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Re: differential form
Posted:
Oct 25, 2006 3:02 PM


On Wed, 25 Oct 2006 17:31:13 +0200, "bluelabel" <bluelabel.invalid.blue@blue.b> wrote:
>Hello, >I have a problem with the integration of the form > >arcsin y dx + y dy > >over the curve > >x=cos t >y=sin t > >with 0 <= t <= Pi. >The point is that the sine function is invertible only in [Pi/2, Pi/2] and >I actually don't know what it happens when t goes fromn Pi/2 to Pi... >However, the solution is 2. What is the interval of integration? >Thank you. >
The interval of integration is t from 0 to Pi as you are given. The arcsin function is defined on its domain [1..1] and the y = sin(t) you are given has its values in that domain. Your only problem is that arcsin( sin(t) ) isn't equal to t for t between Pi/2 and Pi. But just because it isn't equal to t doesn't mean it is undefined. For example, what is arcsin(sin(3Pi/4) and arcsin(sin(2Pi/3)) etc.? There is a simple formula for sin(arcsin(t)) for t in [Pi/2, Pi]. Once you figure it out, just break up your problem into the part from 0 to Pi/2 and from Pi/2 to Pi, and you will get 2.
Lynn



