# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y = 1 + x^2, y = 0, x = 0, x = 2 rotated about the x-axis?

Sep 26, 2015

See explanation below.

#### Explanation:

Here is the region to be revolved about the $x$ axis: If we must use shells, we will need to take our representative slices horizontally and use thickness $\mathrm{dy}$.

We need to rewrite the boundaries as functions of $y$ so the curve becomes $x = \sqrt{y - 1}$.
Note that, in the region $y$ varies from $0$ to $5$. The radius of the shells will involve $y$ values.
For the 'height' of the shell (which is lying on its side), we will use the $x$ on the right minus the $x$ on the left.

For $y = 0$ to $1$, the 'height' is $2 - 0 = 2$, The resulting solid is a cylinder with radius $1$ and height $2$. Its volume is $2 \pi$.
(We could do an integral for this, but we learned the volume of a cylinder in geometry class.)

From $y = 1$ to $y = 5$, the 'height' is $2 - \sqrt{y - 1}$.
The radius of the representative shell is $y$ and the thickness is $\mathrm{dy}$

We integrate:

$\int 2 \pi r h \mathrm{dy} = 2 \pi {\int}_{1}^{5} y \left(2 - \sqrt{y - 1}\right) \mathrm{dy} = \frac{176 \pi}{15}$

Adding the two volumes, we get:

$V = 2 \pi + \frac{176}{15} \pi = \frac{206}{15} \pi$

Note
If we had been allowed to use disks, we would integrate:

$\pi {\int}_{0}^{2} {\left(1 + {x}^{2}\right)}^{2} \mathrm{dx} = \pi {\int}_{0}^{2} \left(1 + 2 {x}^{2} + {x}^{4}\right) \mathrm{dx}$

$= \pi {\left[x + \frac{2 {x}^{3}}{3} + {x}^{5} / 5\right]}_{0}^{2}$

$= \pi \left[2 + \frac{16}{3} + \frac{32}{5}\right] = \pi \left[\frac{30 + 80 + 96}{15}\right] = \frac{206}{15} \pi$

Of course, we get the same answer, but we only need to do one calculation and the integral is (I think) simpler.