Let $\vec{F}(x,y) = \left( 2xy+2\sin\!\left(x\right) \right) \hat{\imath} + \left( x^{2}+7\cos\!\left(y\right) \right) \hat{\jmath}$ be a vector field.
Denote $\displaystyle C = \Big\lbrace (x,y) | \; x^{2} + y^{2} = 1, \; x \geq 0, \; y \geq 0 \Big\rbrace$
Find the vector field's work (counterclockwise) and find the potential function.
My attempt:
Since $\vec{F}(x,y)$ is a conservative vector field ,the vector field's work in closed path is $0$.
The red path is the required path and the blue one is the closed path. Therefore the vector field's work is $\displaystyle \int_{C} \vec{F} \cdot d\vec{r} = {} 0-\int_\frac{\pi}{2}^0 7cos(y)\ dy + \int_0^\frac{\pi}{2} 2sin(x) dx =5 $
The potential function $U(x,y)$:
$U(x,y)=\int \left( 2xy+2\sin\!\left(x\right) \right)dx=x^2y-2cos(x)+C(y)$.
$\frac{dU}{dy}=\left( x^{2}+7\cos\!\left(y\right) \right) \implies 2xy+C'(y)=x^2+7cos(y)\implies U(x,y)=x^2y-xy^2+7sin(y)$
Both answers are wrong ,can't find out where am I wrong, appreciate any help.
$C$ appears to be just the quarter circle (although they need to specify the orientation). You've messed up the potential function at the end. When you get to the $\partial U/\partial y$ step, you should have $x^2+C'(y) = x^2 + 7\cos y$, so $C(y) = 7\sin y + c$. Thus, putting everything together, $$U(x,y) = x^2y -2\cos x + 7\sin y + c.$$ We can of course take $c=0$. Also, avoid using the letter $C$ in the potential function work when $C$ is already used elsewhere in the problem.
Assuming $C$ is oriented counterclockwise, the work done is $$U(0,1)-U(1,0) = (-2 + 7\sin 1) - (-2\cos 1) = -2 + 7\sin 1 + 2\cos 1.$$ When you compute by parametrization, you take $x(t)=\cos t$, $y(t)=\sin t$, $0\le t\le \pi/2$, and \begin{align*} \int_C \vec F\cdot d\vec r &= \int_0^{\pi/2} (2\cos t\sin t + 2\sin(\cos t))(-\sin t) + (\cos^2 t + 7\cos(\sin t))(\cos t)\,dt \\ &= \int_0^{\pi/2} (-2\cos t\sin^2 t+\cos^3 t)dt + \int_1^0 2\sin u\,du + \int_0^1 7\cos u\,du \\ &= 0 + \int_0^1 (7\cos u - 2\sin u)\,du = 7\sin 1 + 2\cos 1 - 2. \end{align*}