## Thomas' Calculus 13th Edition

$\dfrac{\pi \sqrt 5}{2}$
We have $r_r= \cos \theta \space i+\sin \theta \space j-\dfrac{ \sin \theta}{2} \space k ;\\r_{\theta}=-r \sin \theta \space i+r \cos \theta j-\dfrac{r \cos \theta}{2} k$ Also, $|r_r \times r_{\theta}|=\dfrac{\sqrt 5r}{2}$ and $0 \le r \le 1$; $0 \le \theta \le 2\pi$ Now, $Area=\int_0^{2 \pi} \int_0^{1} \dfrac{\sqrt 5r}{2} \space dr \space d \theta \\= \int_0^{2 \pi} [ \dfrac{\sqrt 5r^2}{4}]_0^{1} \space d \theta \\=\dfrac{\pi \sqrt 5}{2}$