I have this line integral:

$\oint 3ydx+x^2dy$

and the path is a line from $(0, 0)$ to $(1, 0)$ (so this is $y=0$), another line from $(1, 0)$ to $(1, 1)$ (so this is $x=1$) and a curve $y=x^2$ from $(1, 1)$ to $(0, 0)$.

Evaluating this integral using the line integral (and anticlockwise = positive):
1) $y=0$ gives $0$
2) $x=1$ gives $0$ as well
Edit: this gives actually $1$
3) $y=x^2$
$dy=2xdx$ and substituing everything in gives $\oint 3x^2dx+x^2*2xdx=\oint 3x^2+2x^3dx$. The limits are from $1$ to $0$, so $\int_1^0 3x^2+2x^3dx=-1.5$
Adding everything gives $-1.5$
Edit: This becomes actually $-0.5$

Now using Green's theorem: Finding the partial derivatives and substituing these into the Green's formula gives: $\int_0^1\int_0^{x^2}(2x-3)dydx=-0.5$

What am I doing wrong because obviously $-0.5 \neq -1.5$? Edit: $-0.5 = -0.5$


  • $\begingroup$ Shouldn't the second integral be for $x=1$, not $x=0$? $\endgroup$ – naslundx Mar 4 '14 at 20:38
  • $\begingroup$ @naslundx Of course. My typo. Thank you! But it still gives $0$ so it doesn't change anything! $\endgroup$ – SomeOne Mar 4 '14 at 20:40

When doing the second part of the line integral, i.e. from $(1,0)$ to $(1,1)$, we have that $\int_{(1,0)}^{(1,1)} x^2dy = \int_0^1 1 dy$ which gives an additional $+1$.

  • $\begingroup$ Of course. Thank you! I didn't notice that mistake. I know that $x=1$ but when I substituted that in I thought that $x=0$ so everything becomes $0$. But now I understand! Thank you! $\endgroup$ – SomeOne Mar 4 '14 at 20:47
  • $\begingroup$ I'm glad that I could help! If you have any similar questions, just post it on this site. People are really friendly here. $\endgroup$ – Zoltan Zimboras Mar 4 '14 at 20:50
  • 1
    $\begingroup$ Did that :) Thank you so much! $\endgroup$ – SomeOne Mar 4 '14 at 20:52

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