Area under curve

C

Chris686

New member
Joined
Sep 1, 2012
Messages
19
I'm trying to set up a problem. I don't see any similar examples in my book, but I believe if I see the problem configured, I could deduce how to put together other problems like it. Here it is:
Find the volume of the solid that results when the region enclosed by y=sqrt(x), y=0, and x=9, when revolved about the line x=9.

Also, I believe the book wants to use the washer method because it's a different chapter for the shell method.
 
Chris686 said:
I'm trying to set up a problem. I don't see any similar examples in my book, but I believe if I see the problem configured, I could deduce how to put together other problems like it. Here it is:
Find the volume of the solid that results when the region enclosed by y=sqrt(x), y=0, and x=9, when revolved about the line x=9.

Also, I believe the book wants to use the washer method because it's a different chapter for the shell method.
Click to expand...

In these types of problems - it is always advised that draw an approximate sketch.

Shade the area under consideration.

Find the point of intersection between y = √x and x = 9 and x = 0. These will give you the limits of integration.

Since you will be rotating around x= 9 - draw your elemental areas appropriately.

Then if you are still stuck - come back and tell us where you are stuck and why....

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
 
First, have you sketched the region to be revolved, along with the axis of rotation?

The easy part of these solids of revolution problems is evaluating the resulting definite integral representing the volume, so it is imperative that you are able to set these up yourself.

I begin by sketching the area and the axis of revolution. This makes it easier to see what we are dealing with. Then, for whatever method I am told to use, or choose to use, I compute the volume of an individual arbitrary part, be it a washer, disk or shell. Finally I add up the parts by integrating, taking care to choose the appropriate limits of integration.

Once you have made the sketch, tell us what you decide is an appropriate differential for the volume.
 
Okay then, allow me to clarify. I have graphed it, or what I think is it. That's as far as I can get.

I went to plug in the equation, though I have no examples to go by.

From 0-3 pi(y^2-9^2)dy
 
Chris686 said:
Okay then, allow me to clarify. I have graphed it, or what I think is it. That's as far as I can get.

I went to plug in the equation, though I have no examples to go by.

From 0-3 pi(y^2-9^2)dy
Click to expand...

I see that you understand you have to integrate along y.

What was the length of your elemental area at an arbitrary height 'y'?

That should be from x = 9 to the curve y = √x. The length should be (9-y2).

Now continue...
 
HOW did you get that? It is not even clear what problem you are doing. You said "I believe the book wants to use the washer method because it's a different chapter for the shell method." If the problem is to rotate around the line x= 9, then, no, there is no "hole" in the center and no "washer". If you are rotating around x= 0, then the washer method would be appropriate. (Not to mention the fact that you titled this "Area under curve".)

If you are rotating around the line x= 9, then each "cross section" would be a circle with radius equal to the distance from x= 9 to y= sqrt(x) so x= y^2. That distance is 9- y^2. The area of a circle with radius 9- y^2 is pi(9- y^2)^2= pi(81- 18y^2+ y^4). That is what you want to integrate.
 
Last edited by a moderator:
You must log in or register to reply here.
Top