Volume of the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0$ I want to find volume the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le  1$ and $z  \ge 0.$
I set it up as $\displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{-\sqrt{\frac{1}{2} - x^2}}^{\sqrt{\frac{1}{2} - x^2}}\int_0^{\frac{1}{\sqrt{2}}} \, dz \, dy \, dx$ - how come that's not the concerned volume?
 A: $x^2+y^2+z^2 \leq 1$ is clearly a ball. And for $z^2 \geq x^2 + y^2$, you can think in this way, let $r^2 = x^2+y^2$, then $x^2+y^2 = r^2 \leq z^2$, which is a cone. The overlap of a cone and a ball is still a cone. So try to use spherical coordinates. 
A: The specified limits of integration are incorrect because $z$ does not depend on $x$ or $y$, whereas the actual region of integration does clearly depend on the value of $x$ and $y$.  For example, when $x = 1/2$ and $y = 0$, the given inequalities become $z \ge 1/2$ and $z \le \sqrt{3}/2$, hence the interval of integration for $z$ is $[1/2, \sqrt{3}/2]$.  But when $x = y = 0$, the interval of integration for $z$ is obviously $z \in [0, 1]$.
To understand what is going on, it helps to sketch the regions corresponding to each condition.  In this particular case, we can observe that each of the inequalities is invariant under rotation about the $z$-axis; that is to say, the cylindrical transformation $$x = r \cos \theta, \quad y = r \sin \theta$$ transforms the first inequality to $z^2 \ge r^2$, the second to $z^2 \le 1 - r^2$, and the third remains the same: $z \ge 0$.  This further simplifies to $$r \le z \le \sqrt{1-r^2}, \quad 0 \le r \le 1.$$  Now we can sketch the cross-sectional region corresponding to these inequalities in the $rz$-plane, and see that this is a sector of a unit circle with angle $\pi/4$, and that the desired volume is a solid of revolution about one of the radii of this sector.
There are many ways to characterize this volume:  for example, $$V = \int_{r=0}^{1/\sqrt{2}} 2\pi r (\sqrt{1-r^2} - r) \, dr$$ is the expression obtained by the method of cylindrical shells.  $$V = \int_{z=0}^{1/\sqrt{2}} \pi z^2 \, dz + \int_{z=1/\sqrt{2}}^1 \pi (1-z^2) \, dz$$ is the expression obtained by the method of disks perpendicular to the $z$-axis.  $$V = \int_{\theta=0}^{2\pi} \int_{r=0}^{1/\sqrt{2}} (\sqrt{1-r^2} - r) r \, dr \, d\theta$$ is the expression obtained by cylindrical coordinates (and is essentially equivalent to the cylindrical shell method).  $$V = \int_{\theta=0}^{2\pi} \int_{\phi=0}^{\pi/4} \int_{r=0}^1 r^2 \sin \phi \, dr \, d\phi \, d\theta$$ is the expression obtained using spherical coordinates.  The Cartesian expression is far too complicated to be worth writing down.
