2.2 Determining Volumes by Slicing [STARTS AT 21:55] INSTRUCTOR: Find the volume of a solid of revolution by revolving the region bounded above by the f of x equals square root of x, below by the graph g of x equals 1/x over the interval 1 to 3, and you're revolving around the x-axis. Now, I even just messed it up, and I've done the problem a couple of times-- on that last one. So read these carefully. It's bounded above by this function and below by this function. So square root of x, bounded above that is our upper function. And it's bounded below by 1/x from 1 to 3. So we're actually looking at potentially a region here. Now, 1, I believe, is a place where they intersect and 3 right there. Now, we are going to integrate this, or revolve at first. First is going to be right here. Second is going to be right here. And we have got an inside function right there and an outside function right there. This actually looks kind of like a bundt pan, if you can kind of see that, looking at it from the side. So our volume is going to be the integral from 1 to 3. I'm going to go and factor my pi out. My outside function is the square root of x squared minus my inside function 1/x squared. So this will be the integral from 1 to 3. Bring that pi down of x minus 1/x squared. So that will be x squared over 2 plus 1/x evaluated from 1 to 3. And we do, of course, have pi there because we have disks here. Plugging in 3, we get 29/6. And then plugging in 1, we get 1/2 plus 1, so that's 3/2. And I have the pi there. So this will be 10/3 pi units cubed. [ENDS AT 24:36]