00:03
The equations you give are y equals the square root of x plus 1, and i think the second equation you have listed is x equals 2, because in there it goes like this, y x equals 2, and i'm thinking you want to adjust x equals 2, so that's the one i'm going to use to answer your question.
00:21
So first let's graph this area.
00:26
So we're starting over here at negative 1, and here we get a 1.
00:36
1, 1 and 2, that's 1, and that's 2, and 2 at 2 is about 1 .7.
00:47
So this will be the square root of x plus 1, and we're cutting it off here at x equals 2.
00:57
So if we're looking to take this area and revolve it around the x -axis, we'll get a mirror image over here like so.
01:06
And we're looking to find that volume.
01:13
Okay, so this would be the disk method, and your disks would look like this.
01:21
So the radius would be from here to here.
01:29
And to find the length of the radius, you take the top curve, which is your square root of x plus one, minus the bottom of it, which is on the x -axis, which is zero.
01:41
So the radius is going to be given by the square of x plus 1.
01:48
Okay, so to do this, you're integrating these disks, which are circles, pi are squared, and the width of each circle or disk is given by dx.
02:01
So i'm going to put pi out, and we know the radius is the square root of x plus 1 squared.
02:09
Now all we need is the boundaries, which are the leftmost x plus 1 squared.
02:13
Here at negative 1 and the right most x here at 2.
02:20
Okay, we're going to simplify that a little bit before integrating...