shell method

• Feb 24th 2011, 01:07 PM
quantoembryo
shell method
I'm having trouble with finding the volume of y=x^2 from x=0, to x=1, revolving about the x axis. I'm having trouble visualizing it with this method. I can execute it with other methods, but not shell.
• Feb 24th 2011, 02:16 PM
Quote:

Originally Posted by quantoembryo
I'm having trouble with finding the volume of y=x^2 from x=0, to x=1, revolving about the x axis. I'm having trouble visualizing it with this method. I can execute it with other methods, but not shell.

Using the washer method, it's easy to visualise them by comparing them to the rings of Saturn.

For the shells, pick a point on the y-axis at about 0.8
and draw a horizontal line until it touches the curve above the positive x-axis.
Rotate this line about the x-axis.
It traces out a cylinder.
You want the curved surface area of this cylinder (the surface area of the shell).
Now imagine doing the same for all the other horizontal lines from y=0 to y=1.
Integrate all the surface areas to get the volume (like layers of an onion).
• Feb 24th 2011, 02:48 PM
quantoembryo
When I envision what you say, it comes up wrong. The shell will have dimensions 2pi*f(x), x, and dxm I get dV=2pi*x^3dx which can't be right. Help?
• Feb 24th 2011, 03:16 PM
Quote:

Originally Posted by quantoembryo
When I envision what you say, it comes up wrong. The shell will have dimensions 2pi*f(x), x, and dx.

I get dV=2pi*x^3dx which can't be right. Help?

Looking more closely at the cylinders.
The "thickness" of the cylinders.... is it "dy" or "dx".
Over which axis do we integrate?
• Feb 24th 2011, 03:37 PM
quantoembryo
I don't know what my issues are with this.... I can now see that it is dy, but that's about it. I keep ending up with 2pi*x^3 and I know that's wrong. My textbook gives one example and I just don't get it..
• Feb 24th 2011, 03:55 PM
Quote:

Originally Posted by quantoembryo
I don't know what my issues are with this.... I can now see that it is dy, but that's about it. I keep ending up with 2pi*x^3 and I know that's wrong. My textbook gives one example and I just don't get it..

Yes, you are integrating over the vertical axis.
Hence you need f(y) instead of f(x).

$y=f(x)=x^2\Rightarrow\ x=\sqrt{y}$

This gives the cylinder heights.

$2{\pi}rh=2{\pi}yx=2{\pi}y\sqrt{y}$
• Feb 24th 2011, 04:17 PM
quantoembryo
Alright, I integrated that and ended up with 4pi/5, however, I did it via splices and got pi/5 which I know is right
• Feb 24th 2011, 04:47 PM
Quote:

Originally Posted by quantoembryo
When I envision what you say, it comes up wrong. The shell will have dimensions 2pi*f(x), x, and dxm I get dV=2pi*x^3dx which can't be right. Help?

Yes, you've calculated the VOR for the region above the curve.

You need the region below the curve.
Hence, the height of each cylinder is 1-x.
• Feb 24th 2011, 07:11 PM
quantoembryo
Thanks a lot for the help. However, my final concern is I don't see why the height is 1-x rather than just x, even from your diagram
• Feb 25th 2011, 03:30 AM
Quote:

Originally Posted by quantoembryo
Thanks a lot for the help. However, my final concern is I don't see why the height is 1-x rather than just x, even from your diagram

All of the cylinders have to rest against the line x=1.
This is the case when we are calculating the VOR of the region
between the curve and horizontal axis using shells.
I've included 2 such cylinders, though there are "infinitely" many.
Therefore we need to subtract x from 1 to get the cylinder heights,
since the lines of length 1-x are being rotated around the horizontal axis.

Hope this helps.

Remember, these cylinders are "resting on their sides", not standing vertically.
• Feb 25th 2011, 05:51 AM
quantoembryo
Yes, I finally am understanding. Thanks for your patience!
• Feb 25th 2011, 06:09 AM