Shoelace formula

Mathematical algorithm for calculating area of a simple polygon

Shoelace scheme for determining the area of a polygon with point coordinates  ( x 1 , y 1 ) , . . . , ( x n , y n ) {\displaystyle (x_{1},y_{1}),...,(x_{n},y_{n})}

The shoelace formula, also known as Gauss's area formula and the surveyor's formula,[1] is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane.[2] It is called the shoelace formula because of the constant cross-multiplying for the coordinates making up the polygon, like threading shoelaces.[2] It has applications in surveying and forestry,[3] among other areas.

The formula was described by Albrecht Ludwig Friedrich Meister (1724–1788) in 1769[4] and is based on the trapezoid formula which was described by Carl Friedrich Gauss and C.G.J. Jacobi.[5] The triangle form of the area formula can be considered to be a special case of Green's theorem.

The area formula can also be applied to self-overlapping polygons since the meaning of area is still clear even though self-overlapping polygons are not generally simple.[6] Furthermore, a self-overlapping polygon can have multiple "interpretations" but the Shoelace formula can be used to show that the polygon's area is the same regardless of the interpretation.[7]

## The polygon area formulas

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Basic idea: Any polygon edge determines the signed area of a trapezoid. All these areas sum up to the polygon area.

Given: A planar simple polygon with a positively oriented (counter clock wise) sequence of points  P i = ( x i , y i ) , i = 1 , … , n {\displaystyle P_{i}=(x_{i},y_{i}),i=1,\dots ,n} in a Cartesian coordinate system.
For the simplicity of the formulas below it is convenient to set  P 0 = P n , P n + 1 = P 1 {\displaystyle P_{0}=P_{n},P_{n+1}=P_{1}} .

The formulas:
The area of the given polygon can be expressed by a variety of formulas, which are connected by simple operations (see below):
If the polygon is negatively oriented, then the result  A {\displaystyle A} of the formulas is negative. In any case  | A | {\displaystyle |A|} is the sought area of the polygon.[8]

### Trapezoid formula

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The trapezoid formula sums up a sequence of oriented areas  A i = 1 2 ( y i + y i + 1 ) ( x i − x i + 1 ) {\displaystyle A_{i}={\tfrac {1}{2}}(y_{i}+y_{i+1})(x_{i}-x_{i+1})} of trapezoids with  P i P i + 1 {\displaystyle P_{i}P_{i+1}} as one of its four edges (see below):  A = 1 2 ∑ i = 1 n ( y i + y i + 1 ) ( x i − x i + 1 ) = 1 2 ( ( y 1 + y 2 ) ( x 1 − x 2 ) + ⋯ + ( y n + y 1 ) ( x n − x 1 ) ) {\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}(y_{i}+y_{i+1})(x_{i}-x_{i+1})\\\&={\frac {1}{2}}{\Big (}(y_{1}+y_{2})(x_{1}-x_{2})+\cdots +(y_{n}+y_{1})(x_{n}-x_{1}){\Big )}\end{aligned}}}

### Triangle formula

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The triangle formula sums up the oriented areas  A i {\displaystyle A_{i}} of triangles  O P i P i + 1 {\displaystyle OP_{i}P_{i+1}} :[9] A = 1 2 ∑ i = 1 n ( x i y i + 1 − x i + 1 y i ) = 1 2 ∑ i = 1 n | x i x i + 1 y i y i + 1 | = 1 2 ∑ i = 1 n | x i y i x i + 1 y i + 1 | = 1 2 ( x 1 y 2 − x 2 y 1 + x 2 y 3 − x 3 y 2 + ⋯ + x n y 1 − x 1 y n ) {\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}(x_{i}y_{i+1}-x_{i+1}y_{i})={\frac {1}{2}}\sum _{i=1}^{n}{\begin{vmatrix}x_{i}&x_{i+1}\\\y_{i}&y_{i+1}\end{vmatrix}}={\frac {1}{2}}\sum _{i=1}^{n}{\begin{vmatrix}x_{i}&y_{i}\\\x_{i+1}&y_{i+1}\end{vmatrix}}\\\&={\frac {1}{2}}{\Big (}x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+\cdots +x_{n}y_{1}-x_{1}y_{n}{\Big )}\end{aligned}}}

### Shoelace formula

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Shoelace scheme, vertical form: With all the slashes drawn, the matrix loosely resembles a shoe with the laces done up, giving rise to the algorithm's name.

