triple integral and limits

Question

$$\iiint\limits_H (x^2+y^2) \, dx \, dy \, dz \\ H=\{(x,y,z) \in R^3: 1 \le x^2+y^2+z^2 \le9, z \le 0 \}$$

I'm using Spherical coordinate system: $$x=r\cos \theta \cos\phi $$ $$y=r\cos \theta \sin\phi $$ $$z=r\sin \theta $$

and $r \in (1,3), \theta \in (0,2 \pi)$ but I don't know how to find set for $\phi$? Could anyone explain me in an easy way how to find $\phi$ [in this case and in general]?

Answer 1

You are integrating in a sphereically symmetric region (from the radius 1 sphere out to the radius 3 sphere). So in these co-ordinates you want $\theta \in \{0,\pi\} $ and $ \phi \in \{0,2\pi\}$. Theta ($\theta$) is the polar angle going from the "north pole" down to the "south pole" -- the lattitude of sorts (only $\pi$ radians). Phi ($\phi$) is the azimuthal angle or the "longitude". That one goes a full circle around the region ($2 \pi$ radians).

Added: Oops I overlooked that $z\le 0$. OK so Longitudinaly you are still good but now you integrate $\theta$ from $0$ to $\pi /2$. The angle $\theta = \pi/2$ represents the equator of your region or $z = 0$.

Answer 2

Hint: should be between $\pi/2$ and $\pi$ for a if you redefine $$z=r \cos(a),x=r \sin(a)\cos(b), y=r \sin(a) \sin(b)$$ and b between 0 2π and r between 1 and 3

Notice the region is below or equal to the plane $z=0$ and the 2 spheres

Answer 3

Since $z\leq 0$ then $\phi$ should be

$$ \pi/2\leq \phi \leq \pi. $$