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Calculus Problem (volume by washer/disk vs volume by shell)


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IM SO CONFUSED (and it doesn't help that my teacher barely knows what she is talking about).

 

I don't know when an object that is rotated around either axis is a washer or when it is a shell. If the problem says "do volume by shell" or "do volume by washer" I can get it right. I can't tell on my own though.

 

http://img297.imageshack.us/my.php?image=ahhcalculus.png

 

For example, get that in #1 it is a disk when rotated around the x axis

 

In #2 though, when you rotate that shaded part, formed by the equation x^3 and some other equation that intersects it at the origin and in the first quadrant somewhere, around the x-axis which one is it? What about around the y-axis? I think I was told that around the x axis it is a washer, and around the y it is a shell. WHY?

 

To me a washer is a shape like #3, where it has an actual length before the inside part is missing (shown where I drew the red lines). However, in #2 when you rotated it around the x axis, it looks more like a cone in that from the circular edge when you move in it immediately starts to slope down. How can that be a washer? And how is it different when it is rotated around the y axis?

 

Does it have something to do with dy or dx changing parallel or perpendicular to the axis you are rotating it around?

 

I'm so confused and I've been grappling with this all day now and I don't get it.

Edited by blackhole123
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"Washer" or "shell" isn't the shape of the volume; it's the name of the method you can use to find the volume.

 

With the washer method it's easy to find the volume of something rotated around the x axis, assuming it's defined in terms of x. If you have a function f(x), its area is:

 

[math]\mbox{Area} = \pi \int_a^b (f(x))^2 dx[/math]

 

What you're doing is moving from left to right (a to b) and finding the area of an infinitely thin circle at each location. The radius of that circle is f(x), since that's how far the line is from the graph, so you'll see the integral is basically of [math]\pi r^2[/math]. Make sense?

 

The shell method can easily find the area of a volume rotated around the y axis when it's in terms of x. That volume can be represented by infinitely thin "shells" going around the y axis (google it and there are some images showing this). I'm not as familiar with the shell method, so I won't try to explain too much.

 

Of course, you can use each method to integrate around the other axis if you really want to. It's just a matter of doing some extra algebra.

 

Make sense?

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Wow thank you so much, that clears it up so much!

 

I was under the impression it was the shape of the volume, but now that I see it is actually just the method it makes so much more sense.

 

So around the x axis in terms of x it is easiest to use the washer method.

 

And around the y axis in terms of x it is easiest to use the shell method.

 

Sooo.... would it be opposite of that when it is in terms of y? Because most of the time I am given "y=..."

I understand I can solve for x, but I'm sure there is an easier way.

 

EDIT: Oh wait, if it is something like y=x^2 it IS in terms of x right? Woops

 

EDIT 2: Wewt, just did started my homework and I GET IT. Checked the answers that I have gotten so far and its actually working. Before I could kind of do it if my hand was held but now I can do it and understand why! Thanks again! Comprehension is a beautiful thing!!!!

 

EDIT 3: Could you explain what they are saying in my book about the different methods being useful for whether the "natural intergral is perpendicular or parallel to the axis of rotation"?

Edited by blackhole123
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  • 2 weeks later...

**I'm not sure if I' too late for this discussion, but I'd like to add**.

 

I concur with Cap't Refsmmat, the washer and shell isn't a shape of a volume but are names of two methods of determining a volume.

 

Although, I'd like to revise the statement of;

" The shell method can easily find the area of a volume rotated around the y axis when it's in terms of x. "

 

Both the washer and shell methods use rotation around the same axis regardless of taking the rotation with respect to x or y axes.

Yes, in your case, the two methods, integrating with respect to x along the x-axis with give a washer volume integration, while integrating along the y-axis (but bear in mind that rotation is still taken around the x-axis, an integration of an infinite amount vertical lines making up the bounded region multiplied by a circumferential distance [math]\bar{y}[/math] will give you the shell method of volume integration.

 

Both methods yield the same answer if you algebraically implement correctly and integrated with correct limits, you’ll be fine.

 

Click on link, I did a quick search on Google also. This will explain more clearly.

http://learning.mgccc.cc.ms.us/math/mathdocs/calc/revol2.pdf

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