The triangle formula is the base of the popular shoelace formula, which is a scheme that optimizes the calculation of the sum of the 2×2-Determinants by hand:  2 A = | x 1 x 2 y 1 y 2 | + | x 2 x 3 y 2 y 3 | + ⋯ + | x n x 1 y n y 1 | = | x 1 x 2 x 3 ⋯ x n x 1 y 1 y 2 y 3 ⋯ y n y 1 | {\displaystyle {\begin{aligned}2A&={\begin{vmatrix}x_{1}&x_{2}\\\y_{1}&y_{2}\end{vmatrix}}+{\begin{vmatrix}x_{2}&x_{3}\\\y_{2}&y_{3}\end{vmatrix}}+\cdots +{\begin{vmatrix}x_{n}&x_{1}\\\y_{n}&y_{1}\end{vmatrix}}\\\\[10mu]&={\begin{vmatrix}x_{1}&x_{2}&x_{3}\cdots &x_{n}&x_{1}\\\y_{1}&y_{2}&y_{3}\cdots &y_{n}&y_{1}\end{vmatrix}}\end{aligned}}}

Sometimes this determinant is transposed (written vertically, in two columns), as shown in the diagram.

### Other formulas

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A = 1 2 ∑ i = 1 n y i ( x i − 1 − x i + 1 ) = 1 2 ( y 1 ( x n − x 2 ) + y 2 ( x 1 − x 3 ) + ⋯ + y n ( x n − 1 − x 1 ) ) {\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}y_{i}(x_{i-1}-x_{i+1})\\\&={\frac {1}{2}}{\Big (}y_{1}(x_{n}-x_{2})+y_{2}(x_{1}-x_{3})+\cdots +y_{n}(x_{n-1}-x_{1}){\Big )}\end{aligned}}} A = 1 2 ∑ i = 1 n x i ( y i + 1 − y i − 1 ) {\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}x_{i}(y_{i+1}-y_{i-1})}

### Exterior Algebra

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A particularly concise statement of the formula can be given in terms of the exterior algebra. Let  v 1 , v 2 , … , v n {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}} be the consecutive vertices of the polygon. The Cartesian coordinate expansion of the outer product with respect to the standard ordered orthonormal plane basis  ( x , y ) {\displaystyle (\mathbf {x} ,\mathbf {y} )} gives  v i ∧ v i + 1 = ( x i y i + 1 − x i + 1 y i ) x ∧ y {\displaystyle \mathbf {v} _{i}\wedge \mathbf {v} _{i+1}=(x_{i}y_{i+1}-x_{i+1}y_{i})\;\mathbf {x} \wedge \mathbf {y} } and the oriented area is given as follows.  A = 1 2 ∑ i = 1 n v i ∧ v i + 1 = 1 2 ∑ i = 1 n ( x i y i + 1 − x i + 1 y i ) x ∧ y {\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}v_{i}\wedge v_{i+1}={\frac {1}{2}}\sum _{i=1}^{n}(x_{i}y_{i+1}-x_{i+1}y_{i})\;\mathbf {x} \wedge \mathbf {y} } Note that the area is given as a multiple of the unit area  x ∧ y {\displaystyle \mathbf {x} \wedge \mathbf {y} } .

Example Horizontal shoelace form for the example.

## Example

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For the area of the pentagon with  P 1 = ( 1 , 6 ) , P 2 = ( 3 , 1 ) , P 3 = ( 7 , 2 ) , P 4 = ( 4 , 4 ) , P 5 = ( 8 , 5 ) {\displaystyle {\begin{aligned}&P_{1}=(1,6),P_{2}=(3,1),P_{3}=(7,2),\\\&P_{4}=(4,4),P_{5}=(8,5)\end{aligned}}} one gets  2 A = | 1 3 6 1 | + | 3 7 1 2 | + | 7 4 2 4 | + | 4 8 4 5 | + | 8 1 5 6 | = ( 1 − 18 ) + ( 6 − 7 ) + ( 28 − 8 ) + ( 20 − 32 ) + ( 48 − 5 ) = 33 A = 16.5 {\displaystyle {\begin{aligned}2A&={\begin{vmatrix}1&3\\\6&1\end{vmatrix}}+{\begin{vmatrix}3&7\\\1&2\end{vmatrix}}+{\begin{vmatrix}7&4\\\2&4\end{vmatrix}}+{\begin{vmatrix}4&8\\\4&5\end{vmatrix}}+{\begin{vmatrix}8&1\\\5&6\end{vmatrix}}\\\&=(1-18)\;+(6-7)\;+(28-8)\;+(20-32)\;+(48-5)=33\\\A&=16.5\end{aligned}}}

The advantage of the shoelace form: Only 6 columns have to be written for calculating the 5 determinants with 10 columns.

