$\iiint_Dz \;dxdydz$ where D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$ I asked a questions about regions and then tried to compute a tripple integral:
$$\iiint_Dz \;dxdydz$$ D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$
I tried, but now I am stuck: how do I calculate the volume of $D_{1b}$? Since I don't have many of these problems left, I decided not to look at the solution photo1, photo2.
Attempt: Here is a photo of my calculations

$$ z>0$$ What implies:$$z\geq\sqrt{2x^2+3y^2-1}$$ 
Volume I need to calculate is the interior of sphere: $$ x^2+y^2+z^2=3$$ and above (where we have orthogonal projection to the xy plane) the hyperboloid. 

Excuse me for lousy picture!
So I have: 
$$ U_{tot}= \int\int\int_{D_1} zdV + \int\int\int_{D_2}z dV=U_1+U_2$$ 
Define orthogonal projection with $z=0$:
$$z=0=\sqrt{2x^2+3y^2-1}\rightarrow 2x^2+3y^2=1$$
Sphere projection is: $x^2+y^2=(\sqrt{3})^2$
Orthogonal projection $D_1$ is now described with the area between the eclipse and a circle, we need to define some parametrization in order to simplify. For example I can try something like this: 
$$x=r\cos\theta,y=r\sin\theta$$
Equation of circle is now: $r=\sqrt{3}$ and equation of eclipse is: $2r^2\sin^2\theta+3r^2\cos^2\theta=1 \rightarrow r=\sqrt{\frac{1}{2cos^2\theta+3\sin^2\theta}}=\sqrt{\frac{1}{2+\sin^2\theta}}$  
On the orthogonal projection $D_1$ we have a volume from hyperboloid to the sphere. 
$$ U_1=\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}}rdr\int^{\sqrt{3-r^2}}_{\sqrt{2r^2\cos^2\theta+3r^2\sin^2\theta-1}=\sqrt{2r^2+r^2\sin^2\theta-1}}zdz =$$
$$=\frac{1}{2}\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} \left(  3-r^2-2r^2-r^2\sin^2\theta+1\right) rdr=\\ =\frac{1}{2}\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} \left(4r-3r^3-r^3\sin^2\theta\right)dr=\\
=\frac{1}{2}\int^{2\pi}_0 \left( 2r^2-\frac{r^4}{4}(3+\sin^2\theta) \right)|^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} d\theta=\\
=\frac{1}{2}\int^{2\pi}_0 \left( 6-\frac{9}{4}(3+\sin^2\theta)\right)-\left( \frac{2}{2+\sin^2\theta}-\frac{3+\sin^2\theta}{4(2+\sin^2\theta)^2}\right)d\theta=...
$$
On the orthogonal projection $D_2$ we have a volume from plane $z=0$ to the sphere.
$$U_2=\int^{2\pi}_0 d\theta \int^{\sqrt{3}}_0 rdr \int^{\sqrt{3-r^2}}_0 zdz =\\
= \frac{1}{2}\int^{2\pi}_0 d\theta \int^{\sqrt{3}}_0 r \left( 3-r^2\right)dr=\\
=  \frac{1}{2}\int^{2\pi}_0 \left( 3r-r^3/3\right)|^{\sqrt{3}}_0d\theta=\\
=  \frac{1}{2}\int^{2\pi}_0 3\sqrt{3}-\sqrt{3} d\theta=\\
=  \sqrt{3}\int^{2\pi}_0 d\theta=2\sqrt{3}\pi.
$$
Sorry if a have mistaken somewhere in the calculus I will check it once again...
