Conservative Vector Fields — Question 5

Question 5 – Solution

Strategy. The right half-plane permits a single-valued angle function with no branch crossing.

Step 1: Choose the angle potential On $x>0$, define $$f(x,y)=\arctan\!\left(\frac yx\right).$$ Using the chain rule, $$f_x=\frac{-y}{x^2+y^2}, \qquad f_y=\frac{x}{x^2+y^2}.$$ Hence $\nabla f=\mathbf F$ on $H$.

Step 2: Evaluate by endpoints The endpoint angles in the branch $(-\pi/2,\pi/2)$ are $$f(A)=\arctan(1)=\frac\pi 4, \qquad f(B)=\arctan\!\left(\frac 1{\sqrt 3}\right)=\frac\pi 6.$$ Therefore $$\boxed{\int_C\mathbf F\cdot d\mathbf r=f(B)-f(A)=-\frac{\pi}{12}}.$$

Verification Both endpoints and the entire allowed path lie in $x>0$, where the selected arctangent branch is continuous. The negative sign matches the decrease in polar angle from $\pi/4$ to $\pi/6$.