1. ## Line integral

To calculate the integral of $ydx+(x^2+y^2)dy$ over the curve $y=\sqrt{4-x^2}$ between (-2,0) and (2,0), can I parameterize the curve to $x=2\cos t$, $y=2\sin t$, $\pi \leq t \leq 2\pi$?

yep

3. ## Re: Line integral

Originally Posted by polarbear73
To calculate the integral of $ydx+(x^2+y^2)dy$ over the curve $y=\sqrt{4-x^2}$ between (-2,0) and (2,0), can I parameterize the curve to $x=2\cos t$, $y=2\sin t$, $\pi \leq t \leq 2\pi$?
Note that the contour is an semicircle in the first and second quadrants in a clockwise direction.
Therefore $y= 0\to 2\to 0~\&~x= -2\to 0\to 2$

So your parametric path will not work. But $x=2\cos(t),~y=-2\sin(t),~\pi\le t\le 2\pi$ will work.

4. ## Re: Line integral

You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.

5. ## Re: Line integral

Originally Posted by romsek
You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.
I taught complex integrals so long that I recognize a standard trick question. The $y=\sqrt{4-x^2}$ is the upper half-circle in I & II.
So $0\le y\le 2~\&~-2\le x\le 2$.

6. ## Re: Line integral

Hey thanks. I did think that the curve went through the third and fourth quadrants because I thought the standard convention of angular displacement was counterclockwise. I failed to notice that the curve, as written, had to be in the first and second quadrant.

7. ## Re: Line integral

Plato keeps us all honest... or tries anyway.

8. ## Re: Line integral

So if I integrate $ydx+(x^2+y^2)dy$ over the curve $y=\sqrt{4-x^2}$ and parameterize the curve as $x=2\cos t, y=2\sin t, 0\leq t \leq\pi$ I get $\int(-4\sin^2 t +8\cos t) dt$ unless I'm doing something wrong. And if I do that integration I get $-2t+8\sin t+\sin 2t+ C$. If I evaluate that from $0$ to $\pi$ I get $-2\pi$. This is not the answer in the text book. Could someone point me in the right direction?

9. ## Re: Line integral

Originally Posted by polarbear73
So if I integrate $ydx+(x^2+y^2)dy$ over the curve $y=\sqrt{4-x^2}$ and parameterize the curve as $x=2\cos t, y=2\sin t, 0\leq t \leq\pi$ I get $\int(-4\sin^2 t +8\cos t) dt$ unless I'm doing something wrong. And if I do that integration I get $-2t+8\sin t+\sin 2t+ C$. If I evaluate that from $0$ to $\pi$ I get $-2\pi$. This is not the answer in the text book. Could someone point me in the right direction?