# Compute $int xy dx +(x+y)dy$ over the curve $Γ$, $Γ$ is the arc $AB$ in the 1st quadrant of the unit circle $x^2+y^2=1$ from $A(1,0)$ to $B(0,1)$.

Mathematics Asked by Promona on August 22, 2020

Compute $$int xydx+(x+y)dy$$ over the curve $$Γ$$, where $$Γ$$ is the arc $$AB$$ in the first quadrant of the unit circle $$x^2+y^2=1$$ from $$A(1,0)$$ to $$B(0,1)$$.

I solved this problem with the help of Green’s theorem. My result is $$frac{pi}{4}-frac{1}{3}$$. But the book I’m following, shows, the result is $$frac{pi}{4}+frac{1}{6}$$. Since I’m a learner, I don’t understand where have I done the mistake.
Can anyone please give the solution?

Your book is correct. First of all, your application of Green's theorem here is incorrect, as indicated in the comment. You can't apply Green's theorem for a curve that's not simple closed. See https://en.wikipedia.org/wiki/Green%27s_theorem. So the correct answer is through the line integral evaluation as follows

$$int_Gamma xydx+(x+y)dy=int_{AB} xydx+(x+y)dy$$ $$=int_{y=0}^1bigg[ysqrt{1-y^2}dbigg(sqrt{1-y^2}bigg)+bigg(sqrt{1-y^2}+ybigg)dybigg]$$ $$=-int_0^1y^2dy+int_0^1sqrt{1-y^2}dy+int_0^1ydy=-frac{1}{3}+frac{pi}{4}+frac{1}{2}=frac{pi}{4}+frac{1}{6}$$

Correct answer by am_11235... on August 22, 2020