- Isaac Clarke
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Solving problems you could have solved by yourself since 1968.
[i]- Veteran Stormtrooper Clarke of the Ordo Malleus[/i]
Life is too short to get irritated because someone's preferences differ from yours.
~ACP Cataclysm
Posted by: Disambiguation
First, let's consider a rectangle. On a coordinate plane, we'll say it has vertices at (0,0), (a,0), (0,b), and (a,b)
Then, an orthogonal parametrization of the boundary can be obtained by taking F(x,y) = (ax, by)
The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}
Our general formula claims that: Area = integral (sqrt [[ Fx*Fx, Fx*Fy][Fy*Fx, Fy*Fy]] over the domain D. The * operation in this case is the vector scalar product. Since the matrix is positive definite, the root of its determinant will never be imaginary so we will always have a positive or zero real area.
The partial derivatives can be easily computed to be
Fx = (a, 0) and Fy = (0, b)
Evaluating the determinant yields a^2*b^2, so that sqrt(a^2*b^2) = a*b
Integrating over [0,1] with respect to x and over [0,1] with respect to y yields Area = (a*b)*1*1 = a*b
So we see that the area of a rectangle is, perhaps not obviously, the product of the lengths of its sides.Posted by: Disambiguation
Then, an orthogonal parametrization of the boundary can be obtained by taking F(x,y) = (ax, by)
The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}
Our general formula claims that: Area = integral (sqrt [[ Fx*Fx, Fx*Fy][Fy*Fx, Fy*Fy]] over the domain D. The * operation in this case is the vector scalar product. Since the matrix is positive definite, the root of its determinant will never be imaginary so we will always have a positive or zero real area.
Posted by: Disambiguation
The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}
Posted by: Disambiguation
F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}
Posted by: Disambiguation
D =
XD
[Edited on 02.03.2011 7:06 PM PST]