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12/10/2010 8:14:51 PM
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Let's learn how to find the area of a rectangle!

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. [Edited on 12.10.2010 12:21 PM PST]
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  • [quote][b]Posted by:[/b] Gman5434 Some one prove that V=(4/3)pi*r^3 for a sphere using Calculus! [/quote]Consider a sphere of radius R centered at the origin, described by the equation x^2 + y^2 + z^2 = R^2. We wish to find its volume. An appropriate coordinate transformation would be: x = r*sin(p)cos(t) y = r*sin(p)(sin(t) z = r*sin(p) With domain D = {0 <= r <= R, 0 <= t <= 2pi, 0 <= p <= pi} The determinant of the Jacobian matrix for this transformation is r^2*sin(p) The volume of the sphere is then given by the triple integral over D of r^2*sin(p)drdpdt. Since the terms are independent, the volume integral can be expressed as the product of simple integrals, resulting in the following: Integral of r^2 from r = 0 to R Integral of sin(p) from p = 0 to pi Integral of dt from t = 0 to 2pi Which evaluates to (1/3)r^3|0,R --> (R^3)/3 - 0 = (1/3)R^3 -cos(p)|0,pi --> -(-1) - (-1) = 2 t|0,2pi --> 2pi Multiplying the results yields V = (4/3)*pi*R^3

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