00:02
For this problem, we are given a curve, x equals 210 of y over 4, and we are bounding a region from 0 to pi over 2, and we're going to revolve this region around the y -axis and find its volume.
00:17
So there are a couple ways we can do this.
00:19
The method that i would recommend would be the method of cylindrical shells.
00:26
So how would that work? so there's a volume formula for the shell method, and it's 2 pi times the integral.
00:35
Of a radius and height of any one of these cylindrical shells.
00:40
So on the diagram, we can draw the radii and the height of any given shell.
00:46
The radius is going to extend from the axis of rotation perpendicularly into a point in the region.
00:56
This is a horizontal distance, which means it will have a distance of x.
01:01
So off to the side, we're going to note that our x, or excuse me, that our r is equal to x.
01:08
And then the height is going to be from one bound of the region to the other.
01:14
It'll be perpendicular to the radius.
01:17
And in this case, we would like to subtract one curve from the other.
01:22
Now, in this case, notice we have our radius in terms of x.
01:27
That means that our height must also be in terms of x.
01:30
Now, in order for that to happen, you actually need to rewrite the curve as a function of x.
01:35
So you'll have to solve for y in the given equation.
01:40
So we'll do that real quick.
01:42
Divide both sides by two.
01:49
And then inverse tan on each side.
01:52
So you'll get y over 4 equals inverse tangent of x over 2 and then multiply the 4.
02:07
There we are.
02:08
So there's the equation for the given curve in terms of x.
02:11
And the bottom curve in this case is the x axis, which is y equals 0.
02:16
You'll do the upper point minus the lower point to get the height.
02:21
So the upper point is our four inverse tangent of x over two minus the lower point, which is minus zero.
02:32
So here we have our radius and height functions given.
02:37
Now we're also then prompted to figure out whether or not this integral has a dx or d .y.
02:42
Notice i didn't write it in our definition for volume.
02:45
Once we see that the radius and height are written in terms of x, that gives us the clearance we need to then write our bounds in terms of x.
02:54
So our volume integral is now going to be written as integral 0 to pi over 2.
03:05
Let's put the constants first, actually.
03:08
So let's put the 2 pi.
03:11
Then our integral from 0 to pi over 2 of our radius times high.
03:18
So we have our radius and height that will be 4x, inverse tangent of x over 2, dx.
03:35
And you can also pull out this constant of 4.
03:51
Okay.
03:53
And now we have this formula here.
03:59
And in order to do this integral, what i would likely do is an integration by parts.
04:05
So in that case, let's try it out.
04:10
Let's say, let's have u equal our maybe inverse trig function.
04:19
Then we'll label dv in this case to be our x, dx.
04:30
Find d .u by taking the derivative of inverse tangent.
04:35
The derivative of inverse tangent is 1 over 1 plus the input squared times the derivative.
04:48
Of x over 2, which is one half.
04:54
And then our v, which will be the integral of x, should be one half x squared.
05:03
All right, let's try this out.
05:04
So our integral then becomes we have 8 pi.
05:10
We have u times v.
05:12
So this is one half x squared times inverse tangent of x over 2.
05:22
Okay, and this is evaluated from 0 to pi over 2 minus the integral of v, du.
05:33
So here's v, and here is du.
05:40
Let's simplify it a little bit.
05:42
That would be x squared over 4.
05:52
Okay, something like this.
05:59
All right, and let's see what we can do now...