Calculate area of hyperbolic paraboloid $z = y^2-x^2$ that lies above the disc $x^2+y^2 \le 1$
My attempt: Using the formula $$\int\int_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}dxdy$$
I get the following $$\int\int_D 2\sqrt{x^2+y^2+\frac{1}{4}}dxdy$$
Plugging these into polar coordinates - Because we're above the disk we get $0 \le \theta \le \pi$ and $0 \le r \le 1$
Replacing the inner square root for $\sqrt{r^2(cos^2(\theta)+sin^2(\theta))+\frac{1}{4}}=\sqrt{r^2+\frac{1}{4}}$
$$\implies \int_0^{\pi}2\int_0^1\sqrt{r^2+\frac{1}{4}}rdrd\theta = \frac{2\pi (r^2+1)^{3/2}}{3}$$
The answer in my book is $\frac{\pi}{5}(5^{5/3}-1^{3/2})$ How did they get this?