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How can I calculate the integral of $f(z) = e^{-z}$ over the surface of a sphere with radius $R$? I tried using cylindrical and spherical systems, both gave an unsolvable integral, suspecting there's a way to change the order of the variables.

Jonathan
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    Is there a physics context to this question? –  Mar 18 '17 at 14:40
  • I've provided an answer to your question but this is appropriate for the maths SE, so I hope it'll get migrated there. – JPhy Mar 18 '17 at 15:04

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Use $\sin\theta d\theta = -d\cos\theta$ to see that $$ I= R^2\int_0^{2\pi} d\phi \int_0^\pi e^{-R\cos\theta} \sin \theta \,d\theta= 2\pi R(e^R-e^{-R}) $$

mike stone
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The parametrisation of a sphere in $\mathbb R^3$ is given by,

$$x = R \sin\phi \cos\theta, \quad y = R \sin\phi \sin\theta, \quad z = R\cos\phi$$

though this is not unique. To perform the surface integral, we need to compute the magnitude of the normal to the sphere, namely

$$\bigg\lvert \frac{\partial \vec r}{\partial \theta} \times \frac{\partial \vec r}{\partial \phi}\bigg\rvert = R^2 \sin\phi$$

though in differential geometry we would think of integrating $\sqrt{\gamma}d^2x$ with the induced metric on the surface. Keeping in mind that $z=R\cos\phi$, one has to compute,

$$R^2\int_0^{\pi} d\phi \int_0^{2\pi}d\theta \, e^{-R\cos\phi}\sin\phi = 2\pi R^2 \int_0^\pi d\phi \, e^{-R\cos\phi}\sin\phi.$$

This is a fairly easy integral despite the appearance (notice the derivative and anti-derivative of a function appear in the integrand) we find as the final result,

$$4\pi R \sinh R$$

as the value of the surface integral of the scalar field over the sphere.

JPhy
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