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]

[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^30,R > (R^3)/3  0 = (1/3)R^3 cos(p)0,pi > (1)  (1) = 2 t0,2pi > 2pi Multiplying the results yields V = (4/3)*pi*R^3

That's awesome.

length*width=area

Some one prove that V=(4/3)pi*r^3 for a sphere using Calculus! [Edited on 02.05.2011 10:59 AM PST]

Base*Perpendicular Height.

[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] Randomhero123 Consider a rectangle of height [i]h[/i] and width [i]w[/i] lying in the XY plane, with one corner at (0,0) and extending in the positive XY quadrant. If we take an element, it will have height h and length dw, giving an area of h*dw. Integrating from 0 [i]w[/i] gives us h*w as the area. My favorite way :)[/quote] It's better if you use an area element dhdw. Of course, the answer is always the same.[/quote]That would involve double integrals, no?

This thread makes me feel dumb. I hate you all.

[quote][b]Posted by:[/b] Randomhero123 Consider a rectangle of height [i]h[/i] and width [i]w[/i] lying in the XY plane, with one corner at (0,0) and extending in the positive XY quadrant. If we take an element, it will have height h and length dw, giving an area of h*dw. Integrating from 0 [i]w[/i] gives us h*w as the area. My favorite way :)[/quote] It's better if you use an area element dhdw. Of course, the answer is always the same.

[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] Dark Martyr 117 [quote][b]Posted by:[/b] Hank You're not as cool as Mister Math.[/quote] Agreed. I love how some of the "higher up" members have been quite intimidated by his intelligence lately though...[/quote] Mister Math, myself and a handful of others are all members of an elite secret society of mathematicians, engineers and scientists.[/quote]This secret society of you are in has 5 days to make a working Minovsky Ultracompact Fusion Reactor.

Consider a rectangle of height [i]h[/i] and width [i]w[/i] lying in the XY plane, with one corner at (0,0) and extending in the positive XY quadrant. If we take an element, it will have height h and length dw, giving an area of h*dw. Integrating from 0 [i]w[/i] gives us h*w as the area. My favorite way :)

I wish I had known this to mess with my grade six math teacher.

[quote][b]Posted by:[/b] Snipers Paradox [quote][b]Posted by:[/b] XxBLU JELLOxX [quote][b]Posted by:[/b] Snipers Paradox Why go through so much trouble just to find the area of a freaking rectangle? [/quote] This so fits [url=http://tf2wiki.net/w/images/3/3a/Your_team_lost.wav]the thread[/url][/quote] My sound card is broken.[/quote] Oh well I guess you can't hear it....

The H times W

[quote][b]Posted by:[/b] XxBLU JELLOxX [quote][b]Posted by:[/b] Snipers Paradox Why go through so much trouble just to find the area of a freaking rectangle? [/quote] This so fits [url=http://tf2wiki.net/w/images/3/3a/Your_team_lost.wav]the thread[/url][/quote] My sound card is broken.

L x W *gasp* I'm smart!

[quote][b]Posted by:[/b] Snipers Paradox Why go through so much trouble just to find the area of a freaking rectangle? [/quote] This so fits [url=http://tf2wiki.net/w/images/3/3a/Your_team_lost.wav]the thread[/url]

Why go through so much trouble just to find the area of a freaking rectangle?

[quote][b]Posted by:[/b] jondoe4362 or lengthXwidth=area on any parallelogram.[/quote]Prove it.

or lengthXwidth=area on any parallelogram. Most of that seemed pretty unnecessary.

MATH POWER!

This thread confuses me...laymen's terms?

[quote][b]Posted by:[/b] The Objectivist [quote][b]Posted by:[/b] LazerSh0t My Physics professor repeatedly tells us how easily understandable the formulas she presents to us are if we knew determinants. We don't, and she lol's at us every time, making us all sadface.[/quote]Why doesn't she teach you determinants? [/quote] Physics honours course in university = not enough time to teach us anything that isn't significantly relevant to the course. We get by fine using integrals; it's not as bad as I made it out to be.

[quote][b]Posted by:[/b] 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.[/quote][quote][b]Posted by:[/b] 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. [/quote][quote][b]Posted by:[/b] Disambiguation The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1} [/quote][quote][b]Posted by:[/b] Disambiguation F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1} [/quote][quote][b]Posted by:[/b] Disambiguation D = [/quote]XD [Edited on 02.03.2011 7:06 PM PST]