MATH 253 WORKSHEET 24
MORE INTEGRATION IN POLAR COORDINATES
xy -plane,
(1) Find the volume of the solid lying above the
the cylinder
2
below the paraboloid
and inside
(x − 1) + y = 1.
(a) We found last time the set of points in the plane lying inside the cylinder is
f (r, θ) describing the height of
Solution: f (r, θ) = x2 + y 2 = r2 .
Find
Solution 1:
D = {(r, θ) | r ≤ 2 cos θ}.
the solid above each such point.
(b) Calculate the volume of the solid, that is
f (r, θ) dA.
D
θ integrate over 0 ≤ r ≤ 2 cos θ.
D is tangent to the y -axis at the origin, so θ ranges from − π2 (downward)
Natural to slice in radial lines, that is for each
The bounding circle of
to
z = x2 + y 2
2
π
2 (upward). The integral is thus
θ=π/2
2
dθ
θ=−π/2
r=2 cos θ
r r dr
=
r=0
θ=π/2
r4
dθ
4
θ=−π/2
=
r=2 cos θ
r=0
θ=π/2
cos4 θ dθ
4
θ=−π/2
2
1
(1 + cos(2θ) dθ
2
θ=−π/2
α=π
1
1 + 2 cos α + cos2 α dα
2 α=−π
1
1
3
1+0+
2π = π .
2
2
2
=
θ=+π/2
4
α=2θ
=
=
cos2 θ = 21 (1 + cos(2θ)) and (2) That the average of
cos α over a full revolution is zero, while the average of cos2 α is 21 (since cos2 α+sin2 α = 1). An
1
2
alternative to 2 is using the half-angle formula again: cos α =
2 (1 + cos 2α) and now integrate
from −π to π .
Solution 2: Let's slice in concentric circles. The largest r value is 2 (when θ = 0 this is
Here we used: (1) The half-angle formula
possible), so we can also write the integral as
r=2
θ=− arccos
3
r dr
r=0
r
2
dθ
θ=− arccos
=
2
r3 arccos
2
r
2
0
r/2=cos u
=
r
2
dr
u=0
cos3 u (u)(− sin u) du
32
u=π/2
by parts
=
u=π/2
8 −u · cos4 u u=0 + 8
=
u=π/2
cos4 u du
u=0
π/2
cos4 u du
8
0
+π/2
cos4 u du).
−π/2
(2) In this problem we will nd the electrical eld due to a sheet of charge. Suppose we have an innite
amd from now on we can continue as above (since
conducting plate in the
xy
plane, containing
σ
cos is even this is equal to 4
units of charge per unit area. The electrical eld due
to the plate must point vertically (why?), and can only depend on the height above the plate.
Date
: 4/11/2013.
1
(a) Consider a small part of the plate of area
∆A
near the point
(x, y, 0).
What is the charge
∆q
in this small part?
Solution 1: ∆q = σ∆A (charge/unit area × area).
(0, 0, z) due to the charge near (x, y, 0) is
k∆q
v
where ~
v
is the vector between the two points. Express the vertical
3 ~
|v|
component of this vector as a function of (x, y).
p
We have ~
v = h−x, −y, zi so |~v | = x2 + y 2 + z 2 and the projection of ~v on the
kσz
vertical axis is z . In other words, we have ∆Ez ≈
∆A.
(x2 +y 2 +z 2 )3/2
(b) Express the electrical eld at (0, 0, z) by an integral.
(a) By the inverse square law, the electrical eld at
given by the vector
Solution:
Solution:
Summing over the contributions from the whole plate, we get
kσz
Ez =
R2
(x2
+ y2 + z2 )
dA
3/2
(c) Evaluate the integral.
Solution:
In polar coordinates
r=∞
r dr
θ=2π
dθ
3/2
θ=0
(r2 + z 2 )
−1/2 r→∞
1
= kσz (2π) (−2) r2 + z 2
2
r=0
1
= kσz (2π) = 2πkσ .
z
= kσz
r=0
Notes: (1) For the whole plane we go over all angles (0
r2 + z 2
(2) The derivative of
As
r→∞ √
1
r 2 +z @
−1/2
is
2r
− 12 (r2 +z
2 )3/2
≤ θ ≤ 2π )
and all radii (0
≤ r < ∞).
which is our integrand up to constant (3)
→ 0.
~ =E
~?
φ(x, y, z) ( Electric potential ) such that −∇φ
~
Solution: E(x,
y, z) = 2πkσ h0, 0, 1i since the by rotational symmetry the eld must point up
~ is perpendicular to the level sets, we see that the level sets must be planes
or down. Since ∇φ
D
E
~ = 0, 0, ∂φ and if this is
parallel to the xy plane, so φ can depend on z alone. Then ∇φ
(d) Can you nd a function
∂z
φ
constant we see that
must be propotional to
z,
(3) In this problem we will nd the area under the bell curve.
