Let $$S = \{ (x,y,z) | x=a+b, y = b+c, z = -b, ac-b^2\ge0, c\ge0, a\ge0\}$$ and $$x^2+y^2+z^2\le 1$$
Compute the volume of S.
My work: $$V=\iiint_R 1 dV = \int\limits_{-1}^{1}\int\limits_{-\sqrt{(1-b^2)-x^2}}^{\sqrt{(1-b^2)-x^2}}\int\limits_{-b}^{\sqrt{1-x^2-y^2}} 1. dzdydx$$
Am i right ???
how to use spherical ?? or cartesian is better here?
Next try:
eliminating a,b,c we get the planes $ x+z\ge0 $ and $ y+z\ge0$, and the hyperboloid $xy+(x+y)z\ge0$ with $x^2+y^2+z^2\le 1$
How to set up now ? $$V=\int\limits_{?}^{?}\int\limits_{?}^{?}\int\limits_{?}^{?} 1. dzdydx$$
Substitute $a=x+z, c=y+z,b=-z$.
The inequalities are $$ac-b^2=xy+yz+xz\geq0\\ \ c=y+z\geq 0\\\ a=x+z\geq 0\\ x^2+y^2+z^2\leq 1$$.
The surface $xy+yz+xz=0$ consists of two cones whose vertices touch each other at the origin. Both cones are symmetric around the line $x=y=z$. All three coordinate axes lie on the cone.
The intersection of the plane $x+z=0$ with the cone $xy+yz+xz=0$ is the $y$ axis since $xy+yz+xz=(x+z)y+xz$ is zero whenever $x=z=0$. Since the plane $x+z=0$ does not slice through the interior of either cone, if one point in the cone is in the region $x+z\geq 0$ then the whole cone is in the region. As it turns out, one of the two cones is in both the region $x+z\geq 0$ and the region $y+z\geq 0$.
All that remains is to find the intersection of this cone with the sphere $x^2+y^2+z^2\leq 1$.
This will be done with a volume integral in spherical coordinates. The region to be integrated over is $r=[0,1]$, $\theta=[0,2\pi]$, $\phi=[0,\cos^{-1}(\frac{1}{\sqrt{3}})]$. The upper bound of $\cos^{-1}(\frac{1}{\sqrt{3}})$ is the angle between any coordinate axis and the line $x=y=z$.
$$\int_0^1\int_0^{2\pi}\int_0^{\cos^{-1}(\frac{1}{\sqrt{3}})}r^2\sin\phi \ \mathrm{d}\phi\mathrm{d}\theta \mathrm{d}r=\frac{2}{3} \left(1-\frac{1}{\sqrt{3}}\right)\pi$$