Применяя формулу:

$$\underset{C}{\int }f\left(x,y\right)dS=\underset{\alpha }{\overset{\beta }{\int }}f\left(x\left(t\right),y\left(t\right)\right)\sqrt{{\left(x^{\prime }\left(t\right)\right)}^2+{\left(y^{\prime }\left(t\right)\right)}^2}dt;$$

найдем $dS$.

$$dS=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=\sqrt{a^2{\mathrm{sh}}^{2}t+a^2{\mathrm{ch}}^{2}t}=a\sqrt{\mathrm{ch}2t}dt$$

$$\begin{gathered} \underset{C}{\int }xydS= \\ =a^3\underset{0}{\overset{t_0}{\int }}\mathrm{ch}t\mathrm{sh}t\sqrt{\mathrm{ch}2t}dt=\frac{a^3}{2}\underset{0}{\overset{t_0}{\int }}\mathrm{sh}2t\sqrt{\mathrm{ch}2t}dt= \\ =\frac{a^3}{4}\underset{0}{\overset{t_0}{\int }}\sqrt{\mathrm{ch}2t}d(\mathrm{ch}2t)=\frac{a^3}{6}\left({\sqrt{{\mathrm{ch}}^{3}2t}=1|}_0^{t_9}\right)= \\ =\frac{a^3}{6}\left(\sqrt{{\mathrm{ch}}^{3}2t_0}-1\right) \end{gathered}$$