# Line integral Along curve C

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!

$\nabla g = (e^{xy}+\sin(x+y))+(e^{xy}+\sin(x+y))$

• Your vector field is conservative. YAY!
– Pedro
Commented Nov 14, 2013 at 4:22
• No :( the only thing I'm getting a hint from in the book is the "fundamental theorem of calculus for line integrals"
– Matt
Commented Nov 14, 2013 at 4:22
• Your contour is not closed. Commented Nov 14, 2013 at 4:29
• Your potential is OK, but there is no need to write it twice. =)
– Pedro
Commented Nov 14, 2013 at 4:29

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 $\gamma$, $$\int_\gamma G=f(B)-f(A)$$

• ill post it in my original
– Matt
Commented Nov 14, 2013 at 4:24
• But it doesn't end where it started? does that matter? The book says the answer is non-zero.
– Matt
Commented Nov 14, 2013 at 4:28
• @Matt Sorry Matt, didn't read that part carefully. You still can apply FTC.
– Pedro
Commented Nov 14, 2013 at 4:30
• Thats what i was trying to do. But i cant get the right answer. I mean it seems straight forward but I cant get the right answer!
– Matt
Commented Nov 14, 2013 at 4:30