Problem:
Find the volume of the solid that lies within the sphere $x^2+y^2+z^2=9$, above the xy plane, and outside the cone $z=4\sqrt{x^2+y^2}$
My attempt:
The triple integral of: ($p^2\sin{\phi}$) dp d$\phi$ d$\theta$
Range of p: $0$ <= p <= 3; as the volume lies above the xy plane.
Range of $\theta$: $0$ <= $\theta$ <= 2$\pi$; as we are dealing with spherical objects.
Range of $\phi$: $0$ <= $\phi$ <= $\arcsin{\frac{1}{\sqrt{17}}}$; by calculating the intersection of the shapes.
Why is $\phi$ wrong? Apparently the range was supposed to be:
$\arcsin{\frac{1}{\sqrt{17}}}$ <= $\phi$ <= $\pi/2$
Which to me doesn't make sense graphically. That bounds $\phi$ weirdly when I draw it out. Or am I doing something wrong here?