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" ( ) or "minus" ( ). 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.

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" ( ) or "minus" ( ). 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.

Integrals

The integral of a real function can be imagined as the signed area between the x -axis and the curve y = f(x) over an interval . The area above the x -axis may be specified as positive ( ), and the area below the x -axis may be specified as negative ( ). The negative area arises in the study of natural logarithm as signed area below the curve y = 1/x for , that is: \ln x = \int_1 ^x \frac{dt}{t} = - \int_x^1 \frac{dt}{t} 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 region is obtained by integration of a differential 2-form: \mu(A) = \int_A dx \wedge dy, where and are the coordinate differential 1-forms. Since the wedge product has the anticommutative property, . The sign is associated with the 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 , the form is positive for , but since the integral \int_1^x \frac{dt}{t} is anchored to , the orientation of the -axis is reversed in the unit interval. For this integral, the opposite orientation yields the opposite sign to the one used for . With this opposite sign 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. \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) , then the area is given by \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 which have an area \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 \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 . Then x \ dy - y \ dx = - \cosh^2 t \ dt + \sinh^2 t \ dt = - dt , so the area of the hyperbolic sector between and is \frac{-1}{2} \int_0^\theta dt = \left. \frac{-t}{2} \right|_0^\theta = -\frac{\theta}{2}, giving a negative hyperbolic angle as a negative sector area. \begin{align} (x,y) &\to (x,\ y + kx), \\ (x,y) &\to (x + ky,\ y) \end{align} for any real number , while a squeeze mapping is (x,y) \to (\lambda x,\ y/\lambda) for any positive real number . An area element is related ( ) 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) = (ka + \ell b, \ k_1 a + \ \ell_1 b) and \delta = k \ell_1 - \ell k_1 \ne 0 , then (a, b) \thicksim (a_1, \ (\frac{1}{\delta}) b_1) . proof: (a,\ b) \thicksim (a + \frac{\ell}{k} b, \ b) shear mapping \thicksim (ka + \ell b, \ \tfrac{1}{k} b) squeeze mapping \thicksim (ka + \ell b, \ \tfrac{1}{k} b + \frac{k_1}{k \delta} (ka + \ell b) ) shear mapping = (ka + \ell b, \ \frac{k_1}{k \delta} (ka + (\frac{\delta}{k_1} + \ell) b) = (ka + \ell b, \ \frac{1}{\delta} (k_1a + \ell_1 b) ) \ \ \text{since}\ \ \ell + \frac{\delta}{k_1} = \frac{k}{k_1} \ell_1 = (a_1, \ \frac{1}{\delta} b_1) .

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