Volume of region bounded by $z=4 - \sqrt{x^2 +y^2}$ and $z=\sqrt{ x^2 +y^2}$ I have to find volume using triple integration of region bounded by $z=4 - \sqrt{x^2 +y^2}$ and $z=\sqrt{ x^2 +y^2}$. I have seen that they are cones intersecting at $z=2$, but my problem is that in $x$-$y$ plane shadow seems to be split up, which I cannot do. Please belp me with this. Thanks


The shaddow in the $xy$ plane is indeed a circle, of radius $2$, so in cartesian:
$$V = \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{4-\sqrt{x^2+y^2}} 1 \ dz\, dy\, dx$$
or in polar (easier):
$$V = \int_{0}^{2\pi}\int_{0}^2\int_{r}^{4-r} 1 \ r\, dz\, dr\, d\theta$$
By symmetry above/below the plane $z=2$ you can calculate the volume of the lower or upper portion and multiply by $2$:
$$V = 2\int_{0}^{2\pi}\int_{0}^2\int_{r}^{2} 1 \ r\, dz\, dr\, d\theta \ \ \ \ \ \text{ or } \ \ \ \ \ \ V = 2\int_{0}^{2\pi}\int_{0}^2\int_{2}^{4-r} 1 \ r\, dz\, dr\, d\theta$$
Geometrically this is the volume of two cones each with base area $B=\pi 2^2$ and height $h=2$. SO the volume should come out to be $$V=2\cdot \dfrac{1}{3}B\cdot h = \dfrac{16\pi}{3}$$
