00:01
Okay, so what we want to do, and this is going to be a pretty long problem, so what we want to find a do, want to do is we want, the very first thing we want to do is to find the area bounded by the ellipse.
00:22
And the ellipse is given by x squared over 4 plus y squared over 1, or just y squared over 1, or just y squared equal to 1.
00:33
And so what that ellipse looks like is it's centered at zero zero.
00:42
Its major axis is, oh, no, it should be a two.
00:47
Its major axis is the x -axis, and its minor axis is the y.
00:55
And so here is that ellipse.
01:00
Okay.
01:01
And we want to find the area that is bounded by this ellipse.
01:06
So the area that's inside this ellipse.
01:09
Okay.
01:10
And so the first thing we need to do is actually rewrite this in terms of y.
01:17
So we want to get y by itself.
01:19
So if you solve for the y, you get plus or minus the square root of 1 minus x squared over 4.
01:29
Or if i multiply, if i factor out a one -fourth, i get a plus or minus one -half times a square root of four minus x squared.
01:45
Okay.
01:47
And so now what i want to do is recognize that these are four symmetrical pieces.
01:52
And so if i can find the area in one, i can multiply the total area by, four and so that's what i want to do.
02:10
I want to go ahead and do that and so area is equal to there's going to be four pieces and i'm going to go from zero to two of the function and the minus the x axis which is going to be zero and it's going to be so that's going to be the square root and i'm just going to be doing the positive side and that's going to be one half four minus x squared and we are integrating with respect to x so this becomes two times the integral from zero to two of the square root of four minus x squared dx okay and then we should start starting start being able to recognize the square root of four minus x squared or the square root of a squared minus x squared.
03:07
And that is going to be 2 times 1 half times x times that square root of 4 minus x squared plus 4.
03:22
4.
03:23
So this would be x times that square root plus a squared times arc sine of x over a.
03:36
And we're evaluating from zero to two.
03:41
And so we recognize that if i put a two in here, i get a zero.
03:49
And if i put a two in here, i get a arc sign of one.
03:59
And so that is going to be pi over two.
04:04
And so then i would have this two times one half.
04:08
So i would have this equal to 4 arc sign of 1.
04:14
And then, of course, if i put 0 in here, i get 0.
04:18
If i put 0 in here, i get arc sine of 0, which is also 0.
04:22
So this is going to be 4 times pi over 2, which is going to be 2 pi.
04:32
So there is my area of the ellipse.
04:39
So now what we want to do for part b, and this is where it gets a little bit longer.
04:45
So for part b, what i want to do is i want to find the volume and the surface area of the solid generated, whoops, by revolving around the major axis, which is going to be the x axis.
05:14
Okay.
05:18
And then part c, we're going to want to do the volume and surface areas by revolving about the minor axis.
05:24
And so let's go ahead and get started with this.
05:27
And so i'm going to quickly draw the graph again.
05:31
And so we have this and opsth.
05:36
And of course, that goes from negative 2 to 2.
05:38
And so we're revolving it about the x axis.
05:43
Okay.
05:44
And so we're going to go ahead and use the x -axis.
05:51
The dis method method for the volume.
05:55
And so that is going to be equal to the volume.
06:01
The volume is equal to 2 pi times integral from 0 to 2.
06:09
And so we're going to be actually just taking this and revolving, yeah, revolving that.
06:19
And so this is going to be equal to what i want to do is this is going to be the square root.
06:32
We're actually squaring it.
06:35
And so i'm going to have to take this, which is the one half times the square root of 4 minus x squared.
06:47
And we're squaring this with respect to x.
06:50
And so remember when you're doing the disk method, it's basically your radius squared.
07:00
And so this becomes one -fourth.
07:05
So we have 2 pi times 1 -4th, the integral from 0 to 2 of now 4 minus x squared dx.
07:17
Okay, and so now what we're going to do is and this i get the two because it really should be pi r squared h basically and so the two is because i'm going to be going from this region to this region.
07:42
But instead of doing negative 2 to 2, i'm going to multiply this by 2 because when i revolve it, when i revolve it this way, i get that kind of like that football.
07:52
Shape, so to speak.
07:54
So that two came from that i'm having to, i'm revolving this one and doubling it because that would also include this region right here.
08:05
I kind of jumped the gun on you guys, so i apologize for that.
08:10
So this is going to be pi over two.
08:13
And then this is going to be 4x minus one -third x cubed and evaluated at upper limit of 2.
08:22
And lower limit of zero.
08:25
And so when i do that, i get 8 pi over 3.
08:30
So there's the volume.
08:32
The volume's not too bad.
08:33
It's going to be the surface area that's going to be the issue.
08:36
So let's go ahead and get the surface area.
08:39
And we know the surface area is actually going to be equal to the surface area, how to find the surface area of the solid generated is actually going to be 2 pi times the integral from a to b of x.
09:07
It's not going to be x.
09:08
It's going to be y.
09:09
And the reason it's going to be y is because you want, or the function, you want the distance from the axis of rotation to the function.
09:20
And so this is going to be the y value, because that's a distance from the axis of rotation to the graph.
09:31
And so times the square root of 1 plus the derivative squared, and we're integrating with respect to x.
09:43
Okay, so first of all, let's kind of step this through.
09:47
We need the derivative.
09:51
And so let's go ahead and take that derivative.
09:56
So let's go back and say y is equal to one half the square root of 4 minus x squared.
10:06
And so if we take the derivative with respect to x, we get dy over dx is equal to 1 4 minus x squared to the negative 1 half times a negative 2x, which leads us to a negative x over 2 times a square to 4 minus x squared.
10:36
Okay, and then we're going to have to square it.
10:39
So we're going to go ahead and square that derivative.
10:45
So we get x squared over 4 times 4 minus x squared, which is going to be x squared over 16 minus 4x squared.
10:56
Okay, so my surface area, my surface area is going to be equal to, and what i'm going to do is, once again, i am going to do twice the surface area from 0 to 2, to include the other half of them.
11:19
And so, and this is going to be the y value.
11:22
And i'm going to go ahead and leave it as the y value, even though i can actually put in this value.
11:28
I'm going to kind of leave it as a y value just to kind of show you guys a little trick.
11:33
And then this is going to be times the square root of 1 plus this x squared over 16 minus 4x squared.
11:42
And we're integrating with respect to x.
11:45
Okay.
11:46
And so this is going to be equal to 4 pi, the integral from 0 to 2 of y...