Find the volume of the solid bounded by $x=\sqrt{y^2+z^2}$ and $x=6-y^2-z^2$ We are given an image:
The left hand cone is $x=\sqrt{y^2+z^2}$, and right hand parabola is $x=6-y^2-z^2$

I see that I can fix $x$. $\sqrt{y^2+z^2}\leq x \leq 6-y^2-z^2$
Should I use cylindrical or spherical coordinates here? Or none? It is not obvious to me if I should be or not. Because the projection on the $yz$ plane has square roots in it, and so I think I'm supposed to be using cylindrical or spherical.
Use a triple integral
 A: Consider when $z = 0$. Then in the $xy$-plane we have a straight line $y = x$ and a parabola $x = 6-y^2$ or $y = \sqrt{6-x}$. These two meet at the point whose $x$ coordinate satisfies
$$ x^2 + x - 6 = 0 \,\, \Rightarrow \,\, x=2$$
(we discard $x = -3$ due to the picture's suggestion that $x>0$).
So we need to revolve $y = x$ for $x \in [0,2]$ and $y=\sqrt{6-x}$ for $x \in [2,6]$ around the $x$ axis for $2\pi$ radians. Thus, by the volume of revolution formula, our answer should be
$$ V = \pi \int_{0}^{2} x^{2} \, dx + \pi \int_{2}^{6} 6-x \, dx $$
A: Your body is a volume of revolution. It looks like it will be easier to compute the volume using the cylindrical-shell method than the washer method Shai proposes.
The cylinder of radius $r\in(0,2]$ has length $(6-r^2)-r$, so you end up simply integrating a single polynomial.
A: Intuition tells you to use cylindrical coordinates, as both surfaces relate $x$ to $y^2+z^2$. So the set up is
$$ x = x, \ y = r\cos \phi, \ z = r\sin\phi $$
Then your problem reduces to
$$ r < x < 6-r^2, \ 0 < r < 2, \ 0 < \phi < 2\pi $$
Then you can place the integral limits as above.

The only non-trivial limit is the upper bound for the radius. This requires you to find the intersection curve (a circle)
$$ r - (6-r^2) = (r-2)(r+3) = 0 $$
Since $r > 0$, we deduce that the intersection must be $r=2$
A: Rotate the figure and place on $z$ axis. The new question will be the volume of the solid bounded by $z=\sqrt{x^2+y^2}$ and $z=6-x^2+y^2$. Use cylindrical co-ordinate system. Equate $z=r=6-r^2$. Find $r$. 
$$r:0 \to 2 $$
$$\theta:0\to 2\pi$$
$$z:r\to 6-r^2$$
$$dxdydz=rdrd\theta dz$$. I think this logic reduce your complications.
