let $D$ = {$(x,0,z) | (x-1)^2 + z^2 \leq 1$} find the volume of the body obtained by revolving $D$ about the $ Z $ axis.
how do i solve this with integrals ( triple / double ) .
intuitive solution ( might be wrong ) is to find the area of $ D $ on the $XZ$ plane which is $\pi$ and to sum the disk around the circule {$ x^2 + y^2 = 1$} right ? so my guess the answer is $ 2\pi ^2 $ .
not sure how to solve it
The volume on revolving $D$ is $8$ times its volume $V$ in the first octant, due to symmetry.
The equation of the surface generated is $(r-1)^2+z^2=1$, where $r=\sqrt{x^2+y^2}$. The centre moves along the circle $r=1,z=0$.
The required integral is $8\displaystyle\int_0^{\pi/2}\int_0^2\sqrt{1-(r-1)^2}\ rdr\ d\theta=2\pi^2$.
Note that you could have arrived at this intuitively. The area of the cross-section is $\pi$. Multiply that with the 'length' of the torus, which is given by $2\pi r_{avg}=2\pi$ and you get $2\pi^2$.