0
$\begingroup$

Calculate $$ I=\iint_{x^2+y^2+z^2=1}{e^{x-y} \mathbb{d}S} $$


Parameterization is not helpful: $$ I=\int_0^{2\pi}{\mathbb{d}\varphi\int_0^\pi{e^{\sin\theta(\cos\varphi-\sin\varphi)}\sin\theta\mathbb{d}\theta}} $$ ... nor is transformation to standard double-integral: $$ I=\int_{-1}^1{\mathbb{d}x\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}{\frac{e^{x-y}}{\sqrt{1-x^2-y^2}}\mathbb{d}y}} $$

$\endgroup$
3
  • $\begingroup$ small trivial point but there is a mixing of $\varphi$ and $\phi$ in your first transformed integral. $\endgroup$
    – asymptotic
    Nov 26, 2020 at 9:01
  • $\begingroup$ @Kevin I'm accustomed to using $\varphi$ here. What's the difference? $\endgroup$ Nov 26, 2020 at 11:21
  • $\begingroup$ I guess there is no difference, you can use either, as long as you choose only one :-) $\endgroup$
    – asymptotic
    Nov 26, 2020 at 11:53

1 Answer 1

3
$\begingroup$

It helps to carry out the $\phi$ integral first, $$I=\int_0^\pi \sin\theta\,d\theta\,\int_0^{2\pi} \,e^{\sin\theta(\cos\varphi-\sin\varphi)}\,d\phi$$ $$\qquad\qquad=2\pi\int_0^\pi I_0\left(\sqrt{2}\sin\theta\right)\,\sin\theta\,d\theta=2^{3/2} \pi \sinh \sqrt{2}.$$

$\endgroup$
6
  • $\begingroup$ Sorry, what does the $I_0$ notation mean? Your result is correct :) $\endgroup$ Nov 26, 2020 at 7:28
  • $\begingroup$ it's a Bessel function $\endgroup$ Nov 26, 2020 at 9:04
  • $\begingroup$ Just to compare: the command of Maple VectorCalculus:-SurfaceInt(exp(x - y), [x, y, z] = Sphere(<0, 0, 0>, 1)) reduces it to the integral $$\int_{0}^{2\,\pi}\!\pi\,{{I}_{1}\left(\cos \left( \theta \right) - \sin \left( \theta \right) \right)}+\pi\,{{\rm \bf L}_{1}\left(\cos \left( \theta \right) -\sin \left( \theta \right) \right)}+2\,{\rm d} \theta, $$ where ${\rm \bf L}_1$ is a Struve function and $I_1$ is a Bessel function. $\endgroup$
    – user64494
    Nov 26, 2020 at 9:40
  • $\begingroup$ @CarloBeenakker Thanks! I'm wondering if there's any elementary solution that sees no use of special functions. $\endgroup$ Nov 26, 2020 at 11:18
  • 2
    $\begingroup$ And you may replace from the beginning $y-x$ in the exponent with $\sqrt2 x$ via a rotation. $\endgroup$ Nov 26, 2020 at 12:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.