00:01
For this problem, we're finding the volume of the solid generated by revolving the region bounded by y equals x -square and the line y -equals 1 about the line y -equals negative point.
00:13
So we are revolving a region around a horizontal line.
00:19
So first we need to draw our, we need to graph our function so we can see what the bounded region looks like.
00:28
So we got x -square.
00:32
Which is a parabola 01124 etc.
00:42
All right, and then we got the line y equals 1, which is horizontal, and it goes through here.
00:51
Okay.
00:53
So you can see that your bounded region is this part.
00:57
And then you're revolving around the line y equals negative 1.
01:06
Okay, that didn't go well.
01:08
Let me do that again.
01:11
Okay, so clearly this will be a washer because this line is not flushed against your bounded region.
01:21
So if we were revolving around the line y equals 1, you would have a disk, but this is a washer.
01:27
Therefore, you need to find your big r and your little r.
01:32
Your big r goes from the axis of revolution, which is the center of rotation all the way to the outside function, the farthest function, which is the horizontal line, y equals 1.
01:44
And your little r goes from your center of revolution to the inside function, which is the x square.
01:53
And since this is a horizontal axis, you don't have to convert your function to be in terms of y.
02:06
Can be the way it is in terms of x so what we need to do is find our bounds and that's easy just if you graft it correctly you can see it's negative 1 1 but you can always just set your equation equal to each other so you got x -way you got x square equal to 1 i got ahead of myself so x x -way equals 1 so x square equals 0 so x square equals oh i mean i okay so i could have done it that way and then just factor it and then get x minus 1 x plus 1 equals 0 so x equals 1 and negative 1 or we can have um square root of both we can swear root both sides and we have x equal plus or minus one that way okay if you take the square root of both size but don't forget the plus or minus so your bounds are negative 1 to 1 so your bounds are you have negative 1 to 1.
03:11
Don't forget the pi.
03:13
So your big r is going to be from negative 1 to the line y equals 1.
03:20
So negative 1 minus 1 square.
03:23
And then minus your little r square, which is from negative 1 to the parabola.
03:29
So negative 1 minus 6 square square.
03:32
Okay.
03:33
And it depends on what you're being asked to do, but you can either put this in a calculator and or solve by hand.
03:42
So it's easy enough to solve by hand.
03:44
Negative 1 minus 1 is negative 2.
03:46
Negative 2 square is 4.
03:49
So 4 minus and then i square out the negative 1 minus x squared so i get 1 minus minus, okay, plus x, plus 2, x square plus x to the 4.
04:05
Okay, so let me simplify some more.
04:08
So i'm going to try to do this by hand and then check with the calculator.
04:14
And hopefully both of those are the same.
04:20
Okay, so simplifying this, i have that.
04:25
Okay, so the anti -derivative is going to be 3x minus 2, x to the 3rd over 3, minus x to the 5th over 5, from negative 1 to 1.
04:39
Okay, so let's plug in your bounds.
04:43
So we have 3 minus 2, 3 3 3 3 3, 3, 3...