# Area functional in two different coordinates systems

I'm studying a problem connected to finding a minimal surface with a given curve as a boundary in the usual hyperbolic space $$\mathbb{H}^5$$ with metric

$$ds^2=\frac{1}{y^2}(dy^2+dx_1^2+dx_2^2+dx_3^2+dx_4^2)$$

The boundary curve is $$y=0$$, $$x_1^2+x_2^2=R^2$$, $$x_3=$$ constant, $$x_4=$$ constant, so it is a circle in the $$x_1-x_2$$ plane of the conformal boundary of $$\mathbb{H}^5$$. The area functional

$$S= \int\limits_{surface} \sqrt{\det(g_{ind})}$$

can be simplified in this case. First, I use polar coordinates in the $$x_1-x_2$$ plane, so $$dx_1^2+dx_2^2=dr^2+r^2d\varphi^2$$, and take $$\varphi$$ and $$r$$ as parameters on the surface. From the symmetry of the problem I can argue that $$y(r,\varphi)=y(r)$$, and the other two coordinates are constants. Therefore, the induced metric is $$g_{ind} : \frac{1}{y^2}((1+y'^2)dr^2+r^2d\varphi^2)$$ and $$S= \int\int \sqrt{\frac{r^2(1+y'^2)}{y^4}}drd\varphi=2\pi\int \frac{r}{y^2}\sqrt{1+y'^2}dr \tag{1}$$

Now, I have considered another coordinate system $$y=\frac{R}{\gamma} \ , \text{where} \ \gamma=\cosh u\cosh\zeta-\cos\theta\sinh u$$ $$r=\frac{R\cosh u \sinh \zeta}{\gamma}$$ $$\tilde{r}=\frac{R\sinh u \sin\theta}{\gamma}$$ where $$\tilde{r}$$ and $$\tilde{\varphi}$$ are polar coordinates in the $$x_3-x_4$$ plane. The hyperbolic space $$\mathbb{H}^5$$ now has the metric $$ds^2=du^2+\cosh^2 u(d\zeta^2+\sinh^2\zeta d\varphi^2)+\sinh^2u ds^2_{S^2} \ , \ ds^2_{S^2}=d\theta^2+\sin^2\theta d\tilde{\varphi}^2$$ I take now $$\zeta$$ and $$\varphi$$ as parameters on the surface. Again, I can argue that $$u(\zeta,\varphi)=u(\zeta)$$, and $$\theta$$ and $$\tilde{\varphi}$$ are constants. Therefore, the induced metric is $$g_{ind} : (u'^2+\cosh^2 u)d\zeta^2+\cosh^2 u\sinh^2\zeta d\varphi^2$$ and $$S=2\pi\int\sqrt{u'^2+\cosh^2 u}\cosh u\sinh\zeta d\zeta\tag{2}$$

Now, I want to relate these two functionals. Explicitly I want to get (2) starting from (1).

My general idea is to change $$dr$$ by $$\frac{dr}{d\zeta} d\zeta$$ so that $$\int \frac{r}{y^2}\sqrt{1+\left(\frac{dy}{dr}\right)^2}dr=\int \sqrt{\frac{r^2}{y^4}\left(1+\left(\frac{dy}{dr}\right)^2\right)}\frac{dr}{d\zeta} d\zeta$$ $$=\int \sqrt{\frac{r^2}{y^4}\left(\left(\frac{dr}{d\zeta}\right)^2+\left(\frac{dy}{dr}\right)^2\left(\frac{dr}{d\zeta}\right)^2\right)} d\zeta$$ Then I have to get the following $$\frac{r^2}{y^4}\left(\left(\frac{dr}{d\zeta}\right)^2+\left(\frac{dy}{dr}\right)^2\left(\frac{dr}{d\zeta}\right)^2\right)=\left(\left(\frac{du}{d\zeta}\right)^2+\cosh^2 u\right)\cosh^2 u\sinh^2\zeta$$ But I didn't succeed. I am having trouble in many steps, for example: Can I say that $$\frac{dy}{dr}\frac{dr}{d\zeta}=\frac{dy}{d\zeta} ?$$

when I calculated $$\frac{dy}{d\zeta}=\frac{d}{d\zeta}(R/\cosh u(\zeta)\cosh\zeta-\cos\theta\sinh u(\zeta) )$$ I got something proportional to $$u'+$$something. So then $$(\frac{dy}{d\zeta})^2$$ will have not only $$u'^2$$ but also $$u'$$.