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




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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:
$$
I=\int\_{-1}^1{\mathbb{d}x\int\_{-\sqrt{1-x^2}}^\sqrt{1-x^2}{\frac{e^{x-y}}{\sqrt{1-x^2-y^2}}\mathbb{d}y}}
$$



A "simple" surface-integral over the unit-sphere
It helps to carry out the $\phi$ integral first,
$$I=\int\_0^\pi \sin\theta\,d\theta\,\int\_0^{2\pi} \,e^{\sin\theta(\cos\varphi-\sin\varphi)}\,d\phi$$
$$\qquad\qquad=2\pi\int\_0^\pi I\_0\left(\sqrt{2}\sin\theta\right)\,\sin\theta\,d\theta=2^{3/2} \pi \sinh \sqrt{2}.$$


