2 sin2 φ,
This reads
z = x2 + y 2 .
Solution: ρ cos φ = ρ2 sin2 φ
that is
|tan φ| = 1,
so
φ=
π
3π
,φ =
4
4
.
(c) The paraboloid
(2) Let
B
1
be the ball of radius
Solution:
The domain is
e−(x
2
+y 2 +z 2 )
so
ρ=
cos φ
.
sin2 φ
ρ≤
3/2
dV
3/2
2
2
2
e−(x +y +z ) dV .
B
1, so the integral factors in spherical coordinates:
about the origin. Evaluate
θ=2π
=
B
sin φ dφ
θ=0
φ=π
dθ
φ=0
! θ=2π
=
φ=π
dθ
ρ=1
3
ρ2 dρe−ρ
ρ=0
! ρ=1
sin φ dφ
θ=0
φ=0
3
ρ2 dρe−ρ
ρ=0
ρ=1
3
1
φ=π
= (2π) [− cos φ]φ=0 − e−ρ
3
ρ=0
−1
4π
1
1−e
=
1−
.
= (2π) (2)
3
3
e
(3) Describe the following regions in words, then set up integration in spherical coordinates:
(a)
E = (x, y, z) | x, y, z ≥ 0, x2 + y 2 + z 2 ≤ 9
Solution:
This is one eighth of the ball of radius
points are in the positive quadrant is equivalent to
the points have
z≥0
is equivalent to
θ=π/2
0≤φ≤
θ=0
(b)
π
2 . In summary, the integral would read
φ=π/2
dθ
3. The ball is dened by ρ ≤ 3. That the
0 ≤ θ ≤ π2 (think polar coordinates). That
ρ=3
ρ2 dρf
sin φ dφ
φ=0
ρ=0
E = (x, y, z) | x2 + y 2 + (z − 1)2 ≤ 1
Solution: This is the ball of radius 1 about (0, 0, 1). The condition is equivalent to x2 + y 2 +
(z 2 − 2z + 1) ≤ 1, that is x2 + y 2 + z 2 ≤ 2z , which reads ρ2 ≤ 2ρ cos φ, or ρ ≤ 2 cos φ. Since the
π
ball is above the xy plane, we have 0 ≤ φ ≤
2 , so the integral is
θ=2π
φ=0
dρ
ρ=2 cos φ
ρ2 dρf .
sin φ dφ
θ=0
If one wants to integrate
φ=π/2
dθ
ρ=0
ρextends from 0 to 2 (the point most
(0, 0, 2)). Then φ must satisfy 0 ≤ φ ≤
rst, then
origin is the north point of the ball, at
so the integral reads
θ=2π
Date
φ=cos−1 (ρ/2)
2
dθ
θ=0
ρ=2
ρ dρ
ρ=0
sin φ dφf .
φ=0
: 27/11/2013.
1
distant from the
ρ
π
2 and cos φ ≥ 2 ,
Cylindrical or Spherical?
(1) Let
E
x2 + y 2 + z 2 = 2
be the dimple inside the sphere
and above the paraboloid
z = x2 + y 2 .
Set
up integration on it in spherical and cylindrical coordinates.
Cylindrical:
p
z =
x2 +
(x, y, z) below the upper hemisphere, that is belog the graph of
2
2 − − y , and above the graph
√ of the paraboloid. In cylindrical coordinates we have
y 2 = r2 so this reads: r2 ≤ z ≤ 2 − r2 . What about r, θ? The problem is clearly invariant
We need points
x2
under rotation around
z -axis, so no constraint on θ.
r, at the origin we have r = 0 and the largest
For
circle in our dimple is at the intersection of the paraboloid and the sphere, that is on the circle of
R2 + (R2 )2 = 2 (plugging in z = r2 into the equation z 2 + r2 = 2 of the sphere). The
2 2
last equation can be rearranged to R
+ R2 − 2 = 0 and factors as R2 + 2 R2 − 1 = 0 which
has the unique positive root R = 1. It follows that 0 ≤ r ≤ 1, so we have
radius
R
where
θ=2π
dθ
√
z= 2−r 2
r dr
θ=0
Cylindrical, other order:
r=1
dzf .
z=r 2
r=0
√z -range is 0 ≤ z ≤ 2 (from origin to north pole of
r ≤ 2 − z 2 . Note that
√ the two constraints point in the
√
same direction, so we take the minimum. If z ≤ 1 then
z ≤ 2 − z 2 (the plane of height z exits
√
√
2
2 − z ≤ z (the plane at height z exits the dimple at the
the dimple at the cone). If z ≥ 1 then
sphere), and given
z
In our dimple the
r≤
we have
√
z
and
sphere). The integral is then
θ=2π
dθ
θ=0
Spherical:
z=1
√
r= z
dz
θ=2π
r drf +
z=0
dθ
r=0
z=2
√
r= 2−z 2
dz
θ=0
z=1
r drf .
r=0
(ρ, θ, φ) is inside the sphere
√ and above the paraboloid.
θ, and to be inside the sphere simply means ρ ≤ 2. To be above the paraboloid
2
cos φ
2
. So we have the same problem as in the second
means z ≥ r so ρ cos φ ≥ (ρ sin φ) or ρ ≤
sin2 φ
cylindrical case: for small φ (near the north pole) the radial line ends on the sphere. For larger φ
(near the xy plane) the radial line ends on the paraboloid instead. The changeover occurs on the
π
circle of intersection, which is at z = 1, r = 1 so at tan φ = 1 and φ =
4 . The integral is then
We need to decide if a point
Nothing depends on
θ=2π
dθ
θ=0
φ=π/4
√
ρ= 2
φ=π/2
dθ
ρ=0
φ=0
θ=2π
ρ2 dρf +
sin φ dφ
sin φ dφ
θ=0
cos φ
ρ= sin
2φ
ρ2 dρf .
ρ=0
φ=π/4
π
For this we also used that 0 ≤ φ ≤
2 since we are above the xy plane
Cylindrical was easiest since we didn't need to break the domain in two.
p
x2 + y 2 and below the plane z = 12 . Set up integration
(2) Let E be the region above the cone 3z =
Disucssion:
on it.
Cylindrical:
θ. Being between the cone
r
1
1
≤
z
≤
. The largest radius is at the base of the cone, when z =
3
2
2 and hence
3
, so the integral reads
2
Symmetry uner rotation means there is no constraint on
and the plane reads
r=
θ=2π
dθ
θ=0
Cylindrical, other order:
also
r=3/2
z=1/2
r dr
r=0
dzf .
z=r/3
We can instead interpret the constraint as
θ=2π
θ=0
z=1/2
dθ
r ≤ 3z ,
so the integral is
r=3z
dz
r drf .
z=0
r=0
r
Points on the cone have tan φ =
z = 3, so being above the cone means
tan−1 (3). The plane z = 12 is ρ cos φ = 12 , so the integral is
Spherical:
θ=2π
dθ
θ=0
φ=tan−1 (3)
1
ρ= 2 cos
φ
sin φ dφ
φ=0
ρ=0
2
ρ2 dρf .
0 ≤ φ ≤
Disucssion:
depend on
Now there is no obviuos advantage to either coordiante system; the choice will
f.
3