Bto Cto A. On the straight-line C 1 where y= 0 and dy = 0 the first line integral has
the value of zero. On the cycloid one finds
x=r(φ−sin φ)
dx =r(1 −cos φ)dφ
y=r(1 −cos φ)
dy =rsin φ dφ
and φvaries from 2 π to zero. By substituting the known values of xand y on the
boundary of the cycloid, the line integral along C 2 becomes
2 A=
∫ 0
2 π
r(φ−sin φ)rsin φ dφ −r(1 −cos φ)r(1 −cos φ)dφ
or 2 A=r^2
∫ 2 π
0
[
1 −2 cos φ+ cos^2 φ+ sin^2 φ−φsin φ
]
dφ
2 A=r^2 [2 φ−2 sin φ+φcos φ−sin φ]^20 π
2 A= 6πr^2
Hence, the area under the curve is given by A= 3πr^2.
Change of Variable in Green’s Theorem
Often it is convenient to change variables in an integration in order to make
the integrals more tractable. If x, y are variables which are related to another set of
variables u, v by a set of transformation equations
x=x(u, v ) y=y(u, v ) (8 .28)
and if these equations are continuous and have partial derivatives, then one can
calculate
dx =∂x
∂u
du +∂x
∂v
dv dy =∂y
∂u
du +∂y
∂v
dv. (8 .29)
It is therefore possible to express the area integral (8.27) in the form
∫∫
R
dx dy =
1
2
∫
C
©x dy −y dx
∫∫
R
dx dy =
1
2
∫
C
©x(u, v )
[
∂y
∂u du +
∂y
∂v dv
]
−y(u, v )
[
∂x
∂u du +
∂x
∂v dv
]
=
1
2
∫
C
©
[
x
∂y
∂u −y
∂x
∂u
]
du +
[
x
∂y
∂v −y
∂x
∂v
]
dv.
(8 .30)
where Ris a region of the x, y -plane where the area is to be calculated.
