Finding volume enclosed by hyperboloid and plane $z = 2$ with double integration

Question

I'm doing this problem andfor the highlighted second I had tried and thought it would be $\int r \,dz=zr$ so then evaluating at $\left.\vphantom{\dfrac 1 1}\right|^2_{\sqrt{r^2+1}}$ would give $2r - \sqrt{r^2+1}\,r$ can someone explain why the $r$ is only on the square root part in the second red boxed section.

Answer 1

There is a typo in the announced solution, we have $$ \int^2_{\sqrt{r^2+1}} dz=\left[z\frac{}{}\right]^2_{\sqrt{r^2+1}}=2-\sqrt{r^2+1} $$ giving $$ \int^2_{\sqrt{r^2+1}} r\:dz=\left(2-\sqrt{r^2+1}\right)r $$ as you have expected.