Question 6 of 7: Line Integral Along the Equatorial Circle of a Hemisphere

Question 6: Line Integral Along the Equatorial Circle of a Hemisphere (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $S$ is the boundary of the solid hemisphere $x^2+y^2+z^2\le9,\ z\ge0$ (the curved hemisphere cap plus the flat equatorial disk); $C$, the intersection of $S$ with $z=0$, is simply the equatorial circle $x^2+y^2=9$, $z=0$.

Find. $\displaystyle\oint_C\mathbf F\cdot d\mathbf r$.

Approach. Rather than invoking Stokes' theorem, parametrize $C$ directly: since $C$ lies entirely in the plane $z=0$, both $z\equiv0$ and $dz=0$ along it, and every component of $\mathbf F$ that survives in the line integral must not depend on carrying a $dz$ factor.

1. Parametrize $C$. $x=3\cos\theta,\ y=3\sin\theta,\ z=0$, $0\le\theta\le2\pi$, so $dz=0$ identically along $C$.
2. Evaluate $\mathbf F$ on $C$. With $z=0$: $F_1=y^2z=0$, $F_2=xz^2=0$, and only $F_3=x^2y$ is generally nonzero — but the line integral is $\oint(F_1\,dx+F_2\,dy+F_3\,dz)$, and $dz=0$ throughout, so the $F_3$ term never contributes regardless of its value.
3. Conclude. $$\oint_C\mathbf F\cdot d\mathbf r=\oint_C(0\,dx+0\,dy+x^2y\cdot0)=\boxed{0}$$