Green's Theorem: exercise problem

[Anonymous No.6193119]

Hi /sci/, I've done a problem using Green's theorem, and I just wanted to check the answer by computing the closed line integral (instead of the double integral), and so I get to different answers. (HELP! :( )

The boundry is the square (0,0), (2,0), (2,2), (0,2) oriented counter-clockwise. The integral is:
<span class="math">\oint_{(x,y)\in C} y^2\, dx + x\, dy=\iint\limits_D (1-2y) dA[/spoiler]
<span class="math">\iint\limits_D (1-2y)dA=\iint\limits_D dA -\iint\limits_D 2y dA=4-\int_{0}^{2}dx \int_{0}^{2}2y dy=4-8=-4[/spoiler]
<span class="math">\oint_{(x,y)\in C} y^2\, dx + x\, dy=-4[/spoiler]

Now, when I calculate the closed line integral directly:
C1: (0,0), (2,0), dy=0, y=0, <span class="math">\int_{C1} 0 =0[/spoiler]
C2: (2,0), (2,2), dx=0, x=1, <span class="math">\int_{C2} dy = \int_{0}^{2} dy=2[/spoiler]
C3: (2,2), (0,2), dy=0, y=1, <span class="math">\int_{C3} dx = \int_{2}^{1} dx=-2[/spoiler]
C4: (0,2), (0,0), dx=0, x=0, <span class="math">\int_{C4} 0 =0[/spoiler]

<span class="math">\oint_{(x,y)\in C} y^2\, dx + x\, dy=0[/spoiler]

I would be very grateful for help

[Anonymous No.6193146]

Here's the rest which got lost in the beginning

[Anonymous No.6193156]

>>6193119
The integrals on C2 and C3 are both wrong. On C2 x=2 not 1. On C3 y=2, gets squared to 4, and the integral should be from 0 to 2. Now you get 0 + 4 + (-8) + 0 = -4.

[Anonymous No.6193158]

>>6193156
*from 2 to 0

[Anonymous No.6193165]

>>6193156
Thank you very much!

[Anonymous No.6193878]

>>6193119 THE JAR CAN RECOGNIZE ITSELF!

[Anonymous No.6195047]

>>6193878
You can clearly see in its behaviour that it matches the pre-defined criteria for claypots to have self-awareness. Do you even mirror test?