I am considering a family of curves indexed by $a\in \mathbb{R}$:
$$\vec{r}_a(t)=\bigg(\frac{2t}{t^2+(t+a)^2+1},\frac{2(t+a)}{t^2+(t+a)^2+1},\frac{t^2+(t+a)^2-1}{t^2+(t+a)^2+1}\bigg).$$ Note that for each $a$, $\vec{r}_a(t)$ is a circular arc on the unit sphere passing through the north pole.
What is the surface area of the patch on the sphere enclosed by $\vec{r}_{a_1}(t)$ and $\vec{r}_{a_2}(t)$?
Assuming $a_2>a_1$, I thought the answer is given by
$$\int_{a_1}^{a_2}\int_{\mathbb{R}} ||\partial_t\vec{r}_a(t)||||\partial_a\vec{r}_a(t)|| dt da$$
I got the above formula by naively looking at the change of the curves with small increments $(a\to a+da)$ and $(t\to t+dt)$ and treating the patch that it covers as a square.
But I don't think this is the right answer. Note that
$$||\partial_t\vec{r}_a(t)||= \frac{2\sqrt{2}}{t^2+(t+a)^2+1}, \quad ||\partial_a\vec{r}_a(t)||= \frac{2}{t^2+(t+a)^2+1}$$
If we integrate over the entire range we should have the surface area of the unit sphere: $4\pi$. However, we get that
$$\int_{\mathbb{R}}\int_{\mathbb{R}} ||\partial_t\vec{r}_a(t)||||\partial_a\vec{r}_a(t)|| dt da=\int_{\mathbb{R}}\int_{\mathbb{R}} \frac{4\sqrt{2}}{(t^2+(t+a)^2+1)^2} dt da=4\sqrt{2}\pi.$$
So, I am not sure what is the right answer.