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. 
 A: When you are doing these problems, the hardest part is parametrizing the domain. Looking at the solution without any explanation can be frustrating because these problems definitely have steps.
First you need to decide on an order of integration. That's right, this is a choice. The person who wrote your solution decided on the order $dzdydx$ so let's stick with that. We will work from the outside in to determine the limits of integration. I prefer to think about it like we are slicing with planes. We want to cut the volume with a plane $x=\text{constant}$, so we need to determine the bounds on $x$. To see these bounds you need to think about the picture that I've attached. The plane $z=2$ meets the sphere in some circle. So, clearly, the radius of this circle determines where we are allowed to slice with an $x-$plane.
The algebraic way to understand "the sphere meets the plane" is to plug $z=2$, the equation for the plane, into the equation for the sphere. This results in $x^2+y^2=5$. So the extreme values for $x$ are $-\sqrt{5}$ to $\sqrt{5}$.

Now slice the volume with some $x-$value. Don't actually pick one, but imagine we chose one at random in the range we've determined for $x$. This is what it looks like after you slice.

But now we need to slice this 2-dimensional shape with $y$. What are the bounds on $y$ in terms of the already chosen $x$? This is given by solving for $y$ in the equation we found above. When we slice with this $y$, we are only left with a vertical segment running from the plane to the sphere. Thus $z$ ranges from $2$ to $z=\sqrt{9-x^2-y^2}$.
A: No because the region of the integration is a circle of radius $5^{0.5}$ and not $3$. Keep in mind that your $x$ and $y$ valued range from the circle of integration that can be projected in the $x-y$ plane.
You are only integrating inside the circle of inner sphere formed by the plane $z=2$ not the whole sphere. To see this project the integration boundaries on the $x-y$ plane.
Also you substitute $z=2$ to set the lower boundary of your $z$ as it cannot come below that amount.
Just draw a picture and project it (it really helps, I can’t stress it enough).
A: 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
$$\int_{-\sqrt{5}}^\sqrt{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1dzdydx$$