R2
e
−x2 −y 2
dA
Solution:
e
Let
(integral over the whole plane).
xy
(a) Using an iterated integral in the
φ(x, y, z) = −2πkσz .
+∞
2
I = −∞ e−x dx, and let J =
specically that
−x2 −y 2
dA
+∞
=
coordinates relate
+∞
dx
R2
−∞
+∞
dye
−∞
dxe−x
=
−x2 −y 2
2
−∞
to
+∞
I.
−x2
=
+∞
dye−y
dxe
−∞
+∞
dye−y
2
2
−∞
= I2 .
−∞
+∞
2
dye−y = −∞
−∞
developing the needed mental exibility.
Note: realizing that
J
+∞
2
dxe−x = I
wasn't easy. We are slowly working on
J.
∞
2
r dre−r = 2π 21 u=0 e−u du = π
(b) Switch to polar coordinates and evaluate
Solution: J =
dθ
θ=0
∞ −u
e du = 1.
0
and used
(c) Given
σ>0
Solution:
to choose
θ=2π
r=∞
r=0
nd a number
+∞ 1
−∞
Z=
√
Ze
2
−x /2σ
Z
2
such that
dx
+∞ 1
−x
Ze
2
/2σ 2
2
u = r2
−∞
√
+∞ −u2
u=x/ 2σ 1 √
=
Z 2σ −∞ e
2πσ 2 .
where we substituted
dx = 1.
√ √
du = Z1 2σ π =
√
2πσ 2
so we need
Z
(4) The electric potential at a point
z
potential at height
Solution:
to the point
Z
due to a charge
σ dA
˛XZ ˛
density is
σ.
the distance
kσ dA
p
[−a,a]2
dA).
kq ˛
is ˛˛−
−
→˛ . Find the electrical
X
2a, if the charge
as {(x, y, 0) | −a ≤ x, y ≤ a}. Then
p
is
x2 + y 2 + z 2 , so the integral is
above the middle of the plate
φ(z) =
(recall that
at the point
above the middle of a square plate of side length
Parametrize the points on the plate
(0, 0, z)
q
x2
+ y2 + z2
is the innitesimal charge at the area element
We convert this integral to polar coordinates. For this we divide the plate into eight triangles
as in the left part of the gure below. Each of those has the same contribution (since any two dier
only by changing the sign of
x or y
or both, or by exchanging
x and y ).
So the total potential is eight
times the potential created by the triangle magnied in the right-hand side. We parameterize the
−1/2
−1/2
(x, y) instead by (r, θ) as in the gure. Then x2 + y 2 + z 2
= r2 + z 2
. Also, we can
see that r is at least zero, and at most the length of the hypotenuse in the right triangle below with
a
π
angle θ and side a. It follows that the triangle can be described as (r, θ) | 0 ≤ θ ≤
4 , 0 ≤ r ≤ cos θ .
point
The integral is then
θ= π
4
8kσ
a
r= cos
θ
dθ
θ=0
r dr
(r2 +
r=0
−1/2
z2)
=
θ= π
4
dθ
8kσ
h
r2 + z 2
1/2 ir=a/ cos θ
r=0
θ=0
=
"r
θ= π
4
8kσ
dθ
θ=0
=
θ= π
4
8kσ
θ=0
r
a2
+ z2 − z
cos2 θ
#
a2
+ z 2 dθ − 2πkσz .
cos2 θ
Remark. The solution up to here was very dicult (many ideas were needed, and using polar coordinates on a triangle would not occur to anyone), but within the scope of 253. Actually calculating
the remaining integral is not.
We now concentrate on the remaining integral.
It seems natural to write the integrand as
√
a2 +z 2 cos2 θ
a2 +z 2 cos2 θ
=
cos
θ
since cos θ dθ = d (sin θ). Changing variables this way we see
2
cos θ
cos θ
that
√
θ= π
4
θ=0
r
a2
+ z 2 dθ =
cos2 θ
√
u=1/ 2
u=0
3
√
a2 + z 2 − z 2 u2
du .
1 − u2
√
u =
We can get rid of the square root by setting
√
a2 + z 2 cos α,
a2 + z 2
z
√
u=1/ 2
so that
p
(a2 + z 2 ) − z 2 u2 =
√
u=1/ 2
2
= z(a + z )
√
u=1/ 2
= z
u=0
For convenience set
A=
cos α
cos α dα
sin2 α
a2 +z 2
z2
1−
u=0
2
u=0
cos2 α dα
z 2 − (a2 + z 2 ) sin2 α
cos2 α
dα .
2
cos2 α − a2a+z2
2
a
a2 +z 2 . We then remove the
a trick
√
u=1/ 2
=z
u=0
cos2 α
from the numerator, in preparation for
cos2 α − A
A
+
dα .
cos2 α − A cos2 α − A
We can do the rst integral, noting that
u=0
corresponds to
α=0
√
u = 1/ 2
and
corresponds to
z
so
2(a2 +z 2 )
√
u=1/ 2
z
1 dα = z arcsin p
u=0
t = tan α
z
2(a2 + z 2 )
.
