Volumes by slicing other than around x or y axis

[hank]

I can do the volumes of slicing with any method around an x or y axis, that's not my main problem.

My problem is what happens when the solid revolves around something other than the x or y axis.

For example, what if the solid revolved around x = 3, or y = -1?

Can someone explain to me how to compensate for that, or point me to a webpage with the information?

 

[skeeter]

easier to explain if you post a specific volume problem where the axis of rotation is not one of the coordinate axes.

 

[hank]

Gotcha.

Example problem:

x=y^2, x=y, y=-1.
Rotate about the y= -1 axis.

Thanks

 

[galactus]

Shells:

\(\displaystyle \L\\2{\pi}\int_{-1}^{0}(y+1)(y^{2}-y)dy\)

Washers:

\(\displaystyle \L\\{\pi}\int_{0}^{1}[(\sqrt{x}+1)^{2}-(x+1)^{2}]dx\)

 

[hank]

Right, but how did you come up with the +1's?

Is that just the distance from wherever the axis is? Does it not matter that y=-1?

If it was y = -2 billion, would it have been the same answer?

 

[skeeter]

well, you sure as **** didn't pick an easy one, did you?

the region described is in quads III and IV ...

in quad III, y = x is above and y = -1 is below
in quad IV, x = y² (or y = -sqrt(x)) is above and y = -1 is below

two ways to do this ...

using disks, you'll have to break it up into two integrals. The radius of the first part is the distance from y = x to y = -1, which is [x - (-1)] = (x + 1). The radius of the second part is the the distance between the lower limb of the parabola y = -sqrt(x) and the line y = -1, which is [-sqrt(x) - (-1)] = [1 - sqrt(x)]. the resulting integral is not pretty, but it will do the job ...
\(\displaystyle \L V = \pi \int_{-1}^0 (x + 1)^2 dx + \pi \int_0^1 (1 -\sqrt{x})^2 dx\)


the easier method (in my opinion) is using cylindrical shells w/r to y.

height of a representative shell (a horizontal strip) is y² - y (right curve - left curve)
thickness of the shell is dy
radius of rotation for each shell is y - (-1) = y + 1

\(\displaystyle \L V = 2\pi \int_{-1}^0 (y+1)(y^2 - y) dy\)

both integrals result in a volume of \(\displaystyle \L \frac{\pi}{2}\).

 

[hank]

"radius of rotation for each shell is y - (-1) = y + 1 "

Ok, this is my question right here.

Does this formula hold true no matter the value for the axis?
So if we would have had to rotate about the y = -2 billion axis, would this have changed to:

y - (- 2 billion) = y + 2 billion

?

 

[skeeter]

correct ... draw a representative horizontal "strip" in the region.

in this case, the radius of rotation is the distance from the strip to the line y = -1 ...

the position of the strip is "y", and its distance from the horizontal line y = -1 is y - (-1) = y + 1. so ... if the axis of rotation was y = -whatever, then r = y - (-whatever) = y + whatever.

finally, please note that we're talking "strips" above the axis of rotation.

if the strip were below the axis of rotation, r = whatever - y.

 

[hank]

I assume the same thing goes for the x-axis?
So if you move left, it's x - ( -a) and to the right it's x - ( a) ?