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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 4

The area element is
$$
dA=f\sqrt{1+f'^2}\,dx\,d\theta.
$$
Consequently
$$
K\,dA
=-\frac{f''}{(1+f'^2)^{3/2}}\,dx\,d\theta
=-d\!\left(\frac{f'}{\sqrt{1+f'^2}}\right)d\theta.
$$
At $x=2-a$, smooth matching to the cylinder gives $f'=0$. At $x=2+a$, the spherical formula gives
$$
f'=\frac{1-a}{\sqrt{2a-a^2}},
\qquad
\frac{f'}{\sqrt{1+f'^2}}=1-a.
$$
The <total Gaussian curvature of a surface-of-revolution strip> is therefore
$$
\boxed{\int_RK\,dA=-2\pi(1-a)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 11G / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 11G / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 11G / b / Solution

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