Search Results for “signed area”

Source: Signed area

In mathematics, the signed area or oriented area of a region of an affine plane is its area with orientation specified by the positive or negative sign, that is "plus" (

+

{\displaystyle +}

) or "minus" (

−

{\displaystyle -}

). More generally, the signed area of an arbitrary surface region is its surface area with specified orientation. When the boundary of the region is a simple curve, the signed area also indicates the orientation of the boundary.

Planar area

= Polygons

= Curved shapes

Integrals

, that is:

ln
⁡
x
=

∫

1

x

d
t

t

=
−

∫

x

1

d
t

t

<
0.

{\displaystyle \ln x=\int _{1}^{x}{\frac {dt}{t}}=-\int _{x}^{1}{\frac {dt}{t}}<0.}

In differential geometry, the sign of the area of a region of a surface is associated with the orientation of the surface. The area of a set A in differential geometry is obtained as an integration of a density:

μ
(
A
)
=

∫

A

d
x
∧
d
y
,

{\displaystyle \mu (A)=\int _{A}dx\wedge dy,}

where dx and dy are differential 1-forms that make the density. Since the wedge product has the anticommutative property,

d
y
∧
d
x
=
−
d
x
∧
d
y

{\displaystyle dy\wedge dx=-dx\wedge dy}

. The density is associated with a planar orientation, something existing locally in a manifold but not necessarily globally. In the case of the natural logarithm, obtained by integrating the area under the hyperbola xy = 1, the density dx ∧ dy is positive for x > 1, but since the integral

∫

1

x

d
t

t

{\displaystyle \int _{1}^{x}{\frac {dt}{t}}}

is anchored to 1, the orientation of the x-axis is reversed in the unit interval. For this integration, the (− dx) orientation yields the opposite density to the one used for x > 1. With this opposite density the area, under the hyperbola and above the unit interval, is taken as a negative area, and the natural logarithm consequently is negative in this domain.

Determinants

Signed areas were associated with determinants by Felix Klein in 1908. When a triangle is specified by three points, its area is:

1
2

|

x

1

y

1

1

x

2

y

2

1

x

3

y

3

1

|

.

{\displaystyle {\frac {1}{2}}{\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}}.}

For instance, when

(

x

3

,

y

3

)
=
(
0
,
0
)
,

{\displaystyle (x_{3},\ y_{3})=(0,0),}

then the area is given by

x

1

y

2

−

x

2

y

1

2

.

{\frac {x_{1}y_{2}-x_{2}y_{1}}{2}}.

To consider a sector area bounded by a curve, it is approximated by thin triangles with one side equipollent to (dx,dy) which have an area

1
2

|

0

0

1

x

y

1

x
+
d
x

y
+
d
y

1

|

=

1
2

(
x

d
y
−
y

d
x
)
.

{\displaystyle {\frac {1}{2}}{\begin{vmatrix}0&0&1\\x&y&1\\x+dx&y+dy&1\end{vmatrix}}={\frac {1}{2}}(x\ dy-y\ dx).}

Then the "area of the sector between the curve and two radius vectors" is given by

1
2

∫

1

2

(
x

d
y
−
y

d
x
)
.

{\displaystyle {\frac {1}{2}}\int _{1}^{2}(x\ dy-y\ dx).}

For example, the reverse orientation of the unit hyperbola is given by

x
=
cosh
⁡
t
,

y
=
−
sinh
⁡
t
.

{\displaystyle x=\cosh t,\quad y=-\sinh t.}

Then

x

d
y
−
y

d
x
=
−

cosh

2

⁡
t

d
t
+

sinh

2

⁡
t

d
t
=
−
d
t
,

{\displaystyle x\ dy-y\ dx=-\cosh ^{2}t\ dt+\sinh ^{2}t\ dt=-dt,}

so the area of the hyperbolic sector between zero and θ is

−
1

2

∫

0

θ

d
t
=

−
t

2

|

0

θ

=
−

θ
2

,

{\displaystyle {\frac {-1}{2}}\int _{0}^{\theta }dt={\frac {-t}{2}}{\big |}_{0}^{\theta }=-{\frac {\theta }{2}},}

giving a negative hyperbolic angle as a negative sector area.

Postnikov equivalence

Mikhail Postnikov's 1979 textbook Lectures in Geometry appeals to certain geometric transformations – described as functions of coordinate pairs

(
x
,
y
)

{\displaystyle (x,y)}

– to express "freely floating area elements". A shear mapping is either of:

(
x
,
y
)

→
(
x
,

y
+
k
x
)
,

(
x
,
y
)

→
(
x
+
k
y
,

y
)

{\displaystyle {\begin{aligned}(x,y)&\to (x,\ y+kx),\\(x,y)&\to (x+ky,\ y)\end{aligned}}}

for any real number

k

{\displaystyle k}

, while a squeeze mapping is

(
x
,
y
)
→
(
λ
x
,

y

/

λ
)

{\displaystyle (x,y)\to (\lambda x,\ y/\lambda )}

for any positive real number

λ

{\displaystyle \lambda }

. An area element is related

(
∼
)

{\displaystyle (\thicksim )}

to another if one of the transformations results in the second when applied to the first. As an equivalence relation, the area elements are segmented into equivalence classes of related elements, which are Postnikov bivectors.
Proposition: If

(

a

1

,

b

1

)
=
(
k
a
+
ℓ
b
,

k

1

a
+

ℓ

1

b
)

{\displaystyle (a_{1},b_{1})=(ka+\ell b,\ k_{1}a+\ \ell _{1}b)}

and

δ
=
k

ℓ

1

−
ℓ

k

1

≠
0
,

{\displaystyle \delta =k\ell _{1}-\ell k_{1}\neq 0,}

then

(
a
,
b
)
∼
(

a

1

,

(

1
δ

)

b

1

)
.

{\displaystyle (a,b)\thicksim (a_{1},\ ({\frac {1}{\delta }})b_{1}).}

proof:

(
a
,

b
)
∼
(
a
+

ℓ
k

b
,

b
)

{\displaystyle (a,\ b)\thicksim (a+{\frac {\ell }{k}}b,\ b)}

shear mapping

∼
(
k
a
+
ℓ
b
,

1
k

b
)

{\displaystyle \thicksim (ka+\ell b,\ {\tfrac {1}{k}}b)}

squeeze mapping

∼
(
k
a
+
ℓ
b
,

1
k

b
+

k

1

k
δ

(
k
a
+
ℓ
b
)
)

{\displaystyle \thicksim (ka+\ell b,\ {\tfrac {1}{k}}b+{\frac {k_{1}}{k\delta }}(ka+\ell b))}

shear mapping

=
(
k
a
+
ℓ
b
,

k

1

k
δ

(
k
a
+
(

δ

k

1

+
ℓ
)
b
)

{\displaystyle =(ka+\ell b,\ {\frac {k_{1}}{k\delta }}(ka+({\frac {\delta }{k_{1}}}+\ell )b)}

=
(
k
a
+
ℓ
b
,

1
δ

(

k

1

a
+

ℓ

1

b
)
)

since

ℓ
+

δ

k

1

=

k

k

1

ℓ

1

{\displaystyle =(ka+\ell b,\ {\frac {1}{\delta }}(k_{1}a+\ell _{1}b))\ \ {\text{since}}\ \ \ell +{\frac {\delta }{k_{1}}}={\frac {k}{k_{1}}}\ell _{1}}

=
(

a

1

,

1
δ

b

1

)
.

{\displaystyle =(a_{1},\ {\frac {1}{\delta }}b_{1}).}