# line integrals

• Aug 6th 2008, 04:11 PM
Dubulus
line integrals
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).

$\displaystyle \int xdy - ydx$
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.
• Aug 6th 2008, 05:25 PM
TwistedOne151
Is the problem the integral of $\displaystyle y\,dx-x\,dy$ or $\displaystyle x\,dy-y\,dx$? 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 $\displaystyle x(t)=\cos{t}$, $\displaystyle y(t)=\sin{t}$, and thus $\displaystyle \frac{dx}{dt}=-\sin{t}$ and $\displaystyle \frac{dy}{dt}=\cos{t}$. Thus we have:
$\displaystyle \int_{C}y\,dx-x\,dy=\int_0^{\pi}\left(y\frac{dx}{dt}-x\frac{dy}{dt}\right)\,dt$
$\displaystyle =\int_0^{\pi}\sin{t}\cdot(-\sin{t})-\cos{t}\cdot\cos{t}\,dt$
$\displaystyle =\int_0^{\pi}-\sin^2{t}-\cos^2{t}\,dt$
$\displaystyle =\int_0^{\pi}(-1)\,dt$
$\displaystyle =-\int_0^{\pi}\,dt$
$\displaystyle =-\pi$

--Kevin C.