Surface Area of Football

We parametrize the surface with $\phi = -\theta/2$ to $\phi = +\theta/2$ from bottom to top. Each infinitesimal step $d\phi$ contributes a circular ring of radius $r(\phi) = R\cos\phi - R\cos \frac\theta 2.$ The height of the ring (measured along the surface) is $dh = R\:d\phi$ . Thus $dA = 2\pi r(\theta)dh = 2\pi R^2\left(\cos \phi - \cos \frac\theta 2\right)\:d\phi.$ We calculate $A = 2\pi R^2 \int_{-\theta/2}^{\theta/2} \left(\cos\phi - \cos \frac\theta 2\right)\:d\phi = 2\pi R^2 \left[\sin\phi - \phi\cos\frac\theta 2\right]_{-\theta/2}^{\theta/2}$ $= 4\pi R^2\left(\sin\frac \theta 2 - \frac \theta 2 \cos\frac\theta 2\right).$ With the given values, $A = 4\pi R^2\left(\frac12\sqrt 3 - \frac \pi 3\cdot \frac 1 2\right) = 1257\cdot 0.3424 = \boxed{430}.$