## Deriving the formulas

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### Trapezoid formula

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Deriving the trapezoid formula

The edge  P i , P i + 1 {\displaystyle P_{i},P_{i+1}} determines the trapezoid  ( x i , y i ) , ( x i + 1 , y i + 1 ) , ( x i , 0 ) , ( x i + 1 , 0 ) {\displaystyle (x_{i},y_{i}),(x_{i+1},y_{i+1}),(x_{i},0),(x_{i+1},0)} with its oriented area

A i = 1 2 ( y i + y i + 1 ) ( x i − x i + 1 ) {\displaystyle A_{i}={\tfrac {1}{2}}(y_{i}+y_{i+1})(x_{i}-x_{i+1})}

In case of  x i < x i + 1 {\displaystyle x_{i}<x_{i+1}} the number  A i {\displaystyle A_{i}} is negative, otherwise positive or  A i = 0 {\displaystyle A_{i}=0} if  x i = x i + 1 {\displaystyle x_{i}=x_{i+1}} . In the diagram the orientation of an edge is shown by an arrow. The color shows the sign of  A i {\displaystyle A_{i}} : red means  A i < 0 {\displaystyle A_{i}<0} , green indicates  A i > 0 {\displaystyle A_{i}>0} . In the first case the trapezoid is called negative in the second case positive. The negative trapezoids delete those parts of positive trapezoids, which are outside the polygon. In case of a convex polygon (in the diagram the upper example) this is obvious: The polygon area is the sum of the areas of the positive trapezoids (green edges) minus the areas of the negative trapezoids (red edges). In the non convex case one has to consider the situation more carefully (see diagram). In any case the result is  A = ∑ i = 1 n A i = 1 2 ∑ i = 1 n ( y i + y i + 1 ) ( x i − x i + 1 ) {\displaystyle A=\sum _{i=1}^{n}A_{i}={\frac {1}{2}}\sum _{i=1}^{n}(y_{i}+y_{i+1})(x_{i}-x_{i+1})}

### Triangle form, determinant form

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Triangle form: The color of the edges indicate, which triangle area is positive (green) and negative (red) respectively.

Eliminating the brackets and using  ∑ i = 1 n x i y i = ∑ i = 1 n x i + 1 y i + 1 {\textstyle \sum _{i=1}^{n}x_{i}y_{i}=\sum _{i=1}^{n}x_{i+1}y_{i+1}} (see convention  P n + 1 = P 1 {\displaystyle P_{n+1}=P_{1}} above), one gets the determinant form of the area formula:  A = 1 2 ∑ i = 1 n ( x i y i + 1 − x i + 1 y i ) = 1 2 ∑ i = 1 n | x i x i + 1 y i y i + 1 | {\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}(x_{i}y_{i+1}-x_{i+1}y_{i})={\frac {1}{2}}\sum _{i=1}^{n}{\begin{vmatrix}x_{i}&x_{i+1}\\\y_{i}&y_{i+1}\end{vmatrix}}} Because one half of the i-th determinant is the oriented area of the triangle O , P i , P i + 1 {\displaystyle O,P_{i},P_{i+1}} this version of the area formula is called triangle form.

### Other formulas

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With  ∑ i = 1 n x i y i + 1 = ∑ i = 1 n x i − 1 y i   {\textstyle \sum _{i=1}^{n}x_{i}y_{i+1}=\sum _{i=1}^{n}x_{i-1}y_{i}\ } (see convention  P 0 = P n , P n + 1 = P 1 {\displaystyle P_{0}=P_{n},P_{n+1}=P_{1}} above) one gets  2 A = ∑ i = 1 n ( x i y i + 1 − x i + 1 y i ) = ∑ i = 1 n x i y i + 1 − ∑ i = 1 n x i + 1 y i = ∑ i = 1 n x i − 1 y i − ∑ i = 1 n x i + 1 y i {\displaystyle 2A=\sum _{i=1}^{n}(x_{i}y_{i+1}-x_{i+1}y_{i})=\sum _{i=1}^{n}x_{i}y_{i+1}-\sum _{i=1}^{n}x_{i+1}y_{i}=\sum _{i=1}^{n}x_{i-1}y_{i}-\sum _{i=1}^{n}x_{i+1}y_{i}} Combining both sums and excluding  y i {\displaystyle y_{i}} leads to  A = 1 2 ∑ i = 1 n y i ( x i − 1 − x i + 1 ) {\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}y_{i}(x_{i-1}-x_{i+1})} With the identity  ∑ i = 1 n x i + 1 y i = ∑ i = 1 n x i y i − 1 {\textstyle \sum _{i=1}^{n}x_{i+1}y_{i}=\sum _{i=1}^{n}x_{i}y_{i-1}} one gets  A = 1 2 ∑ i = 1 n x i ( y i + 1 − y i − 1 ) {\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}x_{i}(y_{i+1}-y_{i-1})}

