I am trying to solve a Green's theorem verify problem. Here is the problem:

$Verify\;Greeen's\;theorem\;in\;the\;plane\;for \int_c\{(xy\;+y^2)dx\;+x^2dy\}, \;where\;C\;is\;the\;closed\;curve\;of\;the\;region\;bounded\;by\;y=x\;and\;y=x^2.$

Here is my solution:

First of all, the point of intersection of $y=x^2\;$and $y=x$ is $(1,1)$, thus, from that perspective,

we know green theorem goes this way,

$\int_C[X\;dx\;+\;Y\;dy]\;=\;\iint[\frac{\partial Y}{\partial x}$-$\frac{\partial X}{\partial y}]\;dx\;dy$

$ LHS:\\[10pt]\int_c[X\;dx+Ydy]\\[10pt]=\int_cXdx+\int_cYdy\\[10pt]=\int_0^1Xdx+\int_0^1Ydy\\[10pt]=\int_0^1(xy+y^2)dx+\int_0^1x^2dy\\[10pt]Solving \;this\;further,\;I\;am\;getting,\\[10pt]=\frac{3y^2+2x^2}{2},\quad putting \;x=1\;and\;y=1, we\;get,\\[10pt]=\frac{5}{2} $

Now, solving, RHS, we get,

$ RHS: \\[10pt]\iint_R[\frac{\partial Y}{\partial x}-\frac{\partial X}{\partial y}]dxdy\\[10pt]=\iint_R\frac{\partial Y}{\partial x}dxdy-\iint_R\frac{\partial X}{\partial y}dxdy\\[10pt]=I_1+I_2\\[10pt]I_1:\\[10pt]\iint_R\frac{\partial Y}{\partial x}dxdy\\[10pt]=\int_0^1\int_0^1\frac{\partial Y}{\partial x}dxdy\\[10pt]=\int_0^1\int_0^1\frac{\partial x^2}{\partial x}dxdy\\[10pt]=\int_0^1\int_0^12x\;dxdy\\[10pt]=2\int_0^1\int_0^1x\;dxdy\\[10pt]=2\int_0^1\vert\frac{x^2}{2}\vert_0^1\;dy\\[10pt]=\int_0^1dy\\[10pt]=1 $

$ I_2:\\[10pt]\iint_R\frac{\partial X}{\partial y}dxdy\\[10pt]=\int_0^1\int_0^1\frac{\partial X}{\partial y}dxdy\\[10pt]=\int_0^1\int_0^1\frac{\partial(xy+y^2)}{\partial y}dxdy\\[10pt]=\int_0^1\int_0^1(x+2y)dxdy\\[10pt]=\int_0^1\int_0^1x\;dxdy+\int_0^1\int_0^12y\;dxdy\\[10pt]=\int_0^1\vert\frac{x^2}{2}\vert_0^1\;dy+2\int_0^1\vert\frac{y^2}{2}\vert_0^1\;dx\\[10pt]=\int_0^1\frac{1}{2}dy+\int_0^1dx\\[10pt]=\frac{1}{2}\int_0^1dy+\int_0^1dx\\[10pt]=\frac{1}{2}+1\;=\;\frac{3}{2} $

Now adding, $I_1+I_2,$ we get $=1-\frac{3}{2}=-\frac{1}{2}$

Thus, as you can see, LHS $\neq$ RHS, if $I_2$ would have be $+\frac{3}{2},$ then in that case, $LHS$ would have been equal to $RHS$. But, right not it is not, so can anyone tell me where it is wrong?

  • $\begingroup$ The points of intersection are $(0,0)$ and $(1,1)$, not $(x,y)$... $\endgroup$ – Ak. Apr 15 '20 at 10:07
  • $\begingroup$ Oh, yeah, silly mistake. $\endgroup$ – peaceHoper Apr 15 '20 at 10:14
  • $\begingroup$ but, i changed now, i have changed the whole process, yet it is coming wrong! $\endgroup$ – peaceHoper Apr 15 '20 at 11:00

This is your page 1

This is your page 2 I have done it on my page. Attached the images.

  • $\begingroup$ helpful! I made some mistakes..but now i understood. $\endgroup$ – peaceHoper Apr 15 '20 at 15:16

Make sure that you are taking the limits correctly. Also in case of line integral, the path of $\ C$ is traversed in a counter-clock-wise sense, i.e in the direction of travel around $\ C$ in which the interior of $\ S$ lies on the left.

Now in case of surface integral, if you consider a small horizontal strip and move it vertically in the region to cover the whole region, then the strip moves from $\ y=x^2$ to $\ y=x$. So the limits of $\ y$ will be $\ x^2$ to $\ x$.

Now if you consider a small vertical strip and move it horizontally to cover the whole region, it will move from $\ x=0$ to $\ x=1$. So the limits of $\ x$ will be $\ 0$ to $\ 1$

  • $\begingroup$ @user8668205 if my answer was helpful , please accept my answer. I need it. $\endgroup$ – Manjoy Das Apr 18 '20 at 9:24

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