Cylindrical Shells

[beau1]

Use the cylindrical shell method to find the volume of the solid generated by revolving the shaded region (the area between the lines x=y^4/4 - y^2/2 and x=y^2/2) about:

a) the x-axis
b) the line y=1
c) the line y=5
d) the line y= -5/8

Sorry, I am very unfamiliar with the cylindrical shell method and do not even know where to start. I know I must integrate at some point but am having trouble setting up the integral.
Thank you for your help.

 

[pka]

Unless you post an image of the “shaded area”, how can anyone help you with this question?

 

[soroban]

Hello, beau1!

If no one is responding, it's because we don't understand the problem.

\(\displaystyle \text{Use the cylindrical shell method to find the volume of the solid generated by revolving the shaded region}\)
\(\displaystyle \left(\text{the area between the curves: }\,x\:=\:\tfrac{1}{4}y^4 -\tfrac{1}{2} y^2\,\text{ and }x\:=\:\tfrac{1}{2}y^2\right)\,\text{ about:}\)

. . \(\displaystyle \text{(a) the }x\text{-axis}\quad \text{(b) the line }y=1\quad \text{(c) the line }y=5\quad \text{(d) the line }y= -\tfrac{5}{8}\)

Although the graph is symmetric to the x-axis, I assume we are using only the upper region.

            |
            |       ............*(2,2)
           .....*:::::::::::*   :
        *:::|:::::::*           :
       :::::|:::*               :
      *:::::|:*                 :
        *:::|                   :
  - - - - - * - - - - - - - - - + -  - -
            |                   2
            |

Since we are revolving the region about a horizontal line, we use the "sideways" formula:

. . \(\displaystyle \displaystyle V \;=\; 2\pi \int^b_a \text{(radius)(height)}\,dy\)



\(\displaystyle \text{(a) about the }x\text{-axis.}\)

\(\displaystyle \text{The radius is }y.\)

\(\displaystyle \text{The height is the horizontal distance between the two curves:}\)
. . \(\displaystyle \tfrac{1}{2}x^2 - \left(\tfrac{1}{4}x^4 - \tfrac{1}{2}x^2\right) \:=\:y^2 - \tfrac{1}{4}y^4\)

\(\displaystyle \text{The volume is: }\;V \;=\;2\pi \int^2_0 y\left(y^2 - \tfrac{1}{4}y^4\right)\,dy\)



(\(\displaystyle \text{b) about }y = 1\)

\(\displaystyle \text{There is no formula for revolving about a line that passes }through\text{ the region.}\)



\(\displaystyle \text{(c) about }y = 5\)

\(\displaystyle \text{The radius is: }5-y\)

\(\displaystyle \text{The height is still: }y^2-\tfrac{1}{4}y^4\)

\(\displaystyle \text{The volume is: }\;V \;=\;2\pi\int^2_0(5-y)(y^2-\tfrac{1}{4}y^4)\,dy\)



\(\displaystyle \text{(d) about }y = -\tfrac{5}{8}\)

\(\displaystyle \text{The radius is: }y + \tfrac{5}{8}\)

\(\displaystyle \text{The height is: }y^2-\tfrac{1}{4}y^4\)

\(\displaystyle \text{The volume is: }\;V \;=\;2\pi\int^2_0\left(y + \tfrac{5}{8}\right)\left(y^2-\tfrac{1}{4}y^4\right)\,dy\)

 

[galactus]

Like Soroban, I too was unsure. But here is how I approached it.

Similar to Soroban though.

Since \(\displaystyle x=\frac{y^{4}}{4}-\frac{y^{2}}{2}, \;\ x=\frac{y^{2}}{2}\),

then we can make a sub and get:

\(\displaystyle x=\frac{y^{4}}{4}-x\)

\(\displaystyle 2x=\frac{y^{4}}{4}\)

\(\displaystyle x=\frac{y^{4}}{8}\)

Using shells. The cross sections are parallel to the axis about which we are revolving.

About the x-axis:

\(\displaystyle 2{\pi}\int_{0}^{2}y\left(\frac{y^{4}}{8}\right)dy\)

If you have the diagram on file, you can post it to the site using the 'Upload Attachments' link directly under

where you type your problem. If possible, it would hep to have a picture of what is going on.