Volume of the intersection of a shifted sphere I want to find the intersection of the sphere $A:x^2+(y-1)^2+z^2=1$ and the sphere $B:x^2+y^2+z^2=1$.
I change to spherical coordinates, with $\theta$ being the horizontal angle and $\phi$ the vertical one.
The sphere $A$ can be written as $x^2+y^2+z^2=2y$ and in spherical coordinates $A:\rho^2=2\rho \sin(\phi)\sin(\theta)$
$\rho$ goes from $0$ to $2\sin(\phi)\sin(\theta)$
Both spheres intersect at the angles $\pi/3$ and $2\pi/3$
I divide the problem into two parts:
The first one is to find the volume enclosed inside the $B$ sphere. As the intersected volume is in the first and second quadrants, the angle $\theta$ goes from $0$ to $\pi$. This can be done with the next integral:
$$(1a)\ \int_{0}^\pi \int_{\pi/3}^{2\pi/3} \int_0^1 \; \rho^2\sin(\phi)\; d\rho\ d\phi\ d\theta \;=\; \pi/3\; =\; 1.0472$$
Now where I'm having trouble is in finding the volume enclosed inside the sphere $A$.
To find the range of the $\theta$ angle, I plotted the graph $\rho(\theta,\phi)=2\sin(\phi)\sin(\theta)\;; \theta=0..2\;\pi,\;\phi=0..\pi/3$

Though $\phi=0..3\pi$,  the part between $\phi=2pi/3..\pi$ it's also plotted, which I don't understand it very well.
Then if the range of $\theta=0..\pi$ I get the next graph:


