I'm trying to find the volume of the solid bounded by $z=x^2 + y^2$ and the sphere of ratio R centered at the origin. My problem is that I can't seem to find the proper limits.
My thoughts: the solid is bounded by $z=x^2 + y^2$ and $R^2=x^2 + y^2 + z^2$, so $0 \le z \le \sqrt{R^2-x^2-y^2} $. I think the best option is to use cylindrical coordinates, so I have: $x^2 + y^2=r^2=z$ and $0\le z \le \sqrt{R^2-r^2} $.
Is this right? If so, what should I do next?
Thanks.
I think cylindrical would be best too. Let's start out with the z-bound's. The solid is bounded below by the paraboloid and above by the sphere. So we need to solve for the z's in terms of r's.
$$z=x^2+y^2=r^2 $$
$$ z=\sqrt{R^2-(x^2+y^2)}=\sqrt{R^2-r^2} $$ This gives us our upper and lower bounds.
Now, we need to find the radial bounds. The radius is at its maximum when the sphere and paraboloid intersect. So we need to solve for the max $z$ and then find $r$.
$$ z^2+(x^2+y^2)=R^2 $$ $$z=x^2+y^2$$ $$ z^2+z-R^2=0 $$ We need to use the quadratic equation. Here we take only the positive solution assuming that $z\geq0$
$$ z=\frac{-1+\sqrt{1+4R^2}}{2} $$ Now we can find the radial maximum. $z=r^2$ $$ r=\sqrt{\frac{-1+\sqrt{1+4R^2}}{2}}=a $$
The theta bounds are trivial and range from $0$ to $2\pi$
Now:
$$ V=\int_0^{2\pi}\!\!\!\int_0^a\!\!\!\int_{r^2}^{\sqrt{R^2-r^2}}rdzdrd\theta $$