Green theorem verifying problem

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?

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

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 $$\ 1$$