Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 3 / b

Parametrize the function graph by $\mathbf{R}(x,y)=(x,y,f(x,y))$. Its two tangent vectors are ${\mathbf{R}}_x=(1,0,f_x)$ and ${\mathbf{R}}_y=(0,1,f_y)$. A small parameter rectangle maps, to first order, to a tangent parallelogram with area $|{\mathbf{R}}_x\times {\mathbf{R}}_y|dxdy$. Sincethe surface area of a graph isFor example, $f\in C^1$ ensures a regular surface chart; more general domains can be handled by the corresponding convergent improper integral.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

The rotation axis is $L=\{(x,1,0):x\in \mathbb{R}\}$. The distance of the generating function graph from this axis is $\sqrt{x}$, so a parametrization isThe surface of revolution therefore has equation $(y-1{)}^2+z^2=x$: it is a circular elliptic paraboloid opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the surface area formula, soThe cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.