I want to find the volume of the region above the xy-plane bounded by the two surfaces:
z = 6 - x^2 - y^2
z = sqrt(x^2 + y^2).
I know the region will look like this: https://gyazo.com/fe44150c89aa3dbf856a074cf772e606
So the region I'm working with is above z = sqrt(x^2 + y^2) and below z = 6 - x^2 - y^2.
I try to convert the expressions into cylindrical coordinates and get:
z = r and z = 6 - r^2
Then I try to find the intersection and get:
r^2 + r - 6 = 0, which yields r = 2 and r = -3.
Now I want to find the boundaries for theta, r and z so I can compute the triple integral. I'm certain the boundary for theta is 0 to 2pi. For r I believe it is 0 to 2. For z I have a hunch that it is r to 6-r^2, but I'm not sure.
Am I on the right lines, or am I going wrong somewhere?
This problem is quite tractable when we realized that these are bodies of revolution. Thus we can consider rotating these curves about the $z$-axis:
$$ z_1=6-x^2\\ z_2=x $$
These curves intersect at $x=2$. Then by the standards for bodies of revolution,
$$ \begin{align} V &=2\pi\int_0^2 x\left(z_1(x)-z_2(x)\right)~dx\\ &=2\pi\int_0^2 x(6-x^2-x)~dx\\ &=2\pi\int_0^2 (6x-x^3-x^2)~dx\\ &=2\pi \left[\frac{6x^2}{2}-\frac{x^4}{4}-\frac{x^3}{3} \right]\biggr|_0^2\\ &=\frac{32\pi}{3} \end{align} $$
I have verified this solution numerically by an independent method.