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 $du=2xdx$ and the bounds change from 0 and ${\sqrt {\pi /2}}$ to 0 and $\pi /2$.

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