Surface integral on a sphere

I find it convenient to work on the sphere $S$ given by $x^2+y^2+z^2=1$ . At some point I will perform a change of variables, $x+y+z=\sqrt{3}\tilde{z}$ (we will not need $\tilde{x}$ and $\tilde{y}$ ).

$\frac{1}{9}\int_S xye^{(x+y+z)/\sqrt{3}}dS=\frac{1}{27}\int_S (xy+xz+yz)e^{(x+y+z)/\sqrt{3}}dS=\frac{1}{54}\int_S ((x+y+z)^2-1)e^{(x+y+z)/\sqrt{3}}dS=\frac{1}{54}\int_S (3\tilde{z}^2-1)e^{\tilde{z}}dS$ $=\frac{2\pi}{54}\int_{-1}^{1}(3\tilde{z}^2-1)e^{\tilde{z}}d\tilde{z}=\left(\frac{2e}{27}-\frac{14}{27e}\right)\pi$

The answer comes out to be $\boxed{20}$ .

A wonderful celebration of symmetry on the sphere! Thank you! I will add this to my repertoire of challenging problems for my calculus classes.

We note that $J \; = \;\iint_{x^2+y^2+z^2=R^2}xye^{x+y+z}\,dS \; = \; R\iint_{x^2+y^2+z^2=R^2}\mathrm{F} \cdot d\mathrm{S} \; = \; R\iiint_{x^2+y^2+z^2 \le R^2}\nabla\cdot\mathrm{F}\,dV \; = \; R\iiint_{x^2+y^2+z^2 \le R^2} ye^{x+y+z}\,dV$ where $\mathrm{F} \; = \; \left(\begin{array}{c} ye^{x+y+z} \\ 0 \\ 0 \end{array}\right)$ Rotating the coordinate system with the transformation $X = \tfrac{1}{\sqrt{3}}(x+y+z)$ , $Y = \tfrac{1}{\sqrt{2}}(x-y)$ , $Z = \tfrac{1}{\sqrt{6}}(x+y-2z)$ , we see that $J \; = \; R\iiint_{X^2+Y^2+Z^2\le R^2}\big(\tfrac{1}{\sqrt{3}}X - \tfrac{1}{\sqrt{2}}Y + \tfrac{1}{\sqrt{6}}Z\big)e^{\sqrt{3}X}\,dXdYdZ \; = \; \tfrac{1}{\sqrt{3}}R\iiint_{X^2+Y^2+Z^2 \le R^2}Xe^{\sqrt{3}X}\,dXdYdZ$ by symmetry, and hence $J \; = \; \tfrac{1}{\sqrt{3}}\pi R\int_{-R}^R X(R^2-X^2)e^{\sqrt{3}X}\,dX$ With $R = \tfrac{1}{\sqrt{3}}$ this becomes $J \; = \; \tfrac13\pi\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} X\big(\tfrac13-X^2\big)e^{\sqrt{3}X}\,dX \; = \; \tfrac{1}{27}\pi \int_{-1}^1 y(1-y^2)e^y\,dy \; = \; \tfrac{1}{27}\pi\big(2e - 14e^{-1}\big)$ which makes $a = \tfrac{2}{27}$ and $b - -\tfrac{14}{27}$ , and hence $18a - 36b = \boxed{20}$ .