00:01
Okay, we're going to find the volume of the region bounded in problem number 22.
00:06
So the bounds are y equals 2, y equal 6, which i've already drawn, and x equal 6, which is also our axis of rotation.
00:13
I kind of wanted to show you how i'm getting this picture.
00:16
So what they give us is x, y, equals 6.
00:20
Now, first and foremost, since my axis of a rotation is defined as x equal 6, i know my strip, when i go to integrate this thing, will be something like that with thickness d .y.
00:34
Importantly, it's going to have thickness d .y.
00:36
So what i want to do is to write this function in the form x equals 6 over y.
00:45
That's ultimately because i'm going to be integrating something in terms of y.
00:50
That's why i want to get a function that is in terms of y.
00:53
Now, if you were to graph this on, say, desmos or something like that, you get something that's kind of going.
00:59
Then you know it's like asymptotically doing something along these lines when it hits six we're actually at the number at y equals one so i think it's something along like this maybe but now that we have this now we got a visual okay our strip is doing something like this less thickness d y that's where i'm solving for x equals six over y now my volume is pi times the integral from a to b of my outer radius squared take away my inner radius squared d y my strip can travel from two up to six y equals two up to six my outer radius is the distance to the top of my strip notice the total distance is six this distance is x which is also called six over y so my outer radius here will be six take away six over y and we got to square that.
02:13
My inner radius is zero.
02:16
So minus zero squared, all with respect to y...