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Mathematical methods II course of the University of Oxford

Past exam of the mathematical methods II course of the University of Oxford

2017 resit

The relevant projection is the disk $x^2+y^2\le9$. The <surface area of a graph> is $\iint\sqrt{1+z_x^2+z_y^2}\,dx\,dy$. Here $z_x=-2x$ and $z_y=-2y$, so <polar coordinates> give
$$
A=2\pi\int_0^3r\sqrt{1+4r^2}\,dr
=\frac\pi6\left[(1+4r^2)^{3/2}\right]_0^3
=\boxed{\frac\pi6\bigl(37\sqrt{37}-1\bigr).}
$$
The boundary circle has zero surface area, so including the points at height zero does not change the result.

Solution

Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 4 / b / Solution

Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 4 / b / Solution

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Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 4 / b / Solution

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