# Volume integral set up

1. May 23, 2013

### 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

2. May 23, 2013

### Staff: Mentor

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...

3. May 23, 2013

### 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

4. May 23, 2013

### Staff: Mentor

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

5. May 23, 2013

### 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

6. May 23, 2013

### Staff: Mentor

Yes and yes.
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.
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 $\pi~ r^2~\Delta y = \pi~(3 - y^2)^2\Delta y$.

Last edited: May 23, 2013
7. May 23, 2013

### 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

8. May 23, 2013

### Staff: Mentor

It's fixed now.
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?
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.

9. May 23, 2013

### 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.

10. May 24, 2013

### Staff: Mentor

Carefully read what I wrote at the end of post #6 and in the middle of post #8. I explained why in both posts.
In the graph for part c of the solutions, what expression represents the length (horizontal) of the thin rectangle in the picture?