Calc2: Solid of revolution about y=2 question

[USN2ENG]

Hi All, I had a question about this problem I was doing and was hoping to see if I did it right.

Homework Statement

Using Disk or Washer method Find volume of the solid generated by revolving the region bounded by y=sqrt(x), y=2, x=0 about the line y=2.

Homework Equations

Disk Method: pi*Integral (R(x))^2dx where R(x) is the distance from the axis of revolution to the regions boundaries.

The Attempt at a Solution

Limits of integration are [0,4]

I was getting caught up with what R(x) is. Since I am revolving up around y=2 I am wondering if the distance is 2-sqrt(x) or sqrt(x)-2

Doing the problem with R(x)=sqrt(x)-2 I get:

8pi/3

Doing the problem with R(x) = 2-sqrt(x) I get:

8pi/3 again...

So what I am wondering is, in the future, which one is the correct R(x)? or does it not matter? I feel like it should be 2-sqrt(x) because that more clearly sticks out to me as the radius but I am not sure.

 

[lineintegral1]

Think of it this way: if you simply shifted the graph down by a 2 units, then you could rotate it about the x-axis as you always have. To do so, of course, you would take \sqrt{x}-2 and rotate it about the x-axis. I think the typical approach, however, is take the radius to be 2 (like you said) and simply analyze 2-\sqrt{x}.

 

[USN2ENG]

Ok, that makes sense, Thanks!