Question

Can somebody check result of line integral?
Integral is ydx-xdy, and 2x/pi<=y<=sinx, x<=0. By Greens theorem I got the result -4+2pi but after solving integral directly I could not get the same result as using Green(result I got solving integral directly is -4). What I'm doing wrong? I have checked integrals calculation via wolfram definite integrator but still something is not as it should be.

Answer 1

Are you there?

Answer 2

yep

Answer 3

Okay, I'm trying to think back to vector calc so I'm going to do the best I can. Using greens theorem:
\[\oint_C (L dx + M dy) = \iint _D \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right)dxdy\]
with L = y, M = -x. Therefore looking at the right side we have we know there are two intersection points (between (-2pi,0]). One is at zero and the other is at some point where:
\[\frac{2 x}{\pi}=\sin(x)\]

Answer 4

Are you sure that's x<=0?

Answer 5

not x>=0?

Answer 6

it is <=0

Answer 7

http://www.wolframalpha.com/input/?i=plot+y+%3D+sin%28x%29%2C+y+%3D+%282%2Fpi%29x

Answer 8

thats the graph i got also

Answer 9

Nvm, I think I got it either way:
\[\int\limits_0^{x_0}\int\limits_{\frac{2 x}{\pi}}^{\sin(x)}dydx=\int\limits_0^{x_0} (\sin(x)-\frac{2 x}{\pi})dx=-\cos(x_0)+1-\frac{x_0^2}{\pi}\]

Answer 10

Does my mental math seem right?

Answer 11

I'm calling x_0 the point where sin(x) = 2 x / pi

Answer 12

maybe boundaries for x i have used are not right

Answer 13

i guess x should go from 0 to the point you suggested, hmm, I trying to figure out upper value...

Answer 14

no idea whats value of x0

Answer 15

Its something you would need to approximate that's why I'm like meh about this question :/

Answer 16

i got it, its -pi/2

Answer 17

Ahh...good call.