For the Vector field, find the line integral along the curve $C$ from the origin to along the $x$-axis to the point $(3,0)$ and then counterclockwise around the circumference of the circle $x^2 + y^2 = 9$ to the point ($\frac 3{\sqrt{2}},\frac 3{\sqrt{2}}$).
$\vec G = (ye^{xy}+\cos(x+y))\vec i+(xe^{xy}+\cos(x+y))\vec j$
I need some help solving this. If It's not too much trouble please show your steps!
$\nabla g = (e^{xy}+\sin(x+y))+(e^{xy}+\sin(x+y))$
Best Answer
Can you find $f$ such that $\nabla f=G$? You can also use that $D_1G_2=D_2G_1$, which we know is equivalent to $G$ being conservative.
ADD You're right: if $f(x,y)=e^{xy}+\sin(x+y)$ then $D_1f=ye^{xy}+\cos(x+y)$ and $D_2f=xe^{xy}+\cos(x+y)$. This means $\nabla f=G$. Now you can apply FTC. That is, if $A$ is the initial point and $B$ is the final point of your curve $\gamma$, $$\int_\gamma G=f(B)-f(A)$$