Integration of the vector field $\mathbf {F } (x,y)=\frac{y}{x^2+y^2}i-\frac{x}{x^2+y^2}j $ over two ellipses Let $\mathbf{F}$ be a vector field defined on $\mathbb R^2 \setminus\{(0,0)\}$ by 
$$\mathbf {F } (x,y)=\frac{y}{x^2+y^2}i-\frac{x}{x^2+y^2}j $$ Let $\gamma,\alpha:[0,1]\to\mathbb R^2$ be defined by 
$$\gamma (t)=(8\cos 2\pi t,17\sin 2\pi t)$$ and 
$$\alpha (t)=(26\cos 2\pi t,-10\sin 2\pi t)$$ 
If $$3\int_{\alpha} \mathbf{F\cdot dr}  -4  \int_{\gamma} \mathbf{F\cdot dr}= 2m\pi,$$ then what is $m$?
How should I approach this question?
Progress
I see that the parametrization of ellipses are given already. For evaluating say first integral,  I need to substitute given parametrization of ellipse in vector field. The parameter $t$ will vary from $ 0$ to $2\pi$. Am I correct?
 A: You are not expected to actually compute these line integrals using the parameterizations; this would be a rather painful procedure. The question appears to be testing your knowledge of the winding number. The  curve $\gamma$ has winding number $1$ about the origin, since it travels once counterclockwise. The curve $\alpha$ has winding number $-1$, being clockwise. 
Since $\mathbf F$ is irrotational in the punctured plane, the integral of this field over a closed loop depends only on the winding number about the origin. (Alternatively, you can the relation with arctangent pointed out by BaronVT to reach the same conclusion.) 
Since the integral of $\mathbf F$ over the unit circle is $-2\pi$ (easy direct calculation), it follows  that the integral over a closed  curve with winding number $w$ is $-2\pi w$. Hence,
$$
\displaystyle 3\int_{\alpha} \mathbf{F\cdot dr} -4\displaystyle \int_{\gamma} \mathbf{F\cdot dr} = 3(-2\pi)(-1) - 4(-2\pi)(1)
$$
A: Hint 1: This vector field is essentially $\nabla \theta$ (i.e. $\nabla \arctan(y/x)$). It has trouble at the origin, so the line integrals aren't zero, but you can at least use the fact that it's "almost" conservative to set up easier contours for integration (you are given ellipses, circles would be much easier though).
Hint 2: You can show that the integral around an ellipse is the same as the integral around a circle (up to the choice of clockwise vs counterclockwise). The integral around a circle is easy to compute (you are integrating the change in angle around a circle).
Hint 3: If you have a line integral of a conservative vector field enclosing a simply connected region, the integral is zero. The region I've shaded in this picture: is simply connected, and so the two "main" integrals must be equal.
