Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 10 / c / Solution

A scalar line integral is ${\int }_{C}fds=\int f(r(t))\Vert r^{\prime }(t)\Vert dt$: it weights arc length by a scalar field and is unchanged by reversing orientation. A vector-work line integral is ${\int }_{C}F\cdot dr=\int F(r(t))\cdot r^{\prime }(t)dt$: 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 $(1,0)$.

On the upper arc, use $x=\cos t$, $y=\sin t$, $0\le t\le \pi$. Thenso 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 $-1\le t\le 1$. Here $x^2-y=1$ andThereforeAs 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.