MHB Volume about x axis using cylindrical shells

[karush]

Use the method of cylindrical shells to find the volume
of the solid obtained by rotating the region bounded by the given curves about the x axis.

$x=1+{y}^{2}$, $x=0$, $y-1$, $y=2$

https://www.physicsforums.com/attachments/4237

The answer is $\frac{21\pi}{2}$ but I couldn't get it using $V=2\pi\int_{a}^{b} \,xf(x)dx$

 

[Prove It]

Use the method of cylindrical shells to find the volume
of the solid obtained by rotating the region bounded by the given curves about the x axis.

$x=1+{y}^{2}$, $x=0$, $y-1$, $y=2$
The answer is $\frac{21\pi}{2}$ but I couldn't get it using $V=2\pi\int_{a}^{b} \,xf(x)dx$

Washers work best in this case. Try $\displaystyle \begin{align*} V = \pi \int_0^2{ 2^2 - 1^2\,\mathrm{d}x } + \pi \int_2^5{ 2^2 - \left( \sqrt{x - 1} \right) ^2 \,\mathrm{d}x } \end{align*}$.

 

[karush]

got it, I forgot to square the equations

 

[MarkFL]

Use the method of cylindrical shells to find the volume...

The volume of an element is:

$$dV=2\pi y\left(y^2+1\right)\,dy$$

Thus, the total volume is:

$$V=2\pi\int_1^2 y^3+y\,dy$$

Computing the definite integral will give you the desired result. :D