= Solution
A scalar <line integral> is $\int_Cf\,ds=\int f(r(t))\|r'(t)\|\,dt$: it weights arc length by a <scalar field> and is unchanged by reversing orientation. A vector-work <line integral> is $\int_CF\cdot d r=\int F(r(t))\cdot r'(t)\,dt$: it measures the tangent component of a <vector field> and changes sign under reversal. Both are invariant under a regular orientation-preserving reparametrization.
For the prescribed counterclockwise curve, traverse the upper semicircle from $(1,0)$ to $(-1,0)$, then the lower parabola from $(-1,0)$ back to $(1,0)$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit-semicircle-parabola-contour.png]
{title=Counterclockwise contour formed by the upper unit semicircle and the lower parabolic arc.}
{description=The upper arc runs from right to left; the lower arc runs from left to right. The enclosed area is the sum of a semicircle and a parabolic segment.}
On the upper arc, use $x=\cos t$, $y=\sin t$, $0\leq t\leq\pi$. Then
$$
F\cdot d r=\big[-3\sin^2t+\cos^3t-\sin t\cos t\big]dt,
$$
so its integral is $-3\pi/2$: the last two terms integrate to zero by symmetry or direct integration. On the parabolic arc, use $x=t$, $y=t^2-1$ for $-1\leq t\leq1$. Here $x^2-y=1$ and
$$
F\cdot d r=(3t^2-3+2t)dt,
\qquad\int_{-1}^1(3t^2-3+2t)\,dt=-4.
$$
Therefore
$$
\boxed{\oint_CF\cdot d r=-\frac{3\pi}{2}-4.}
$$
As an independent check, <Green's theorem> gives the integrand $\partial_x(x^2-y)-\partial_y(3y)=2x-3$. The region is symmetric in $x$, so its odd part vanishes. Its area is $\pi/2+4/3$, giving $-3(\pi/2+4/3)$, exactly the same value.
Back to article page
