Application of integrals - volume

[michi]

Hey all. I'm having a bit of trouble with this problem

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=1/x^3, y=0, x=4, x=6;

about y=-1

Here is what I know or in the case that I am wrong...think

I think that the equation needs to be set in terms of x instead of y since we're rotating about the y axis, so the equation will become

x= y^(-1/3)

Then, we have the function pi[y^(-1/3)]^2, correct?, and then we take the antiderivative of that which would be 3*(y)^1/3?

I'm not sure if that's correct for the function. I think, that's not how I'm supposed to go about doing it. I'm also having a lot of trouble figuring out what my boundaries are as I don't know what I'm doing. How do I figure out my inner and outer function. I think that would be the best way to set this equation up, but I don't know how as the bounds of this graph confuses me.

Help please!

 

[galactus]

You are revolving thin vertical strips which create a washers about y=-1. It has

outer radius \(\displaystyle \frac{1}{x^{3}}\) and inner radius 1.

It's like finding the area of a cylindrical ring: \(\displaystyle {\pi}R^{2}-{\pi}r^{2}\)

Only you're adding up the area of a whole lot of washers and thus arriving at the

volume over the region. Try graphing \(\displaystyle y=\frac{1}{x^{3}\) and y=-1. You

should be able to see it is a mighty small area you're revolving. Try picturing it

revolving around y=-1. It'll make it easier.

\(\displaystyle {\pi}\int_{4}^{6}((\frac{1}{x^{3}}+1)^{2}-(0-(-1))^{2})dx\)

 

[soroban]

Hello, michi!

Find the volume of the solid obtained by rotating the region about the specified axis.

\(\displaystyle \;\;y\,=\,\frac{1}{x^3},\;y\.=\.0,\;x\,=\,4,\;x\,=\,6\;\)about \(\displaystyle y\,=\,-1\)

It's obviious that you didn't make a sketch . . . well, not a good one, anyway.

          |*
          |
          | *
          |  *
          |    *
          |    |::::*
          |    |::::|       *
    ------+----+----+------------
          |    4    6
          |
    - - - + - - - - - - - -   y = -1
          |

You can use "Washers" . . .

The outer volume is; \(\displaystyle \L\:V_1\;=\;\pi\int^{\;\;\;6}_4\left(\frac{1}{x^3}\,+\,1\right)^2\,dx\)

The inner volume is: \(\displaystyle \L\: V_2\;=\;\pi\int^{\;\;\;6}_41^2\,dx \;= \;2\pi\)

The answer is: \(\displaystyle \L\:V_1\,-\,V_2\)

 

[michi]

thank you!

this made more sense to me than what I was trying to do. I understand what you guys did here. Thanks so much.