Surface integral on a sphere

Question

I want to evaluate $$\int_S e^{y_{2}+y_{1}}sin(y_{1}) ~dS,$$where $S$ is the sphere $y_{1}^2+y_{2}^2+y_{3}^2=R^2$. I've tried spherical coordinates I but don't see how from there. Could you give me a hint how to start?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffd,10px]{\ds{\iint_{r\ =\ R\ >\ 0} \expo{y + x}\sin\pars{x}\,\mathrm{dS}}} = \Im\iint_{r\ =\ R\ >\ 0}\expo{y + x}\expo{\ic x}\,\mathrm{dS} = \Im\iint_{r\ =\ R\ >\ 0}\exp\pars{\bracks{1 + \ic}x + y}\,\mathrm{dS} \\[5mm] = &\ 4\pi\,\Im\iint_{r\ =\ R\ >\ 0}\exp\pars{\braces{\bracks{1 + \ic}\hat{x} + \hat{y}}\cdot\vec{r}}\,{\mathrm{dS} \over 4\pi} = 4\pi\,\Im\pars{\sinh\pars{\root{1 + 2\ic}R} \over \root{1 + 2\ic}R} \\[5mm] = &\ 4\pi\,\Im\pars{\sinh\pars{5^{1/4}\expo{\ic\varphi/2}R} \over 5^{1/4}\expo{\ic\varphi/2}R}\,,\qquad\varphi \equiv \arctan\pars{2}. \end{align}

Answer 2

I would start with solving the simpler problem of evaluating the integral $$I(\mathbf{a})\stackrel{\text{def}}{=}\int_{S^2}\mathrm{e}^{\mathbf{a\cdot \mathbf{x}}}=\int_{S^2} \mathrm{e}^{a\cos \theta}\mathrm{d}\phi\,\mathrm{d}(\cos\theta)$$ over the unit sphere $S^2$: $x_1^2+x_2^2+x_3^2=1$.

Your integral is expressible as a linear combination of special values of $I(\mathbf{a})$.