1
1
cos2 α and recall that cos2 α
For the second integral, we divide through by
setting
sin α
leading so the integral equals
=
α = arcsin √
a2 +z 2
z
= 1 + tan2 α =
d(tan α)
, so
dα
we have
√
u=1/ 2
z
u=0
A dα
cos2 α − A
=
√
u=1/ 2
z
1
1
A
√
u=1/ 2
u=0
=
zA
=
u=0
1
cos2 α
dα
cos2 α
1
dt
− (1 + t2 )
1
A
u=0
√
u=1/ 2
z
−
1
dt .
− t2
1−A
A
Now,
2
1 − a2a+z2
1−A
z2
=
.
=
2
a
A
a2
a2 +z 2
This integral therefore equals
u=1/√2 a
1
= z
+
z
2z u=0
a +t
u=1/√2
z
+
t
a
=
log az
.
2
a − t u=0
√u = 0 we have α = 0
u = 1/ 2 we have sin α = √
Now at
and thus
t = tan 0 = 0,
z
a
1
−t
dt
at which point the logarithm vanishes. At
z
so
2(a2 +z 2 )
s
cos α =
Thus at
u=
p
2
1 − sin α =
z2
1−
=
2
2(a + z 2 )
√1 we have
2
t= √
z
.
2a2 + z 2
4
r
2a2 + z 2
.
2a2 + 2z 2
In conclusion,
√
u=1/ 2
√
z
a
a2 + z 2 − z 2 u2
du = z arcsin p
+ log
1 − u2
2(a2 + z 2 ) 2
z
a
z
a
+
z
2a2 +z 2
√ z
2a2 +z 2
√
−
√
z
a
2a2 + z 2 + a
= z arcsin p
+ log √
2a2 + z 2 − a
2(a2 + z 2 ) 2
u=0
and the potential is therefore
φ(z) = 8kσz arcsin p
√
2a2 + z 2 + a
+ 4kσa log √
− 2πkσz .
2a2 + z 2 − a
2(a2 + z 2 )
z
The following is entirely unrelated to MATH 253.
Let's use Taylor expansions + physics intuition to check this kind of complicated answer. We'll examine
the behaviour as
z→0
and when
(1) As
z → 0, z arcsin
expansion
4kσa log
√
z →∞
z
2(a2 +z 2 )
√
√2+1
2−1
and check if it looks reasonable.
and
+ Cz 2 + · · ·
log
√
2
2
√2a +z +a are even functions of
2a2 +z 2 −a
C . It follows that for
√
2+1
− 2πkσz + · · ·
φ(z) ≈ 4kσa log √
2−1
for some
z,
small
so jointly have a Taylor
z,
our potential looks like
In this approximation, the closer we are to the plate the more it looks innite, and indeed the nonconstant term matches the answer to 2(d). What about the constant term? Recall that
φ(z)
in 2(d)
was determined by its gradient, so we could put in it whatever constant we wanted so in fact the
answers match, except for higher-order corrections (terms in Taylor series).
(2) As
z → ∞,
z
−1/2
a2
1
1 a2
1+ 2
≈√ − √ 2.
z
2 2 2z
√
1
1
x = √2 is √
= 2, we make
√
2
1
p
=√
2
2
2
2(a + z )
Since the derivative of
see that as
arcsin x
at
1−(1/ 2)
a linear approximation to
z → ∞,
π 1 a2
4kσa2
≈ 8kσz
−
.
8kσz arcsin p
=
2πkσz
−
4
2 z2
z
2(a2 + z 2 )
z
For the second term,
√
2a2 + z 2 + a
√
2a2 + z 2 − a
2a2
z2
1+
2a2 1/2
z2
=
≈
1/2
1+
1+
a
z
a
z
1−
≈
1+
2
a
z2
2
+ az2
a 2
+
+
a
z
−
a
z
1+
≈
1−
a
z
a
z
=
1+
1−
z
1
1
to rst order in
z (that is, ignoring terms like z 2 or higher). Taking
Adding up
a 2
z
a2
z2
log we
√
2
2
2a + z + a
a
a
4kσa log √
≈ 8kσa log 1 +
≈ 8kσa .
2
2
z
z
2a + z − a
everything we see that as z → ∞
φ(z) ≈
=
4kσa2
8kσa2
+
− 2πkσz
z
z
σ(2a)2
4kσa2
=k
.
z
z
2πkσz −
5
see that
(2a)2 is
z → ∞, φ(z)
Now
the area of the plate, so
σ(2a)2
is the total charge of the plate.
In other words, as
is approximately the potential due to a charge equal to the total charge of the plate,
placed at the origin.
This is what we expect (from very far away, the plate really looks like a
point-like object).
6