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