# 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=6x^2, y=6sqrtx rotated about the y-axis?

Jul 5, 2015

The normal revolution method calls for:

$V = \pi {r}^{2} h = {\sum}_{a = {a}_{0}}^{b} \pi {\left(r \left(x\right)\right)}^{2} \Delta x = \pi {\int}_{a}^{b} {\left(r \left(x\right)\right)}^{2} \mathrm{dx}$

where one stacks circle analogs of varying radius $r \left(x\right)$ across an orthogonal interval $\Delta x$.

In contrast, the shell method calls for a volume formula as such:

$V = {\int}_{a}^{b} 2 \pi x r \left(x\right) \mathrm{dx}$

where the way $r \left(x\right)$ varies determines the height and shape of the revolved solid. Here, $\Delta x$ is instead the thickness of the shell and $x$ is its radius. This is sometimes easier, especially if you are rotating about the y-axis instead of the x-axis.

Let's see how this looks.

$y = 6 {x}^{2}$:
graph{6x^2 [-2, 2, -1, 1]}

$y = 6 \sqrt{x}$:
graph{6sqrtx [-2, 2, -2, 6]}

If you layer these graphs on top of each other, you can see that they intersect to form a "stretched lemon" of sorts. Let's find where they intersect to determine our $\Delta x$.

Besides $x = 0$:

$6 {x}^{2} = 6 \sqrt{x}$
${x}^{2} = \sqrt{x}$
${x}^{4} = x$
${x}^{3} = 1$
$x = 1$

Thus, the practical interval is $\left[0 , 1\right]$, as you can see here.

So, taking the area between the two curves as the difference between the top $\left(6 \sqrt{x}\right)$ and bottom $\left(6 {x}^{2}\right)$ curves and using it as $r \left(x\right)$:

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

$= 2 \pi {\int}_{0}^{1} x \left[6 \sqrt{x} - 6 {x}^{2}\right] \mathrm{dx}$

$= 12 \pi {\int}_{0}^{1} x \left[\sqrt{x} - {x}^{2}\right] \mathrm{dx}$

$= 12 \pi {\int}_{0}^{1} {x}^{\frac{3}{2}} - {x}^{3} \mathrm{dx}$

$= 12 \pi \left[\frac{2}{5} {x}^{\frac{5}{2}} - {x}^{4} / 4\right] {|}_{0}^{1}$

$= 12 \pi \left[\left(\frac{2}{5} {\left(1\right)}^{\frac{5}{2}} - {\left(1\right)}^{4} / 4\right) - \left(\frac{2}{5} {\left(0\right)}^{\frac{5}{2}} - {\left(0\right)}^{4} / 4\right)\right]$

$= 12 \pi \left(\frac{2}{5} - \frac{1}{4}\right)$

$= 12 \pi \left(\frac{8}{20} - \frac{5}{20}\right)$

$= 12 \pi \left(\frac{3}{20}\right)$

$= \frac{36}{20} \pi$

$= \textcolor{b l u e}{\frac{9}{5} \pi \approx 5.6549 {\text{u}}^{3}}$