# How do you find the volume of the solid generated by revolving the region bounded by y=2x^2, y=0, x=2, revolving on y=8?

Aug 19, 2015

It looks like this:

graph{(y-2x^2)(y-8)(x-2)(sqrt(x)/sqrtx) <= 0 [-0.5, 5, -0.35, 8.0]}

I'm assuming it's also in the domain of $x \ge 0$, which was not stated. (This will make the volume finite.)

Let's make it easier to work with though; due to the symmetry of the revolved solid, if you just take the portion of the revolved solid that would be above the line $y = 8$ and shift it down 16 units to touch the bottom edge of the portion below $y = 8$, the solid will change from a circular-base bowl to a curved "cone".

Different solid---same volume.

And no need to do any subtractions to specify a range for the graph---we can just use $y = 2 {x}^{2}$ as-is since now we're revolving around $y = 0$. The volume is written like this:

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

...since you are essentially stacking "circles" with radii that vary according to a function to generate a solid. These "circles" are perpendicular to the revolution axis. The "circle" is defined by $r \left(x\right) = y = 2 {x}^{2}$ bounded below by the x-axis. Your horizontal bounds are of course said to be $\left[0 , 2\right]$. Therefore, you just have:

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

$= \pi {\int}_{0}^{2} 4 {x}^{4} \mathrm{dx}$

$= \pi {\left[\frac{4}{5} {x}^{5}\right]}_{0}^{2}$

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

color(blue)(= (128pi)/5 "u"^3) ~~ 26.808 "u"^3