## Calculus (3rd Edition)

$\rho=1$ , $0\leq \phi\leq \pi/2$, $0\leq\theta\leq \pi/2$.
The set $x^2+y^2+z^2=1$, $x\geq0$, $y\geq 0$, $z\geq 0$ is the one-eighth sphere in $R^3$, centered at the origin. Since \begin{aligned} &x=\rho \sin \phi \cos \theta\\ &y=\rho \sin \phi \sin \theta\\ &z=\rho \cos \phi, \end{aligned}we have $$x^2+y^2+z^2=\rho^2(\sin^2\phi\cos^2\theta+\sin^2\phi\sin^2\theta+\cos^2\phi)\\ =\rho^2$$ That is, $\rho^2= 1$ and since $\rho$ is non-negative then $\rho=1$ and since $z\geq 0$, then $\rho\cos \phi \geq0$ i.e, $0\leq \phi\leq \pi/2$ and for $x\geq0$, $y\geq 0$, which leads to $0\leq\theta\leq \pi/2$.