Find limits of integration for the interior region of sphere with center $(a,0,0)$ and radius $a$ using spherical coordinates I am asked to find limits of integration for the interior region of sphere with center $(a,0,0)$ and radius $a$ using spherical coordinates. How can one do that?
I know that one may use
$$
x = r \cos(\theta) \sin(\phi)\\
y = r \sin(\theta) \sin(\phi)\\
z = r \cos(\phi)
$$
Is it possible do to the same with cylindrical coordinates?
Thank you.

you have a few choices.
rectangular:
$(x-a)^2 + y^2 + z^2 = a^2\\
x^2 + y^2 + z$ = 2ax$
Spherical... since x is the "special one", I would suggest.
$$x = r \cos(\phi)\\
y = r \sin(\theta) \sin(\phi)\\
z = r \cos(\theta) \sin(\phi)$$
Plug these into your equation for the sphere and,
$r^2 = 2a\,r\cos\phi $
$r$ will range from $0$ to $2a\cos\phi, \theta$ from $0$ to $2\pi, \phi$ from $0$ to $\pi/2$
If you went with the traditional.
$$x =r \cos(\theta) \sin(\phi)\\
y = r \sin(\theta) \sin(\phi)\\
z =  r \cos(\phi)$$
Then $r$ will range from $0$ to $2a\cos\theta\sin\phi, \theta$ from$-\pi/2$ to $\pi/2$
how about...Taking the traditional and translating it.
$$x =r \cos(\theta) \sin(\phi) + a\\
y = r \sin(\theta) \sin(\phi)\\
z =  r \cos(\phi)$$
And $r$ goes from $0$ to $a.$
Cylindrical.
$$x = x+a\\
y = r \sin(\theta)\\
z = r \cos(\theta)$$
x from $-\sqrt {a^2-r^2}$ to $\sqrt {a^2-r^2}$
or,
$$x = x\\
y = r \sin(\theta)\\
z = r \cos(\theta)$$
x from $a-\sqrt {a^2-r^2}$ to $a+\sqrt {a^2-r^2}$ 
etc.
