Chapter 16: Integrals and Vector Fields - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 989: 3

$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$ and $0 \le r \le 6$; $0 \le \theta \le \dfrac{ \pi}{2}$

Work Step by Step

We have $z=\dfrac{\sqrt{x^2+y^2}}{2} \implies z=\dfrac{\sqrt{r^2}}{2}=\dfrac{(r^{1/2})^2}{2}=\dfrac{r}{2}$ Now, $r(r, \theta)=xi+yj+zk ;\\ r=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$; and $0 \le z \le 3;\\ 0 \le \dfrac{r}{2} \le 3 \implies 0 \le r \le 6$ Hence, $r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$ and $0 \le r \le 6$; $0 \le \theta \le \dfrac{ \pi}{2}$

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