Computing volume of $ D = \{ (x,y,z) \in \mathbb{R}^{3} : x^2 + y^2 + z^2 \leq 1, y^2 + z^2 \leq x^2 \} $ I want to compute the volume given by: $$ D = \{ (x,y,z) \in \mathbb{R}^{3} : x^2 + y^2 + z^2 \leq 1, y^2 + z^2 \leq x^2 \} $$
Which acording to Geogebra looks like: Plot of D
(Sorry guys, I tried to add some fancy plot of $ D $, but it looks like I am not allowed to show images).
In other words, I want to compute the 3D integral: $$ \int_{D} 1 dA $$
So, I used spherical coordinates and setup the integral in the following way: 
$$ 2 \int_{0}^{1} \int_{0}^{\pi/2} \int_{0}^{\pi/2} r^2 \sin\theta \, d\varphi \, d\theta \, dr \tag{*} $$
Where $ 0 \leq \varphi \leq 2\pi $ and $ 0 \leq \theta \leq \pi $.
My explanation:
First, I reduce the problem to compute only one volume and duplicate it, to get the whole volume of $ D $. Then: the integral depeding of $ r $ is clear because of the sphere. The integral depending of $ \theta $ and the integral depending of $ \varphi $ have to have the same interval because the conic surface is simetric in $ y $ and $ z $ coordinates. At first, I tried to write it down just like: $$ 2 \int_{0}^{1} \int_{-\pi/4}^{\pi/4} \int_{-\pi/4}^{\pi/4} r^2 \sin\theta \, d\varphi \, d\theta \, dr $$ but then I remembered that both angles have to be positive. So I think that the volume of the function $ f(x,y,z) = 1 $ in $ [-\pi/4,\pi/4] $ is the same that in $ [0,\pi/2] $. Is it right? $ \tag{**} $

My questions:
  
  
*
  
*Is the integral $ (*) $ right? Does it really compute the volume of $ D $?
  
*Is my explanation correct?
  
*Is $ (**) $ true?

At the moment, I just want to know if everything is fine, so I can compute the value of the integral. But my dificulties are the ones I presented before, so I appreciate if you can answer my questions.
 A: May what i did it isn't such a complex solutions as you are looking for, but i hope it helps.
Let's analyze what is our domain $ D = \{ (x,y,z) \in \mathbb{R}^{3} : x^2 + y^2 + z^2 \leq 1, y^2 + z^2 \leq x^2 \}$, which gives us two restrictions,
1.-That's the sphere or ratio r=1 $$   x^2 + y^2 + z^2 \leq 1$$
And
2.-The cone $$  y^2 + z^2 \leq x^2  $$
On the other hand, we can compute a solid's volume by Solid of revolution th, 
$$ V=\pi\int_a^b f(x)^2 dx$$.
The volume of the cone can be compute using the formula below taking $f(x)=x, a=-1, b=1$ .Furthermore we can compute the volumen of the sphere taking $g(x)=\sqrt{1-x^2}, a=-1, b=1$ (I can´t add photos to this anwer, but please try to draw this two functions).
Now, let's calculate where these to functions take the same value, ie
$$f(x)=g(x), x=\sqrt{1-x^2}$$ after some computations we obtain $$x=+1\sqrt{2}, x=-x=+1\sqrt{2}$$, Finally, considering the symmetry of the function, it's easy to see that que volume we are looking for is $$V=2(\pi\int_{0}^{1/\sqrt{2}} f(x)^2dx+\pi\int_{1/\sqrt{2}}^{1} g(x)^2dx);$$
$$$V=2(\pi\int_{0}^{1/\sqrt{2}}x^2dx+\pi\int_{1/\sqrt{2}}^{1} ( \sqrt{1-x^2} )^2dx)=2[\pi\frac{(1/\sqrt{2})^3}{3}+ \pi (1-\frac{1}{3})- \pi(\frac{1}{ \sqrt{2}}-\frac{(\frac{1}{\sqrt{2}})^3}{3})]=2[\frac{2\pi}{3}-\frac{\pi}{\sqrt{2}}]$$
