compute $\int_\gamma \sin(y-x)\ dx +2xy + \sin(x-y) \ dy$

Question

The question is $$ \int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy, \quad \gamma:y=\sqrt{x}, \ \ 0\leq x\leq 1$$ I've tried to solve it like this: $$\int_{0}^{1}\sin(\sqrt{t}-t) +t+ \sin(t-\sqrt{t}) \ \frac{1}{2\sqrt{t}}\ dt \quad\text{with} \ \ \gamma:r(t)=(t,\sqrt{t}) $$

How should i proceed from here ? Any suggestion would be great, thanks

Answer 1

$\int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy$

$= \int_\gamma \sin(y-x)\ dx +\left(\sin(x-y)\right) \ dy \ + \ \int_\gamma 2xy \ dy$

But please note that $\big(\sin(y-x), \sin(x-y)\big) = \nabla f(x,y) \ $ where $\ f(x,y) = \cos(y-x)$ and $f(1,1) - f(0,0) = 1 -1 = 0$.

So the line integral for part of the vector field is zero and all you are left with is to find -

$\displaystyle \int_{\gamma} 2xy \ dy = \int_0^1 t \ dt = \frac{1}{2}$

Answer 2

hint

$$\int \sin(t-\sqrt{t})(\frac{1}{2\sqrt{t}}-1)dt=$$ $$\cos(t-\sqrt{t})$$