MATH 253 WORKSHEET 31
CYLINDRICAL COORDINATES
(1) Express the following surfaces in cylindrical coordinates.
(a) The cylinder of radius
Solution: r = 2.
2
about the
z = x2 + y 2 .
Solution: z = r2 .
drill bit of diameter a is used to
z -axis.
(b) The paraboloid
(2) A
drill a hole through a ball of radius
a.
What is the volume of the
remaining object?
Solution:
z -axis
We work in cylindrical coordinates, with the
being in the middle of the drilled
cylinder (1) the problem has an axis of symmetry, and (2) we want to walk up and down the axis.
We rst nd the equations of the bounding surfaces: the sphere around the ball has the equation
x2 + y 2 + z 2 = a2 , so in our coordinates r2 + z 2 = a2 . The cylinder in the middle has the equation
r = a2 (note we are given the diameter!).
• Slicing method 1: Projecting from the top to the xy plane we get the an annula shadow,
a
specically
a z -line extending from
2 ≤ r ≤ a. Above each point (r, θ) in the xy plane we see
√
√
the bottom hemisphere to the top hemisphere, that is from z = − a2 − r 2 to z = + a2 − r 2 .
The volume is then
V
=
=
u=a −r
=
r=a
dθ
θ=0
2
θ=2π
r dr
r=a
r dr2
r=a/2
u=0
2
√
z=− a2 −r 2
r=a/2
(2π)
(2π)
√
z=+ a2 −r 2
dz · 1
p
a2 − r2
√
(− du) u
u= 34 a2
=
u= 43 a2
(2π)
√
u=0
=
•
(2π)
2
3
3 2
a
4
u du
!
=
Slicing method 2: Slice in planes parallel to the
z , θ,
√
3/2
xy
3π 3
a .
2
plane (that is, planes of xed
an r -line is a ray extending perpendicular to the
z -axis.
z ).
For each
It will begin at the cylinder
√
a
z 2 − r2 . The z -range
2 ≤ r ≤
a
begins and ends at the meeting points of the cylinder and the sphere, that is where r =
2 and
and end at the sphere, so the range for the
Date
: 25/11/2013.
1
r-integral
will be
r2 + z 2 = a2
√
V
z=+
=
z=−
=
√
3
2 a
dz
√
3
2 a
dθ
(2π)
√
3
2 a
√
z=−
dz
3
2 a
r dr
r= a
2
√ 2 2
2 r= a −z
θ=0
z=+
3
2 a. The volume is thus also
r=√a2 −z2
θ=2π
z=±
that is where
r
2
r= a
2
√
3
2 a
z=+
a2
2
2
= π
dz a − z −
√
4
z=− 23 a
√
3
z=+ 2 a
3
z3
= π a2 z −
4
3 z=− √3 a
2
"
" √
√ 3#
√
√ 2# √
2 3πa3
3 1 3
3
3π 3
3 3
=
3−
=
·
−
a .
= 2πa
4 2
3 8
8
3
2
(3) Where is the center of mass of a right circular cone? Suppose the base has radius
has height
R
and the cone
H.
Solution:
z -axis is the axis of symmetry of the cone, and
(r, θ, z) is on the
√ the point
√ side of the cone then
(R, θ, H) is also on the side, and the right triangles with sides z, r, z 2 + r2 and H, R, H 2 + R2 are
z
r
similar. The equation of the side of the cone is then
H = R (note that if we put z = 0 at the base,
r
H−z
the equation would read
H = R which is ne, but complicates the algebra later). The equation
of the base is simply z = H . We now set up integrals using either slicing method:
• The shadow on the xy plane is the disc r ≤ R (coming from the base). A vertical z -line over
a point (r, θ) on the xy plane will begin on the cone, and end on the base. And integral over
such that
We set our coordinate system so that the
z=0
at the
apex.
the cone is then
This way, if the point
θ=2π
V
=
=
=
=
=
dz · 1
z= H
Rr
r=0
r=R
H
r dr H − r
R
r=0
2
3 r=R
r
r
πR2 H
2πH
−
=
.
2
3R r=0
3
1
V
θ=2π
θ=0
1
(2π)
V
r=R
dθ
=
z=H
r dr
(2π)
and
z̄
r=R
dθ
θ=0
=
θ=2π
=
dzf (r, θ, z) .
z= H
Rr
r=0
z=H
r dr
θ=0
In particular:
r=R
dθ
z=H
dz · z
r dr
z= H
Rr
r=0
r=R
r dr
r=0
z2
2
z=H
z= H
Rr
π r=R
H2
r H 2 − 2 r2 dr
V r=0
R
2
r=R
2
4
πH r
r
−
V
2
4R2 r=0
πH 2
R2
3
= H.
2
πR H/3 4
4
2
(amusingly this doesn't depend on
(on the
•
z -axis),
R).
The center-of-mass must also be on the axis of symmetry
1
3
4 H away from the apex, so 4 H above the base.
R
is a disc of radius
H z . The integral over the cone is then
so we see that it is a point
Each slice of constant
z
z=H
r=0
z=H
z=0
=
and
=
=
=
r dr · 1
θ=0
r=0
2
R2 z
H2 2
πR2 z=H 2
z dz
H 2 z=0
πR2 H 3
πR2 H
·
=
.
H2
3
3
=
=
R
r= H
z
dθ
dz (2π)
=
θ=2π
dz
=
z=0
z=H
z̄
r drf (r, θ, z) .
θ=0
V
R
r= H
z
dθ
dz
z=0
In particular,
θ=2π
1
V
1
V
z=H
θ=2π
z=0
2
2
R
r= H
z
dθ
r dr · z
r=0
2 2
R z
dzz (2π)
H2 2
dz
z=0
z=H
θ=0
z=H
πR /H
z 3 dz
2
πR H/3 z=0
3
3 H4
·
= H.
H3 4
4
3