00:01
All right, for this problem, we want to figure out the surface area of an ellipse when rotated about the x -axis for part a and then about the y -axis for part b.
00:11
So to begin, if we're trying to rotate about the x -axis, that means that we want to have y as a function of x, and we want to have a less than or equal to x, less than or equal to b.
00:22
Where the a and b are unrelated to those a's and b there.
00:27
So, for us here, we'll have trying to find y as a function of x we can rearrange first of all subtracting x squared over a squared from both sides so we'll have y squared over b squared equals 1 minus x squared over a squared then multiply both sides by b squared we get y squared equals b squared minus b squared x squared over a squared then lastly we can take the square root of both sides to get y equals the square root of b squared minus b squared x squared over a squared.
01:03
And we can also simplify this a little bit by putting everything in that radical over a common denominator.
01:10
So we can say that it's going to be the square root of a squared b squared minus b squared x squared all over a squared.
01:21
Alternatively, we can then factor out the b squared from the two terms on top.
01:26
We get b squared times, i guess it would be b squared times a squared minus x squared up top then we have b squared over a squared out front which means that we can actually simplify this down a little bit further of b over a times the square root of that's square root of just a squared minus x squared inside okay now that we have that we want to figure out the derivative which we can do just by applying the chain rule so we'll take the derivative of the inside with respect to the outside or sorry outside with respect to the inside multiplied by the derivative of the inside is just going to be negative 2x so we can write that out front we'll have negative 2b x over a then we'll have times one half the square root or rather one over two times the square root of a squared a squared minus x squared so the two on top and bottom will cancel each other out so we get negative bx over a and z square root of a squared minus x squared then we'll also want to figure out y prime squared so that will be b squared x squared over a squared times a squared minus x squared having that we can now plug everything into our surface area formula where we have s equals the integral uh will address what the boundaries on that integral will be in just a moment so we have an integral with unspecified boundaries thus far of 2 pi times y.
03:14
So i'll have 2 pi times b over a times the square root of a squared minus x squared times the square root of 1 plus y prime squared.
03:26
So times the square root of 1 plus b squared x squared over a squared times a squared minus x squared, which we can simplify down a little bit by getting everything in that radical overcome denominator.
03:44
So it will become a squared times a squared minus x squared plus b squared x squared divided by a squared times a squared minus x squared dx.
04:03
Dx.
04:05
One moment.
04:07
Now the range that x will vary over is going to be from negative a up to positive a.
04:17
We are rotating about the x -axis, so we'll actually just want from zero up to a.
04:25
No, i take that back, negative a up to positive a.
04:29
So next up, we will want to look at what sort of simplifications that we can do.
04:36
So first of all, we can represent this second radical here, as first of all just separates the top from the bottom.
04:46
The top is its own expression.
04:48
Then on the denominator, we'd have the square root of a squared, which is just going to turn into a, times the square root of a squared minus x squared.
04:58
Now the square root of a squared minus x squared can multiply out with that, a squared of a squared minus x squared to give us a one there.
05:06
And then we'd have this b over a multiplied by a, so we'll get a b over a squared.
05:12
So our simplified form at this point is going to be negative a up to a...