Question

Find the integral field of the vector field given by X(p)=(p,2(y-b),-2(x-a) through the point (a+r,b) and f(x,y)=(x-a)^2+(y-b)^2 of a circle C=f^-1(r^2)

Answer 1

1. We point out that the vector field can be represented as $\nabla f=(\frac{\partial f}{\partial p},\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ . Indeed, if we integrate the first component with respect to p, we receive: $f=\frac{1}{2}p^2+\alpha(x,y)$ . It implies $\alpha_x=2(y-b)$, $\alpha_y=-2(x-a)$ . It can be easily checked that the latter system has no common solution. In particular, it follows from the fact that $\alpha_{xy}=2$ from the first equation and $\alpha_{yx}=-2$ from the second one.

Threfore, we correct the statement: $X(p)=(-2(x(p)-a),2(y(p)-b)),$ where $x(p)=p$ , $y(p)=p$ . Now we solve the equation: $-2(p-a)=a+r$ . The latter implies: $p=\frac{a-r}{2}$ . The second equation then yields: $2(\frac{a-r}2-b)=b$ . The latter yields $a-r=3b$ . Now we will find an integral $\int_{x_0}^{x_1} ds$ , where s is given by the curve $X(p)$ and $x_0=X(0),x_1=X(\frac{a-r}{2})$ . $\int_{x_0}^{x_1}ds=\int_{0}^{\frac{a-r}{2}}|X'(p)|dp=\int_{0}^{\frac{a-r}{2}}\sqrt{4(1-a)^2+4(1-b)^2}dp=\sqrt{4(1-a)^2+4(1-b)^2}\frac{a-r}{2}$

2.As it is stated, the aim is to find $f(C),$ where C is a circle with a radius r and center (a,b). We receive an obvious answer: $f(C)=r^2$