i need to find the volume of the region bounded by

y = 4x^2 and y = 16 rotated about the line y = 16

do i use disk method? V = pi integral(R(y))^2 dy?

sorry tried to make as easy as possible to understand

if anyone can clarify to me some sort of distinct way to differentiate between using these 3 methods please lmk :X

This post was edited by Duckling on Oct 4 2013 06:02pm

Did i set this integral up correct?

volume between (about the x-axis)

y = 1/x
y = x^2
x = 2

V = pi integral [0,2] [(1/x)^2 - (x^2)^2] dx

?
isnt this the washer method?

This post was edited by Duckling on Oct 4 2013 06:18pm

The volume of a disk is given by the formula:

volume of disk = surface area of disk (well I mean the area of the circle on one side of the disk) * thickness of disk.
the surface area is just a circle and its area is pi*r^2 so restating:
volume of disk = pi * r^2 * thickness

In my drawing the green lines are the radius of some of the possible disks. The length of any given green line is 16 - 4x^2, if this is not obvious, you need to think about it, it is really key to setting it up (imo). So if we substitute that for r in our equation:
volume of disk = pi *(16 - 4x^2)^2 * thickness

the thickness of a disk is just dx, so now we have:

volume of disk = pi *(16 - 4x^2)^2 * dx

We want the sum of the volume of every possible disk between x = -2 and x = 2, that is:

Looks a little scarey but you can expand it and break it into simple pieces.

I realize this is not exactly what you were asking for. But I hope showing you my thought processes on how I would do the problem might help you.

This post was edited by Azrad on Oct 4 2013 07:02pm

this is the best way

disk method is a good idea. easiest way imo for this one. so the formula is (pi)r^2*y (dy)

So you would have the integral from 0 to 16 of (pi)*(4x^2)^2(16-y) dy

multiply out. take the integral and then obv use your bounds from 0 to 16

This post was edited by tidus_ffx2 on Oct 4 2013 07:42pm

If you want to use the shell method (I call it the "skinny castle wall method").


the volume of a thin castle wall (shell) is:
volume of shell = length of wall * height of wall * thickness of wall

length of the wall: is the radius it is being rotated around (for lack of a better phrase). The circumference of this circle we are rotating around is 2*pi*r. r is simply 16-y. Which gives us:
length of wall = 2*pi*(16 - y)

height of the wall: is the green lines in the drawing. Height of the wall = x. Now of course we need to change that to y, so we have to solve for x:
y = 4x^2
(1/4) y = x^2
x = +/- (1/2)*y^(1/2)
We will discard the negative solution and only work in the first quadrant (this is why i only drew the green lines in the first quadrant), and to correct this later we will multiple our result by 2. So
Height of wall = (1/2)*y^(1/2)

Thickness of wall: is easy, it is just:
Thickness of wall = dy

So putting that all together:
volume of shell = 2*pi*(16 - y) * (1/2)*y^(1/2) * dy

We want the sum of all possible shells from 0 to 16 so (and the 2 outside the following integral is to fix our little problem with the height of the wall above):



This expression and the integral from the first method have the same value. I think the disk method was easier in this problem, but flavor to taste!

This post was edited by Azrad on Oct 5 2013 12:31am

i do have a hard time wrapping my head around the "sheet idea" or pulling out a sheet to find the thickness
when my prof goes over it in class i feel like full retard mode

anyway i believe i solved that first 1 but post #2 i could still use help on my bounds

I don't have much time right now, hopefully this helps:



To find the value of a:

x^2 = 1/x

x = 1 (therefor a is one).

/e oops made a mistake and I'm out of time, I think what is below is correct:

the integral is from 1 to 2 of:

pi*(x^2)^2*dx - pi*(1/x)^2*dx





This post was edited by Azrad on Oct 5 2013 02:03pm