Volume Integral Set Up for Rotating Objects - Calculus Problem Explained

[Jbreezy]

hi

http://www.calcchat.com/book/Calculus-ETF-5e/
part b they want you to rotate it about the y-axis and in part c about the line x = 3.
I don't understand this difference in writing for part b... 3^2 - (y^2)^2

And in part c they write (3-y)^2 I don't get it.
It is chapter 7 section 2 question 11.
Thanks

 

[Mark44]

hi

http://www.calcchat.com/book/Calculus-ETF-5e/
part b they want you to rotate it about the y-axis and in part c about the line x = 3.
I don't understand this difference in writing for part b... 3^2 - (y^2)^2

And in part c they write (3-y)^2 I don't get it.
It is chapter 7 section 2 question 11.
Thanks

What are they using for typical volume elements in each case? If you can answer that, you'll be on your way to answering your questions.

BTW, do not delete the homework template when you post a problem. It's there for a reason. Fair warning...

 

[Jbreezy]

They are using a disk? That is what you mean by volume element? So, there calculating the radius? But I don't understand why 3^2 - (y^2)^2 for rotation around x-axis and (3-y)^2 for rotation about the line x = 3.
I feel like nothing has changed in the two but yet it is written differently.

I'm not arguing the template is no big deal but why can't I delete it? I don't like to use it. I will if it is a issue but just asking why?
thx

 

[Mark44]

They are using a disk? That is what you mean by volume element? So, there calculating the radius? But I don't understand why 3^2 - (y^2)^2 for rotation around x-axis and (3-y)^2 for rotation about the line x = 3.

I feel like nothing has changed in the two but yet it is written differently.

Of course something has changed. In the first problem (b part), the region is being rotated around the y-axis. In the other problem (c part), the region is being rotated around the line x = 3. The solids of revolution for these problems look very different, and the typical volume elements are different.

For each problem, sketch the solid that is produced when the region is rotated around the relevant axis, and sketch the typical volume element (disk/washer/shell). You're probably not doing this, which is why you are asking the questions.

I'm not arguing the template is no big deal but why can't I delete it? I don't like to use it. I will if it is a issue but just asking why?
thx

It's part of the rules, which you agreed to abide by when you joined. See the section titled "Homework Help".

NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums.

 

[Jbreezy]

OK, I drew each one out. I guess I"m still having a hard time deciding which solid to use. Like for part b they used the washer method. For part c they use the disk method? Why do they have R(y) = ( 3 - y^2) and r(y) = 0?
I don't understand why they even mention r(y) = 0 if your using the disk method.
When you write ( 3 - y^2) you are saying from the line x = 3 back to the origin?
THx

 

[Mark44]

OK, I drew each one out. I guess I"m still having a hard time deciding which solid to use. Like for part b they used the washer method. For part c they use the disk method?

Yes and yes.

Why do they have R(y) = ( 3 - y^2) and r(y) = 0?
I don't understand why they even mention r(y) = 0 if your using the disk method.

They seem to be trying to do things in a more generic fashion, considering that a disk is a washer that has a hole with radius 0.

When you write ( 3 - y^2) you are saying from the line x = 3 back to the origin?

No, back to the curve. The disk here has a radius of 3 - y², so the volume of the typical volume element (a disk) is $\pi~ r^2~\Delta y = \pi~(3 - y^2)^2\Delta y$.

 

[Jbreezy]

You forgot to square the whole thing:)
I guess I don't understand why the equation is 3 - y^2 maybe that is silly questions but I don't.
It is really the set up I'm having isssues with it is no problem once i get it set up.
Thanks

 

[Mark44]

You forgot to square the whole thing:)

It's fixed now.

I guess I don't understand why the equation is 3 - y^2 maybe that is silly questions but I don't.

That's not an equation - an equation has '=' in it.

To get the volume of the disk typical element, you need the radius of this disk and its thickness. The radius R is 3 - y², so the volume of a typical disk is $\pi R^2~\Delta y$.

Have you drawn a sketch of the solid that is formed and a typical disk?

It is really the set up I'm having isssues with it is no problem once i get it set up.

That's usually the case. Being able to analyze the problem and figure out what the integrand is going to be is often much more difficult that actually doing the integration.

That's why it's so important to draw some pictures. In my experience, lots of students think they can save time or effort by not bothering to draw some good pictures, but whatever time they save is outweighed by the time it takes for them to figure out where they went wrong.

 

[Jbreezy]

OK, forget the integral ha. Why is the expression 3-y^2
That is what I want to know.
Thank you. Yeah I have numerous drawings I always draw if I can.

 

[Mark44]

OK, forget the integral ha. Why is the expression 3-y^2
That is what I want to know.

Carefully read what I wrote at the end of post #6 and in the middle of post #8. I explained why in both posts.

Thank you. Yeah I have numerous drawings I always draw if I can.

In the graph for part c of the solutions, what expression represents the length (horizontal) of the thin rectangle in the picture?