Evaluating integral on surfaces

[Goklayeh]

Hello everyone! Could someone tell me how to evaluate the integral
$$\int_{B_{\delta}(p_0)}{\frac{1}{(1 + d^2(p_0, p))^2}\mathrm{d}v_g(p)}$$
where $B_{\delta}(p_0) \subset M$ and $(M, g)$ is a generic compact surface and $\delta > 0$ can be made small as one wishes (as usual, $d$ is the intrinsic metric)? Thanks you in advance

P.S. I strongly suspect that the value should be $\pi$, but I'm a newbie in integration-on-manifold without any idea about how to proceed.

 

[pat_connell]

The integral can be evaluated using the divergence theorem. Since the integrand is a function, the divergence theorem states that \int_{B_{\delta}(p_0)}{\frac{1}{(1 + d^2(p_0, p))^2}\mathrm{d}v_g(p)} = \int_{\partial B_{\delta}(p_0)}{\frac{1}{(1 + d^2(p_0, p))^2}\mathrm{d}s_g(p)}where \partial B_{\delta}(p_0) is the boundary of the ball and \mathrm{d}s_g(p) is the area element on the boundary. Now, since the boundary of the ball is a circle, the integral can be evaluated using the formula\int_{\partial B_{\delta}(p_0)}{\frac{1}{(1 + d^2(p_0, p))^2}\mathrm{d}s_g(p)} = \int_0^{2\pi}{\frac{1}{(1 + d^2(p_0, p))^2}\delta\,\mathrm{d}\theta}where \theta is the angle around the circle. The integral can then be evaluated explicitly:\int_0^{2\pi}{\frac{1}{(1 + d^2(p_0, p))^2}\delta\,\mathrm{d}\theta} = \pi\deltaTherefore, the value of the integral is \pi.