I know the volume of a unit sphere is $\frac{4\pi}{3}$ (using integration) but I don't really know how to evaluate the volume of a tetrahedron with vertices: $$(0,1,1), (0,-1,1), (-\pi,0,-1), (\pi,0,-1)$$
what is the way to approach this?
I think I found an answer, but not sure about it: I can divide the tetrahedron to two equal halves. One of them with vertices: $$(-\pi,0,-1), (\pi,0,-1), (0,0,1), (0,1,1)$$
The area of one cross section of that tetrahedron (parallel to the xz plane) is $y^2\cdot 2\pi$ at height $y$.
Thus, the volume of he entire half is given by the following integral: $$\int_{0}^{1}y^2\cdot 2\pi \cdot dy=2\pi\int_{0}^{1}y^2 \cdot dy=2\pi\cdot (\frac{1^3}{3}-\frac{0^3}{3})=\frac{2\pi}{3}$$
and because of that the volume of the tetrahedron is $\frac{4\pi}{3}$
is this correct?