Alternatively, this is a special case of Green's theorem with one function set to 0 and the other set to x, such that the area is the integral of xdy along the boundary.

## Manipulations of a polygon

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A ( P 1 , … , P n ) {\displaystyle A(P_{1},\dots ,P_{n})} indicates the oriented area of the simple polygon  P 1 , … , P n {\displaystyle P_{1},\dots ,P_{n}} with  n ≥ 4 {\displaystyle n\geq 4} (see above).  A {\displaystyle A} is positive/negative if the orientation of the polygon is positive/negative. From the triangle form of the area formula or the diagram below one observes for  1 < k < n {\displaystyle 1<k<n} :  A ( P 1 , … , P n ) = A ( P 1 , … , P k − 1 , P k + 1 , … , P n ) + A ( P k − 1 , P k , P k + 1 ) {\displaystyle A(P_{1},\dots ,P_{n})=A(P_{1},\dots ,P_{k-1},P_{k+1},\dots ,P_{n})+A(P_{k-1},P_{k},P_{k+1})} In case of  k = 1 or n {\displaystyle k=1\;{\text{or}}\;n} one should first shift the indices.

Hence:

1. Moving  P k {\displaystyle P_{k}} affects only  A ( P k − 1 , P k , P k + 1 ) {\displaystyle A(P_{k-1},P_{k},P_{k+1})} and leaves  A ( P 1 , . . . , P k − 1 , P k + 1 , . . . , P n ) {\displaystyle A(P_{1},...,P_{k-1},P_{k+1},...,P_{n})} unchanged. There is no change of the area if  P k {\displaystyle P_{k}} is moved parallel to  P k − 1 P k + 1 ¯ {\displaystyle {\overline {P_{k-1}P_{k+1}}}} .
2. Purging  P k {\displaystyle P_{k}} changes the total area by  A ( P k − 1 , P k , P k + 1 ) {\displaystyle A(P_{k-1},P_{k},P_{k+1})} , which can be positive or negative.
3. Inserting point  Q {\displaystyle Q} between  P k , P k + 1 {\displaystyle P_{k},P_{k+1}} changes the total area by  A ( P k , Q , P k + 1 ) {\displaystyle A(P_{k},Q,P_{k+1})} , which can be positive or negative.

Example:

P 1 = ( 3 , 1 ) , P 2 = ( 7 , 2 ) , P 3 = ( 4 , 4 ) , {\displaystyle P_{1}=(3,1),P_{2}=(7,2),P_{3}=(4,4),}
P 4 = ( 8 , 6 ) , P 5 = ( 1 , 7 ) ,   Q = ( 4 , 3 ) {\displaystyle P_{4}=(8,6),P_{5}=(1,7),\ Q=(4,3)}
Manipulations of a polygon

With the above notation of the shoelace scheme one gets for the oriented area of the

* blue polygon:  A ( P 1 , P 2 , P 3 , P 4 , P 5 ) = 1 2 | 3 7 4 8 1 3 1 2 4 6 7 1 | = 20.5 {\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&7&4&8&1&3\\\1&2&4&6&7&1\end{vmatrix}}=20.5}
* green triangle:  A ( P 2 , P 3 , P 4 ) = 1 2 | 7 4 8 7 2 4 6 2 | = − 7 {\displaystyle A(P_{2},P_{3},P_{4})={\tfrac {1}{2}}{\begin{vmatrix}7&4&8&7\\\2&4&6&2\end{vmatrix}}=-7}
* red triangle:  A ( P 1 , Q , P 2 ) = 1 2 | 3 4 7 3 1 3 2 1 | = − 3.5 {\displaystyle A(P_{1},Q,P_{2})={\tfrac {1}{2}}{\begin{vmatrix}3&4&7&3\\\1&3&2&1\end{vmatrix}}=-3.5}
* blue polygon minus point  P 3 {\displaystyle P_{3}} :  A ( P 1 , P 2 , P 4 , P 5 ) = 1 2 | 3 7 8 1 3 1 2 6 7 1 | = 27.5 {\displaystyle A(P_{1},P_{2},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&7&8&1&3\\\1&2&6&7&1\end{vmatrix}}=27.5}
* blue polygon plus point  Q {\displaystyle Q} between  P 1 , P 2 {\displaystyle P_{1},P_{2}} :  A ( P 1 , Q , P 2 , P 3 , P 4 , P 5 ) = 1 2 | 3 4 7 4 8 1 3 1 3 2 4 6 7 1 | = 17 {\displaystyle A(P_{1},Q,P_{2},P_{3},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&4&7&4&8&1&3\\\1&3&2&4&6&7&1\end{vmatrix}}=17}

