Line Integrals of Vector Fields — Question 10

Question 10 – Solution

Strategy. Write the field as $\langle x,y\rangle+2\langle-y,x\rangle$; on a circle the first part is normal and the second is tangent.

Step 1: Decompose $$\mathbf F=\underbrace{\langle x,y\rangle}_{\text{radial}} +\underbrace{2\langle-y,x\rangle}_{\text{counterclockwise tangential}}.$$ The radial part is perpendicular to the circle’s tangent, so it contributes zero work.

On the radius-$3$ circle, $\langle-y,x\rangle$ has magnitude $3$ in the positive tangent direction. Hence the doubled part has tangential magnitude $6$. Multiplying by the circumference $6\pi$ gives $$\oint_C\mathbf F\cdot d\mathbf r=0+6(6\pi)=\boxed{36\pi}.$$

Step 2: Direct verification Set $\mathbf r(t)=\langle 3\cos t,3\sin t\rangle$, $0\le t\le 2\pi$. Then $$\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=18,$$ because the radial terms cancel and the tangential terms sum to $18(\sin^2t+\cos^2t)$. Therefore $$\int_0^{2\pi}18\,dt=36\pi,$$ confirming both the magnitude and the positive orientation.