Moment of Inertia of a semi ellipsoid? 
I understand that mass is simply found by taking the integral of $\rho$ and differential volume.
I just do not understand how you can take the del of $I_x$ and where that equation $y^2\text{d}m/2$ comes from. I could not find the specific moment of inertia of an ellipsoid in a table.
I am assuming that the integral equation for $I_x$ comes from the inertia of a disk which is $\frac{1}{2} m r^2$? I just wanted to confirm my understanding of inertia.
 A: I note that you are not looking for the answer, but you are having difficulty understanding some of the steps that led to it.
For a solid of revolution such as this, the moment of inertia about the symmetry axis may be calculated by regarding it as being composed of a large number of thin disks. Your first formula
$$
\text{d}I_x = \tfrac{1}{2}\text{d}m\,y^2
$$
is the formula for the moment of inertia of a thin disk of mass $\text{d}m$ and radius $y$. So I'm hoping that you actually do understand this formula, despite your uncertainty in the question.
For both the mass, and the moment of inertia, we will need to integrate with respect to $x$. So we need to link $\text{d}m$ with $\text{d}x$. The volume of the thin disk is $\pi y^2 \text{d}x$, so 
$$
\text{d}m=\rho \pi y^2 \text{d}x, 
$$ 
and if we insert the formula for $y^2=b^2(1-x^2/a^2)$, and integrate, we get your expression for $m$. No problem there, I think.
We follow the same procedure for the moment of inertia. Using the expression for $\text{d}I_x$ we get
$$
\text{d}I_x = \tfrac{1}{2}\text{d}m\,y^2 = \tfrac{1}{2}\rho \pi y^4 \text{d}x ,
$$
and again we need to insert $y^2=b^2(1-x^2/a^2)$.
So we can see that your integral formula for $I_x$ is wrong: the term $(1-x^2/a^2)$ should be squared. Yet you got the correct answer on the next line, so I presume this was a simple slip in writing down the integral.
Some other points may help understand the situation here. The relation $I_x=\frac{2}{5}mb^2$ is the same for a full ellipsoid of revolution as for this semi-ellipsoid (both $I_x$ and $m$ would be twice as large) and indeed it is the same as the moment of ellipsoid of a sphere of radius $b$, since uniformly stretching a solid body along the $x$ axis will not change its mass, or its moment of inertia about the $x$ axis. So actually, you can find it in standard tables. If you want to look further, Wikipedia provides the moments of inertia of a general ellipsoid with three different semi-axis lengths; again, if you divide such an ellipsoid in two along a plane through the centre perpendicular to one of the principal axes, both the mass, and the moment of inertia about that particular axis, of the semi-ellipsoid will simply be halved.
Hopefully that's all you need to know, to understand what's going on. 
