substituting spherical coordinates to evaluate an integral. I have to evaluate
$$\int^1_{-1} \int^{ \sqrt {1-x^2}}_{-\sqrt {1-x^2}} \int^1_{-\sqrt{x^2+y^2}} \, dz \, dy \, dx$$
using spherical coordinates.
This is what I have come up with
\begin{align}
& \int^1_0 \int^{2\pi}_0 \int^{3\pi/4}_0 r^2\sin\theta \, d\theta \, d\phi \, dr \\[10pt]
= {} & \int^1_0 r^2 \, dr \int^{2\pi}_0 d \phi \int^{3\pi/4}_0 r^2\sin\theta \, d\theta \\[10pt]
= {} & \frac 1 3 \times 2\pi \times \left[-\cos\theta\vphantom{\frac 1 1}\right]^{3\pi/4}_0
\end{align}
by a combination of sketching and substituting spherical coordinates. 
After evaluating I obtain this integral to equal 3.57.
where as the first one evaluates to 5.236.
EDIT:
A bit of thought shows me that the above integral gives a spherical volume. We need to restrict $r$
As $x^2 + y^2 = 1 \implies \rho = \csc \theta$
$$\int^{3\pi/4}_{\pi/4} \int^{2\pi}_0 \int^{\csc \theta}_1 r^2\sin\theta \, dr d\phi \, d\theta $$
However
This, yet again, does not give me what I want.
 A: To compare various answers:
As described in the comments, the region is a enclosed by the cylinder $x^2+y^2 = 1$, with top $z=1$, and bottom the cone $z=-\sqrt{x^2+y^2}$.
Sol 1:
The region has volume $V$ = that of a cylinder minus the cone, i.e.,
$$ V = 2 \pi 1^2 - 1/3 \cdot \pi 1^2 \cdot 1 = 5 \pi/3.$$
(The volume of the cone can most easily be done by Pappus's theorem, aka, looking it up, or by the disc method of integration.)
Sol 2: Using cylindrical coordinates - as in Michael Hardy's solution:
$$V = \int_{\phi=0}^{2\pi}\int_{r=0}^1\int_{z=-r}^1\, dz\, r dr \, d\theta = 2\pi \int_{r=0}^{1}(1+r)r\,dr= 2\pi\left(1/2 + 1/3\right). $$
Sol 3: Using spherical coordinates, as desired. 
Notation: if I have understood the comments correctly: 


*

*$\phi$ sweeps counter-clockwise in the $x/y$ plane, starting from the $x$-axis, 

*$\theta$ sweeps upwards from the $z$-axis, so that $\theta=0$ corresponds to due South;  

*$\rho$ is the distance from the origin.


Then the volume element $dV = \rho^2 \sin \theta \, d\theta \, d\phi\, d\rho$.
Walking on a line from the origin (i.e., holding $\theta$ and $\phi$ fixed), we start at $\rho=0$, but eventually encounter the boundary of the region. 


*

*If $\pi/4 \le \theta \le 3\pi/ 4$, the line encounters the exterior of the cylinder ($x^2+y^2 =1 $); call this region $R_1$ - here $$\rho_{\rm max} = \csc \theta,$$
as you have it in your edit above.

*If $3\pi/4 \le \theta \le \pi $, the line encounters the top disk, where $z= 1$; call this region $R_2$ - here $$\rho_{\rm max } = -\sec \theta,$$
the negative sign because $\cos \theta$ is negative in this region, with this (unusual?) convention for $\theta$...
Calculate $V = V_1 + V_2$:


*

*$V_1$:
$$\begin{align} V_1 &= \int \int \int_{R_1} \,dV = \int_{\phi=0}^{2\pi} \int_{\theta = \pi/4}^{3\pi/4}\int_{\rho=0}^{\csc \theta} \rho^2 \,d\rho\, \sin \theta \, d\theta \,d\phi \\
&= 2 \pi \int_{\theta =\pi/4}^{3\pi/4} 1/3 \csc^3\theta \sin \theta \, d\theta \\
& ={2\pi\over 3} \int_{\theta =\pi/4}^{3\pi/4}\csc^2\theta \,d\theta \\
& ={2\pi\over 3}( -\cot \theta )\mid_{\pi/4}^{3\pi/4} \\
&={4 \pi \over 3}.\\
\end{align}$$

*$V_2$:
$$\begin{align} V_2 &= \int \int \int_{R_2} \,dV = \int_{\phi=0}^{2\pi} \int_{\theta = 3\pi/4}^{\pi}\int_{\rho=0}^{-\sec \theta} \rho^2 \,d\rho\, \sin \theta \, d\theta \,d\phi \\
&= 2 \pi \int_{\theta =3\pi/4}^{\pi} (-1/3) \cos^{-3}\theta \sin \theta \, d\theta \\
& =-{2\pi}\cdot{1/6}\, \cos^{-2}\theta \,\mid_{3\pi/4}^{\pi} \\
&={ \pi \over 3}.\\
\end{align}$$
So $V= 5\pi/3$. 
I hope I didn't mess anything up... In any case, spherical coordinates are perhaps not the best choice for this problem - right?
Note: I'm not sure I got your convention for $\theta$ correctly... In any case, why don't you calculate the volume of the cone using spherical coordinates? - it has to come out to $\pi/3$... If you are still confused, probably you should do the problem 'live' in front of your instructor...
A: $$
\int^1_{-\sqrt{x^2+y^2}} \, dz = 1+\sqrt{x^2+y^2}.
$$
Therefore
\begin{align}
& \int^1_{-1} \int^{ \sqrt {1-x^2}}_{-\sqrt {1-x^2}} \int^1_{-\sqrt{x^2+y^2}} \, dz \, dy \, dx = \int^1_{-1} \int^{ \sqrt {1-x^2}}_{-\sqrt {1-x^2}}  \left(1+\sqrt{x^2+y^2}\,\right) \,dy\,dx \\[20pt]
& \text{Now switch to polar coordinates:} \\[10pt]
= {} & \underbrace{\int_0^{2\pi} \quad \overbrace{\int_0^1 (1+r)\, r\,dr}^\text{No “$\,\theta\,$'' appears here.} \quad d\theta = 2\pi \int_0^1 (1+r)\, r\,dr}_\text{These are equal because “$\,\theta\,$'' does not appear in the inside integral.} = \cdots \cdots
\end{align}
