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 x=y^2, y=0, and y=sqr2 rotated about the x axis?

Jul 28, 2015

It's unclear what your solid is, but the only one that makes sense is $y = \sqrt{2}$ and not $y = 0$ because $y = 0$ is redundant when you are rotating about the x-axis. Maybe you meant to emphasize that it is $y = \sqrt{x}$ and not $y = - \sqrt{x}$. Assuming that:

$x = {y}^{2} \to y = \sqrt{x}$
You only need the positive one since $y = 0$ is a boundary, and is also your axis of rotation. Thus, you are looking at this:

Since the shell method implies you are rotating about the y-axis (which is inconvenient for the regular revolution method), we can rewrite this for that.

$x = {y}^{2} \to y = {x}^{2} \in \left[0 , \infty\right)$ (only positive $x$)
$y = 0 \to x = 0$
$y = \sqrt{2} \to x = \sqrt{2}$

Now we really just have $y = {x}^{2}$ in the first quadrant rotated around the y-axis, stopped at $x = \sqrt{2}$, intersecting at $\left(\sqrt{2} , 2\right)$. Which is:

The shell method is:

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

where:
$2 \pi$ is the radian measure for the "circumference" of the shape
$x$ is the radius of the shell
$f \left(x\right)$ is the height of the shell

$V = 2 \pi {\int}_{0}^{\sqrt{2}} x \left({x}^{2}\right) \mathrm{dx}$

$= 2 \pi \left({x}^{4} / 4\right) {|}_{0}^{\sqrt{2}}$

$= 2 \pi \left({\left(\sqrt{2}\right)}^{4} / 4 - {0}^{4} / 4\right)$

$= 2 \pi \left(1\right)$

$= \textcolor{b l u e}{2 \pi \text{rad}}$