One checks, that the following equations hold:  A ( P 1 , P 2 , P 3 , P 4 , P 5 ) = A ( P 1 , P 2 , P 4 , P 5 ) + A ( P 2 , P 3 , P 4 ) = 20.5 {\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})=A(P_{1},P_{2},P_{4},P_{5})+A(P_{2},P_{3},P_{4})=20.5} A ( P 1 , P 2 , P 3 , P 4 , P 5 ) + A ( P 1 , Q , P 2 ) = A ( P 1 , Q , P 2 , P 3 , P 4 , P 5 ) = 17 {\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})+A(P_{1},Q,P_{2})=A(P_{1},Q,P_{2},P_{3},P_{4},P_{5})=17}

## Generalization

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In higher dimensions the area of a polygon can be calculated from its vertices using the exterior algebra form of the Shoelace formula (e.g. in 3d, the sum of successive cross products): A = 1 2 ‖ ∑ i = 1 n v i ∧ v i + 1 ‖ {\displaystyle A={\frac {1}{2}}\left\|\sum _{i=1}^{n}v_{i}\wedge v_{i+1}\right\|} (when the vertices are not coplanar this computes the vector area enclosed by the loop, i.e. the projected area or "shadow" in the plane in which it is greatest).

This formulation can also be generalized to calculate the volume of an n-dimensional polytope from the coordinates of its vertices, or more accurately, from its hypersurface mesh.[10] For example, the volume of a 3-dimensional polyhedron can be found by triangulating its surface mesh and summing the signed volumes of the tetrahedra formed by each surface triangle and the origin: V = 1 6 ‖ ∑ F v a ∧ v b ∧ v c ‖ {\displaystyle V={\frac {1}{6}}\left\|\sum _{F}v_{a}\wedge v_{b}\wedge v_{c}\right\|} where the sum is over the faces and care has to be taken to order the vertices consistently (all clockwise or anticlockwise viewed from outside the polyhedron). Alternatively, an expression in terms of the face areas and surface normals may be derived using the divergence theorem (see Polyhedron § Volume).

Proof

Apply the divergence theorem to the vector field  v ( x , y , z ) = ( x , y , z ) {\displaystyle \mathbf {v} (x,y,z)=(x,y,z)} and the polyhedron  P {\displaystyle P} with boundary  ∂ P {\displaystyle \partial P} consisting of triangular faces  F i {\displaystyle F_{i}} :

∇ ⋅ v = ∂ v ∂ x + ∂ v ∂ y + ∂ v ∂ z = 3 {\displaystyle \nabla \cdot \mathbf {v} ={\frac {\partial \mathbf {v} }{\partial x}}+{\frac {\partial \mathbf {v} }{\partial y}}+{\frac {\partial \mathbf {v} }{\partial z}}=3} So  1 3 ∫ P ∇ ⋅ v d V = V ( P ) {\displaystyle {\frac {1}{3}}\int _{P}\nabla \cdot \mathbf {v} \,dV=V(P)}

For each triangular face  F {\displaystyle F} with vertices  v 1 , v 2 , v 3 {\displaystyle v_{1},v_{2},v_{3}} , denote the outward normal vector by  n {\displaystyle \mathbf {n} } , denote the area by  A {\displaystyle A} .

n A = ( v 3 − v 2 ) ∧ ( v 1 − v 2 ) {\displaystyle \mathbf {n} A=(v_{3}-v_{2})\wedge (v_{1}-v_{2})} is the normal vector of  F {\displaystyle F} with magnitude  A {\displaystyle A} .

