Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 12

Put $R=|b|$ so that the radial range respects the stated allowance $b\ne 0$. With the azimuth measured from the starting meridian, the cylindrical coordinates of the curved surface satisfyAt $r=0$ all azimuths describe the same apex. The surface is a sector of an elliptic paraboloid.

The volume element in cylindrical coordinates is $rdzdrd\theta$. Integrating from the base to the curved surface givesHenceFor a complete rotation this is half the volume of the enclosing cylinder.

Use the embedded surface parametrizationIts tangent vectors have cross productwhose magnitude is $r\sqrt{1+4a^2{r}^2/b^4}$. Thus the surface area isWith $u=1+4a^2{r}^2/b^4$, the radial integral becomes $b^4(8a^2{)}^{-1}\int u^{1/2}du$. Evaluating its endpoints givesThis is the scalar-area part of the paraboloid sector surface measures.

For $a/|b|\ll 1$, apply the binomial series to the result in (c):ThereforeFor $\alpha =2\pi$, the leading term is $\pi b^2$, the area of the base disk. A very shallow cap is nearly planar; its first area excess is $\pi a^2$.

The upward orientation of ${\mathbf{X}}_r\times {\mathbf{X}}_{\theta }$ corresponds to traversing the boundary from the apex to the starting rim, around the rim in increasing azimuth, and back to the apex. This is the indicated $ABCA$ orientation and is outward from the solid. Integrate that vector area element directly:The radial integrals are $2aR/3$ in the horizontal components and $b^2/2$ in the vertical component. Hence the vector area isThe absolute value is needed only because the original statement permits negative $b$. Reversing the boundary orientation reverses the entire vector. For a full rotation the horizontal components cancel, leaving $\pi b^2\mathbf{k}$, the signed projected area of the base.