Volume of Revolution of a Skew Line Segment

The segment can be parametrized as $(x,y,z)=t(-\frac{5}{2},\frac{5\sqrt{3}}{2},15)+(10,0,0)$ , then

$\displaystyle s(t)=(-\frac{25}{2}t+10,\frac{5\sqrt{3}}{2}t,15t)$ , $t\in\mathbb{R}$ , $t\in[0,1]$ . The volume of a solid of revolution can be considered as the sum of the infinite-small thick disks the solid is cut in. The volume of a disk is $\pi R^2 h$ , where $h$ is the hight of the disk. In this case, the radius of the disk is the distance from a generical point of $s(t)$ to the $z$ -axis. It can be written as

$\displaystyle dist(s(t),z)= \sqrt{(10-\frac{25}{2}t)^2+\frac{75}{4}t^2}$ .

So, the infinitesimal volume of the solid is

$\displaystyle dV= dist(s(t),z)^2\pi dz(t)$

eventually,

$\displaystyle V=\pi\int_{0}^{1} (10-\frac{25}{2}t)^2+\frac{75}{4}t^2 dz(t)=15\pi\int_{0}^{1} (10-\frac{25}{2}t)^2+\frac{75}{4}t^2dz=500\pi$