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

Use $s$ for cylindrical radius, to distinguish it from spherical radius $r$. In cylindrical coordinates $(s,\varphi ,z)$, the azimuthal angle $\varphi$ is measured from positive $x$ in the horizontal plane:Here $0\le \varphi <2\pi$ and $z\in \mathbb{R}$.

In spherical polar coordinates $(r,\theta ,\varphi )$, take $\theta$ as the polar angle from positive $z$ and use the same azimuthal angle $\varphi$:The ranges are $r\ge 0$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The radial direction has horizontal projection of length $r\sin \theta$, which explains the extra factor in $x,y$. The azimuth is undefined on the polar axis, and both angles are nonunique at the origin; these are coordinate degeneracies rather than geometric singularities.

On the unit upper hemisphere, use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. The surface mass density is $\sigma =z=\cos \theta$, while the surface area element is $dS=\sin \theta d\theta d\varphi$. Thus the total mass isThe answer is in the mass units implicit in the specified unit radius and density law. A useful geometric check is that $zdS$ on the unit hemisphere equals the area element of its vertical projection onto the horizontal unit disc, whose area is $\pi$.

This is a scalar surface integral, not an oriented flux. Split the whole closed surface into the bottom disc, curved side and tilted top; shared edge curves have zero surface area.

On the bottom, $z=0$ and the integrand is $y$. Its integral over the symmetric disc $x^2+y^2\le 3$ vanishes by oddness:On the top, $z=4-y$, so $y+z=4$. The graph has $z_x=0$, $z_y=-1$, giving $dS=\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$. HenceFor the side, let $R=\sqrt{3}$ and parametrize by $(R\cos \varphi ,R\sin \varphi ,z)$, with $0\le z\le 4-R\sin \varphi$. The surface area element is $Rdzd\varphi$. ThusAdding all three pieces givesThe tilted top remains above the bottom everywhere because $4-y\ge 4-\sqrt{3}>0$. The calculation is an instance of a scalar surface integral over a cylinder with one tilted cap; reversing a surface orientation would not change any of its terms.