Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^3: x^2 + y^2+z^2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$ Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^ 3: x ^ 2 + y ^ 2 + z ^ 2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$
I have to graph the region and calculate the volume. But I am having trouble visualizing the region. Is a kind of 1/2 cone formed?
What I have so far considering its a cone:
\begin{align*}
z^2=\sqrt{3} y\\
 \phi = arctg(\frac{1}{\sqrt3})\\ 
\theta=\frac{\pi}{6}
\end{align*}
Using polar coordinates:
\begin{align}
0 \leq r\leq1\\0\leq\phi\leq\frac{\pi}{6}\\0\leq\phi\leq 2\pi\\
Vol(S)=\int_{0}^{2 \pi}\int_{0}^{\frac{\pi}{6}}\int_{0}^{1}sin\phi dr d\phi d\theta\\
Vol(S)=\theta\mid_{0}^{2\pi}\frac{r^3}{3}_{0}^{1}\int_{0}{\frac{\pi}{6}} sin\phi d\phi = \frac{2\pi}{3}(-cos \phi)\mid_{0}^{\frac{\pi}{6}}=-\frac{2\pi}{3}((\frac{\sqrt3}{2})-1)
\end{align}
$VS= \frac{2\pi}{3}(1-\frac{\sqrt3}{2})=0.2806 u^3$
 A: Projection of intersection of sphere $x ^ 2 + y ^ 2 + z ^ 2 \leqslant 1$ with cone $z ^ 2 = 3 (x ^ 2 + y ^ 2)$ gives circle $x ^ 2 + y ^ 2 = \frac{1}{4}$ on $OXY$ plane, so volume can be found by integral
$$4\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{\sqrt{\frac{1}{4}-x^2}}\int\limits_{\sqrt{3 (x ^ 2 + y ^ 2)}}^{\sqrt{1-x ^ 2- y ^ 2}}dzdydx$$.
Using cylindrical coordinates $x=\rho \cos \phi, y = \cos \phi, z=z$ with Jacobian $J=\rho$ gives "sphere" $\rho^ 2+z^2=1$ and "cone" $z \geqslant 3 \rho$, so, volume can be found as
$$4\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{1}{2}}\int\limits_{3\rho}^{\sqrt{1-\rho^ 2}}\rho dzd\rho d\phi$$
and for spherical coordinates $x=r\sin\phi \cos\theta, y=r\sin\phi \sin\theta, z=r\cos\phi$ with $r \geqslant 0, \theta \in [0, 2\pi],  \phi \in [0, \pi]$ and with Jakobian $J=r^2 \sin{\phi}$ we should have "sphere" $r \leqslant 1$ and "cone" $\tan \phi = \frac{1}{\sqrt{3}}$
$$4\int\limits_0^{\frac{\pi}{2}} \int\limits_0^{\frac{\pi}{6}} \int\limits_0^1 r^2 \ \sin{\phi}  \ dr \ d\phi \ d\theta$$
A: It is a half cone with a bulge on the top.
Condition I: $x ^ 2 + y ^ 2 + z ^ 2 \leq 1$
Condition II: $z ^ 2 \geq 3 (x ^ 2 + y ^ 2)$
Condition III: $z \geq 0$
Condition I represents the set ($S_1$) of points inside and on a sphere with radius 1. Condition II represents the set ($S_2$) of points inside and on the circular cone. Finally, condition III is the set ($S_3$) of points on the positive $z$-axis including origin. Note that for $z \in [0, 1]$ $S_3 \subset S_2 \subset S_1$ become. With condition I and II we get a half cone as $S_2 \subset S_1$. Also note the volume generated by the intersection of circular cone and sphere on the positive $z$-axis as the two meet at a $z$-axis less than $1$, which can be found by solving the equations $x ^ 2 + y ^ 2 + z ^ 2 = 1$, $z ^ 2 = 3 (x ^ 2 + y ^ 2)$ for $z$.
