Triple integral for the volume between $z=2$ and the top of the sphere $x^2 + y^2 + z^2 = 9$ Apparently we do something like this: 
$z=2$, so $x^2 + y^2 + z^2 = 9$ tells us that $x^2+y^2=5$, and so our integral is: 
$$\int_{-\sqrt{5}}^{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1\,dz\,dy\,dx$$
That solution makes very little sense to me. Why do we substitute the formula for the plane $z=2$ into the formula for the sphere? What exactly does that accomplish? The innermost integral makes sense, but the limits on the first two do not. 
Shouldn't the outer two integrals simply be something like $$\int_0^3\int_{-\sqrt{9-x^2}}^\sqrt{9-x^2}$$ 
since the radius of the sphere is $3?$
Any help is appreciated. 

You want to project your solid onto the $xy$-plane first (Hint: look at the last two variables where you integrate). What does it look like?    
Essentially, when $$z=2,$$ the solid has a region $R$ bounded by $$x^2+y^2=5,$$
a circle, which you pointed out already. From this region, we start to look at $y$ first and where it tends to and from (Hint: look again at the last two variables where you integrate:$\,dy$ is the outermost, so we look at $y$ first).  
From the equation of the region $R$ above, we can see that the radius is equal to $\sqrt{5}$. Here, we can finally say that $$-\sqrt{5}≤y≤\sqrt{5}$$
Now we look at $x$, where you just manipulate the equation of the same region
$$x^2+y^2=5,$$
$$\implies x=±\sqrt{5-y^2}$$
Hence, the bounds for $x$ is
$$-\sqrt{5-y^2}≤x≤\sqrt{5-y^2}$$
For $z$, since the solid is bounded below by the plane $z=2$ and above by the sphere, we have
$$2≤z≤\sqrt{9-x^2-y^2}$$
Finally, we can set up the iterated triple integral
$$\int_{-\sqrt{5}}^\sqrt{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1dzdydx$$
