Chapter 2
Application Background of Circle
Integrals
On many occasions we repeatedly emphasize that application of engineering
mathematics has the same importance as mathematics itself. Here we start the
discussion from circle integral, e.g.,
Z1
0
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
ð2:1Þ
Circle integral is related to capacitance C of conducting circle disk. We first
consider a conducting circle disk with an infinitely small thickness shown in
Fig. 2.1.
Assume charge density rðr Þ on the conducting circle disk. According to the
symmetry of the disk, rðr Þ is irrelevant to azimuthal angle u. Charge density on the
periphery of the disk is distributed in the form of inverse square root, i.e.,
r0
rðr Þ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi
1 ar
ð2:2Þ
in which r0 denotes a maximum magnitude of the charge density. As shown in
Fig. 2.2, on the periphery of the disk the charge density has the inverse square root
distribution.
The capacitance of the circle disk C is denoted as
C ¼ QV
© Science Press, Beijing and Springer Nature Singapore Pte Ltd. 2017
C.-H. Liang, Notes on the Ellipsoidal Function,
DOI 10.1007/978-981-10-2908-0_2
ð2:3Þ
11
12
2 Application Background of Circle Integrals
Fig. 2.1 A conducting disk
with an infinitely small
thickness
z
σ
o
a
y
x
σ (r )
Fig. 2.2 The distribution of
the charge density over a
circle disk
σ0
0
a
r
in which Q is total charge of the disk, and V is the electric potential. Hence we have
ZZ
rðrÞrdrdu
Q¼
ð2:4Þ
s
ZZ
V¼
rðrÞ
rdrdu
4per
ð2:5Þ
s
Substituting Eq. (2.2) into Eq. (2.4), we have
ZZ
r0
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r 2ffi rdrd/
1
s
a
h
i
a
Z 1 d 1 r 2
2
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 2pr0 a2
r 2ffi
1
0
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi0
2
1
1 ar 2
¼ 2pr0 a2
1
2
Q¼
a
¼ 2pr0 a2
ð2:6Þ
2 Application Background of Circle Integrals
13
Considering Eq. (2.6) and charge distribution shown in Fig. 2.2, we can completely think that the average charge density on the circle disk is
r ¼ 2r0
ð2:7Þ
Q ¼ rS ¼ rpa2 ¼ 2r0 pa2
ð2:8Þ
We rewrite Eq. (2.6) as
On the other hand, the electric potential V can be expressed as
ZZ
r0 rdrd/
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r 2ffi
4per
1
s
a
r Za
d a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼2pr0 a
r 2ffi
4pe
1
0
a
a
2pr0 a 1 r sin
¼
4pe
a 0
pr0 a
¼
4e
V¼
ð2:9Þ
According to Eq. (2.3), we can obtain the capacitance as follows:
C ¼ QV ¼ 8ea
ð2:10Þ
Further, if we express the electric potential V in an integral form, we have
Q ¼ 2pr0 a2
2pr0 a
V¼
4pe
Z1
0
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
ð2:11Þ
ð2:12Þ
Therefore, we can get
C ¼ R 14peadx
pffiffiffiffiffiffi
2
0
ð2:13Þ
1x
It can be clearly seen that capacitance
of the conducting circle disk is tightly
R 1 .pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2 .
related to the circle integral 0 dx
In addition, according to the above discussion, we can know that there are not
pffiffiffiffiffiffiffiffiffiffiffiffiffi
circle integrals of the first and the second kinds. This is because factor 1 x2 is
14
2 Application Background of Circle Integrals
always in the denominator of the studied problem. The second important application of the circle integral is to calculate the perimeter of the circle, i.e.,
Z1
L ¼ 4a
0
p
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 4a
¼ 2pa:
2
1 x2
ð2:14Þ
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