line integrals

**Dubulus** · August 6th 2008, 04:11 PM

ok...Compute the line integral bounded by C of ydx - xdy (i dont know if that was a typo if thats normal, ive never seen an integral like that)
where C is the upper semicircle of radius 1 oriented from (1; 0) to (-1; 0) (i.e. counter-clockwise).

$\int xdy - ydx$
when i did the problem i got 0 as the answer, but my teachers answer page says -Pi. any ideas?

also i used C = <cos t, sin t> from 0 to Pi.

**TwistedOne151** · August 6th 2008, 05:25 PM

Is the problem the integral of $y\,dx-x\,dy$ or $x\,dy-y\,dx$ ? In your stating words, you have the first, but for your formula you wrote the second. These two have opposite signs, and will give opposite answers. I think you mean the first, as that gives the answer you gave:

Your parametrization gives $x(t)=\cos{t}$ , $y(t)=\sin{t}$ , and thus $\frac{dx}{dt}=-\sin{t}$ and $\frac{dy}{dt}=\cos{t}$ . Thus we have:
$\int_{C}y\,dx-x\,dy=\int_0^{\pi}\left(y\frac{dx}{dt}-x\frac{dy}{dt}\right)\,dt$
$=\int_0^{\pi}\sin{t}\cdot(-\sin{t})-\cos{t}\cdot\cos{t}\,dt$
$=\int_0^{\pi}-\sin^2{t}-\cos^2{t}\,dt$
$=\int_0^{\pi}(-1)\,dt$
$=-\int_0^{\pi}\,dt$
$=-\pi$

--Kevin C.

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