The flux of  v {\displaystyle \mathbf {v} } through  F {\displaystyle F} is  ∫ F v ⋅ n d A {\displaystyle \int _{F}\mathbf {v} \cdot \mathbf {n} \,dA}

For each point  ( x , y , z ) {\displaystyle (x,y,z)} on  F {\displaystyle F} ,  v ( x , y , z ) ⋅ n {\displaystyle \mathbf {v} (x,y,z)\cdot \mathbf {n} } is the projection of the vector  ( x , y , z ) {\displaystyle (x,y,z)} onto the unit normal vector  n {\displaystyle \mathbf {n} } , which is the height  h {\displaystyle h} of the tetrahedron formed by  v 1 , v 2 , v 3 {\displaystyle v_{1},v_{2},v_{3}} and  ( 0 , 0 , 0 ) {\displaystyle (0,0,0)} . So the integrand is constant  h {\displaystyle h} on  F {\displaystyle F} .

∫ F v ⋅ n d A = A ⋅ h = ± 1 2 ‖ v 1 ∧ v 2 ∧ v 3 ‖ {\displaystyle \int _{F}\mathbf {v} \cdot \mathbf {n} \,dA=A\cdot h=\pm {\frac {1}{2}}\|v_{1}\wedge v_{2}\wedge v_{3}\|}

where  ‖ v 1 ∧ v 2 ∧ v 3 ‖ {\displaystyle \|v_{1}\wedge v_{2}\wedge v_{3}\|} is 6×the volume of the tetrahedron formed by  v 1 , v 2 , v 3 {\displaystyle v_{1},v_{2},v_{3}} and  ( 0 , 0 , 0 ) {\displaystyle (0,0,0)} . The sign of the pseudoscalar value  v 1 ∧ v 2 ∧ v 3 {\displaystyle v_{1}\wedge v_{2}\wedge v_{3}} represents the orientation of our area, and needs to be taken into account to calculate the total flux.

The total flux is the sum of the fluxes through all faces:

V ( P ) = 1 3 ∫ ∂ P v ⋅ n d A = 1 6 ‖ ∑ F v 1 ∧ v 2 ∧ v 3 ‖ {\displaystyle V(P)={\frac {1}{3}}\int _{\partial P}\mathbf {v} \cdot \mathbf {n} \,dA={\frac {1}{6}}\left\|\sum _{F}v_{1}\wedge v_{2}\wedge v_{3}\right\|}

## See also

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* Planimeter
* Polygon area
* Pick's theorem
* Heron's formula

## External links

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* Mathologer video about Gauss's shoelace formula

## References

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1. ^ Bart Braden (1986). "The Surveyor's Area Formula" (PDF). The College Mathematics Journal. 17 (4): 326–337\. doi:10.2307/2686282. JSTOR 2686282. Archived from the original (PDF) on 29 June 2014.
2. ^ a b Dahlke, Karl. "Shoelace Formula". Retrieved 9 June 2008.
3. ^ Hans Pretzsch, Forest Dynamics, Growth and Yield: From Measurement to Model, Springer, 2009, ISBN 3-540-88306-1, p. 232.
4. ^ Meister, A. L. F. (1769), "Generalia de genesi figurarum planarum et inde pendentibus earum affectionibus", Nov. Com. Gött. (in Latin), 1: 144.
5. ^ Max Koecher, Aloys Krieg: Ebene Geometrie, Springer-Verlag, 2013, ISBN 3662068095, 9783662068090, p. 116
6. ^ P.W. Shor; C.J. Van Wyk (1992), "Detecting and decomposing self-overlapping curves", Comput. Geom. Theory Appl., 2 (1): 31–50, doi:10.1016/0925-7721(92)90019-O
7. ^ Ralph P. Boland; Jorge Urrutia (2000). Polygon Area Problems. 12th Canadian Conference on Computational Geometry. pp. 159–162.
8. ^ Antti Laaksonen: Guide to Competitive Programming: Learning and Improving Algorithms Through Contests, Springer, 2018, ISBN 3319725475, 9783319725475, p. 217
9. ^ Mauren Abreu de Souza, Humberto Remigio Gamba, Helio Pedrini: Multi-Modality Imaging: Applications and Computational Techniques, Springer, 2018, ISBN 331998974X, 9783319989747, p. 229
10. ^ Allgower, Eugene L.; Schmidt, Phillip H. (1986). "Computing Volumes of Polyhedra" (PDF). Mathematics of Computation. 46 (173): 171–174\. doi:10.2307/2008221. ISSN 0025-5718. JSTOR 2008221.