Put $h(r,\theta)=\sqrt{G(r,\theta)}$. The <Jacobi equation in geodesic polar coordinates> and the initial conditions from part (b) are
$$
h_{rr}=-Kh,
\qquad h(0,\theta)=0,
\qquad h_r(0,\theta)=1.
$$
Inside a geodesic polar coordinate ball, $h>0$. If $K\leq0$, then $h_{rr}\geq0$, so $h_r\geq1$ and therefore $h(r,\theta)\geq r$. The <area element of a surface>[Riemannian area element] is $h\,dr\,d\theta$, whence
$$
\operatorname{Area}B(p,\varepsilon_0)
=\int_0^{2\pi}\int_0^{\varepsilon_0}h(r,\theta)\,dr\,d\theta
\geq\int_0^{2\pi}\int_0^{\varepsilon_0}r\,dr\,d\theta
=\pi\varepsilon_0^2.
$$
\b[Thus nonpositive <Gaussian curvature> makes such a ball at least as large as the <Euclidean disk> of the same radius.]

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 25I / c / Solution

{title2=$h_{rr}+Kh=0$}

For $h(r,\theta)=\sqrt{G(r,\theta)}$, the angular <Jacobi field> equation on a surface becomes
$$
h_{rr}+K(r,\theta)h=0,
\qquad h(0,\theta)=0,
\qquad h_r(0,\theta)=1.
$$
The <area element of a surface>[Riemannian area element] is $h(r,\theta)\,dr\,d\theta$.

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 Incoming links: Jacobi equation in geodesic polar coordinates