Computing Triple Integral Using Spherical Coordinates

I just have trouble finding the bounds for $\phi$ in this problem. So far I found $\pi < \theta < 3\pi/2$ and $0 < \rho < \sqrt{10}$

Use spherical coordinates to calculate the triple integral of $f(x,y,z) = y$ over the region $$x^2+y^2+z^2= 10$$ and $$x,y,z ≤ 0$$

At the moment I am trying to give an answer to this old question, but I am not entirely sure. The discussion in the comments mentions boundaries $\pi/2\leq\varphi\leq\pi$ and $\pi\leq\theta\leq \frac32\pi$ which I do not understand.

Can you enlighten me, or spot my mistake? Thanks in advance. (I will of course edit, or delete my answer if necessary)

$\rho=\sqrt{10}$

We get:

$$\int_{\pi}^{\frac32\pi}\int_{\pi/2}^{\pi}\int_0^{\sqrt{10}} \rho\sin(\varphi)\sin(\theta)\cdot\rho^2\sin(\varphi)\,d\rho\,d\varphi\,d\theta$$

Using the formula for cylindrical coordinates

$$\iiint_B f(\rho,\theta,\varphi)\cdot \rho^2\sin(\varphi)\,d\rho\,d\varphi\,d\theta$$

And the change of coordinates:

$x=\rho\sin(\varphi)\cos(\theta)$

$y=\rho\sin(\varphi)\sin(\theta)$

$z=\rho\cos(\varphi)$

Which then evaluates to $$-\frac{25}{4}\pi$$ which kind off makes no sense.