Help with shell method around x-axis

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rir0302

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Sep 11, 2019
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"Use the shell method to calculate the volume of rotation about the x‑axis for the area underneath the graph y=(x−4)^(1/3)−2 on [12, 31]."

First I rewrote the formula to x=(y+2)^3+4
Then I did the integral of 2pi(y)((y+2)^3+4) on [0, 1]
I got the answer (117pi)/5, but it's not correct. What did I do wrong?
 
I think you found the volume of the region bounded by that curve, the axes, and y=1 revolved about the x axis.

Check your sketch of the region, and look at the element of area you are using. How high (long) is your shell, really?
 
Thanks for your response. So I got the volume of the area above the graph instead of below? I'm confused as to how to write the equation correctly though.
 
Not above the graph -- to the left of the graph.

Have you drawn the region you are working with? Please show what you drew.
 
So where is the shell? What are its left and right ends, and how long is it?

Don't just graph the function; sketch the region. That is what I am emphasizing (and is a problem with depending on technology rather than a pencil).
 
1568586702479.png
Like this? Or would this be the volume, not the shell?
But the radius is y, and the height is the function of y, right?
I just learned about the shell method, so I'm still confused... sorry.
 
Here's a slightly easier to read diagram:

fmh_0078.png

I wuld agree that the radius of each shell is \(y\), but the height isn't \(x\)...look at the graph again...do you see what it actually is?
 
rir0302 said:
View attachment 13647
Like this? Or would this be the volume, not the shell?
But the radius is y, and the height is the function of y, right?
I just learned about the shell method, so I'm still confused... sorry.
Click to expand...
The shell will be a cylinder within the region you drew, whose side is generated by a horizontal line from the left to the right end of the region. I've drawn this on MarkFL's graph:
FMH118022.png

That's what will be revolving around the axis to form a cylindrical shell (which is like the label of a soup can).

How long is this, for a given value of y?
 
Yes. Just express x in terms of y, and stick that in your integral where you just had x itself.
 
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