In order to find the part of volume of intersection inside the sphere $A$ I do the next integral:
$$(1b)\ 2\int_0^\pi \int_0^{\pi/3} \int_0^{2\sin(\theta)\sin(\phi)}\; \rho^2\sin(\phi)\;d\rho\ d\phi\ d\theta\ = \ \frac{289\pi}{840}\ = 1.08086 $$
I don't understand very well this integral. I found the integration bounds mostly using the graph. And I don't understand why it is wrong.
Note: I found the volume of the intersection of the spheres $C:x^2+y^2+(z-1)^2=1$ and $D:x^2+y^2+z^2=1$.
I find it much easier to understand and grasp the idea.
The angle of intersection of both spheres was $$\frac{\pi}{3}$$
The volume of intersection enclosed inside sphere $B$ was:
(2a) $$\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^1\ \rho^2 \sin(\phi)\ d\rho\ d\phi\ d\theta \;=\; \pi/3\; =\; 1.0472 $$
Equal to equation $(1a)$
For the volume of intersection inside sphere $C$ I solved  the next integral:
$$ (2b) \int_0^{2pi}\int_{\pi/3}^{\pi/2}\int_0^{2 \cos(\phi)}\ \rho^2 \sin(\phi)\ =\ \frac{\pi}{12}\ =\ 0.261799 $$ This outcome is right and is different than the result of (1b).
 A: We can swap the definitions of $y$ and $z$ to obtain
$$\begin{cases}x = \rho\sin\phi\cos\theta \\ y = \rho\cos\phi \\ z = \rho\sin\phi\sin\theta\end{cases} \implies \rho = 2\cos\phi$$
then we can do the integral with $\phi$ first to do the integral in one piece
$$\int_0^{2\pi} \int_0^1 \int_0^{\cos^{-1}\left(\frac{\rho}{2}\right)}\rho^2\sin\phi\:d\phi\:d\rho\:d\theta = 2\pi \int_0^1 \rho^2-\frac{\rho^3}{2}\:d\rho = \frac{5\pi}{12}$$
This gives us intuition on how to approach the original problem without swapping the $y$ and $z$ coordinates. We should approach the problem with the integration order $\phi,\theta,\rho$. First, as a function of $\phi$ the first bounds will be
$$\rho = 2\sin\phi\sin\theta \implies \phi = \sin^{-1}\left(\frac{\rho\sin\theta}{2}\right),\pi-\sin^{-1}\left(\frac{\rho\sin\theta}{2}\right)$$
(given by the $\sin(\pi-z) = \sin(z)$ reflection formula). Next, the integral projects on to the $\phi = \frac{\pi}{2}$ surface (the $xy$ plane since this is the widest slice possible perpendicular to the $\phi$ direction) and the exact same principle applies to find the next set of bounds
$$\rho = 2 \sin \frac{\pi}{2}\sin \theta \implies \theta = \sin^{-1}\left(\frac{\rho}{2}\right),\pi-\sin^{-1}\left(\frac{\rho}{2}\right)$$
Now we have enough information to set up the integral
$$\int_0^1 \int_{\sin^{-1}\left(\frac{\rho}{2}\right)}^{\pi-\sin^{-1}\left(\frac{\rho}{2}\right)} \int_{\sin^{-1}\left(\frac{\rho\sin\theta}{2}\right)}^{\pi-\sin^{-1}\left(\frac{\rho\sin\theta}{2}\right)}\rho^2\sin\phi\:d\phi\:d\theta\:d\rho  $$
A: Now reading various comments and seeing your edit, I have changed my answer accordingly to focus on mistakes in your working and what should be the correct integrals.
In fact, both $1(a)$ and $1(b)$ are incorrect.
While $1(a)$ gives you the same answer but that is just a coincidence. Limits of $\phi$ is not between $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. Also on sphere $B$ in the given region, $\theta$ cannot be between $0$ and $\pi$.
At intersection of both sphere,
$\rho = 2 \sin \theta \sin \phi = 1$
$ \implies \phi = \arcsin \left(\frac {\csc\theta}{2}\right)$
Similarly below z-axis, $\phi = \pi - \arcsin \left(\frac {\csc\theta}{2}\right)$
Also at intersection $\theta$ is min when $\sin\phi$ is max, which is $1$.
So, $\sin\theta = \cfrac{1}{2} \implies \theta = \cfrac{\pi}{6}, \cfrac{5\pi}{6}$.
So the integral for $1(a)$ is,
$ \displaystyle \int_{\pi/6}^{5\pi/6} \int_{\arcsin{((\csc\theta)/2)}}^{\pi - {\arcsin{((\csc\theta)/2)}}} \int_0^1 \rho^2 \sin\phi \ d\rho \ d\phi \ d\theta = \frac{\pi}{3}$
Now coming to $1(b)$, please note that for $\frac{\pi}{6} \leq \theta \leq \frac{5\pi}{6}$, above z-axis we have, $0 \leq \phi \leq \arcsin \left(\frac {\csc\theta}{2}\right)$ and below z-axis, $\pi - \arcsin \left(\frac {\csc\theta}{2}\right) \leq \phi \leq \pi$
You should also note that for,
$0 \leq \theta \leq \frac{\pi}{6}$ and then again for $\frac{5\pi}{6} \leq \theta \leq \pi$, $0 \leq \phi \leq \pi$.
So $1(b)$ is to be split into two integrals for evaluation. Using the symmetry, the integrals are,
$ \displaystyle 2 \int_{\pi/6}^{5\pi/6} \int_0^{\arcsin{((\csc\theta)/2)}} \int_0^{2\sin\theta \sin\phi} \rho^2 \sin\phi \: d\rho \: d\phi \: d\theta = \frac{(3 \sqrt3 - 5) \pi}{4} $
$ \displaystyle 2 \int_0^{\pi/6} \int_0^{\pi} \int_0^{2\sin\theta \sin\phi} \rho^2 \sin\phi \: d\rho \: d\phi \: d\theta = \left(\frac{4}{3} - \frac{ 3 \sqrt3}{4}\right) \pi $
Adding all three of them, you get the volume of the region as $\cfrac{5\pi}{12}$.

However if you are at ease with spherical coordinates, you could convert this into one integral by going in the order such that the outermost integral is wrt $\rho$.
For $\rho \leq 1$ in the given region, note that the bounds of $\phi$ and $\theta$ are on sphere $A$. So, $\rho = 2 \sin \theta \sin \phi \implies \phi = \arcsin \left(\cfrac{\rho \csc\theta}{2}\right)$ and similarly below z-axis, $\phi = \pi - \arcsin \left(\cfrac{\rho \csc\theta}{2}\right)$
and then, $\theta = \arcsin \left(\cfrac{\rho}{2}\right) \ $ and for $ \ x \leq 0 \ $, $\theta = \pi - \arcsin \left(\cfrac{\rho}{2}\right)$
So the integral is,
$\displaystyle \int_0^1 \int_{\arcsin (\rho / 2)}^{\pi - \arcsin (\rho / 2)} \int_{\arcsin ((\rho \csc\theta) / 2)}^{\pi - \arcsin ((\rho \csc \theta) / 2)} \rho^2 \sin \phi \ d\phi \ d\theta \ d\rho$
