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#?

Question

Charles Arocha · Accepted Answer

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 >= 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 = 2x^2# as-is since now we're revolving around #y = 0#. The volume is written like this:

#V = sum_a^b pi(r(x))^2Deltax#

...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(x) = y = 2x^2# bounded below by the x-axis. Your horizontal bounds are of course said to be #[0,2]#. Therefore, you just have:

#V = int_0^2 pi(2x^2)^2dx#

#= piint_0^2 4x^4dx#

#= pi[4/5x^5]_(0)^(2)#

#= pi[(4/5 (2)^5) - cancel(4/5 (0)^5)]#

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

Scarlett Allison · Answer

undefined To find the volume of the solid generated by revolving the region bounded by $y = 2x^2$, $y = 0$, $x = 2$, and revolving on $y = 8$, you can use the method of cylindrical shells.

The formula to find the volume using cylindrical shells is:

$$V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx$$

where $h(x)$ is the height of the shell and $a$ and $b$ are the bounds of integration.

In this case, $h(x)$ is the distance between $y = 8$ and the curve $y = 2x^2$, which is $8 - 2x^2$. The bounds of integration are from $x = 0$ to $x = 2$.

$$V = 2\pi \int_{0}^{2} x \cdot (8 - 2x^2) \, dx$$

Now, integrate the function $x \cdot (8 - 2x^2)$ with respect to $x$ from 0 to 2, then multiply the result by $2\pi$ to get the volume.