User:Editeur24/diskintegration

Disc integration

Consider the horn shape created by rotating the region bounded by the   line x = 0, the line $$x = 2$$, and the curve $$y =  (x+1)^2 $$  in the x-y-plane, around the x-axis into the z-dimension. We can divide this shape into disks. The area of each disk is   $$ \pi \cdot radius^2$$, which is  $$ \pi y^2$$ in this context. The thickness of the disk is dx, so in three  dimensions the volume of the disk is  $$ \pi   [(x+1)^2]^2 dx$$. Adding up all the disks as x changes, we come out with
 * $$   volume = \int_0^2   \pi(x+1)^4  dx =  \Big|_0^2  \pi(x+1)^5 = \pi (3^5 -1)= \frac{242}{5}   $$

where this integral has been solved by the method of substitution setting $$u = x^2$$ so $$du = 2x dx$$ and the bounds change from 0 and $$\sqrt{\pi/2}$$ to 0 and $$\pi/2$$. -