calculus on finding the volume Find the volume defined by the following:
$$x^2+y^2+z^2\leq1$$ and $$x^2+y^2\leq\frac{(z^2)}{3}$$
Can roughly draw the picture but not sure where to start.
 A: Just realize that the first equation is redundant (just adds the limits) as the second space is always in the first sphere. Now calculate the volume enclosed by second cone

 It can be dissected into small discs of width $dz$ and radius $z/\sqrt{3}$ , calculate volume of this and integrate for $z \epsilon [-1,1] $.

A: Below is the graph of $z = \sqrt{3}.\sqrt{x^2 + y^2}$. Similarly you can draw if for the negative $z$'s.
Since the volume has identical 2 parts, (one part for positive $z$'s, and one for negative $z$'s), finding the volume for only positive $z$'s is enough.
From the graph, you can see that the volume is bounded below by $z = \sqrt{3}.\sqrt{x^2 + y^2}$, and bounded above by $ z = \sqrt{1-x^2-y^2}$.
Thus when you form the triple integrand, you first find the intersection points of these 2 curves, that is to find a circle $x^2 + y^2 = r^2$. Then you can form your integral by
$\int_{-r}^r \int_{-r}^r \int_{\sqrt{3}.\sqrt{x^2 + y^2}}^{\sqrt{1-x^2-y^2}} 1.dz.dy.dz$

EDIT: It is easy to see $r^2 = \frac{3}{4}$
