00:01
In this problem, we're going to compute the surface area of kosh x from minus l and 2 to ln2 when it's revolved about the x axis.
00:11
So we get that the surface area is the integral from minus ln2 to ln2 of 2 pi y, which is kosh x, times the square root of 1 plus y prime squared, which is cinch squared x, dx.
00:42
And from here, we can use a few identities, or a couple identities, kosh squared x minus inch squared x is equal to 1, and also kosh squared x is equal to 1 half times the quantity 1 plus kosh of 2x.
01:24
So we get that this is the integral from minus ln2 to ln2 of 2 pi of 2 pi times kosh x but one plus cent squared x is kosh squared x so this is a square root of kosh squared x which is kosh x so we get that this is the integral of two pi times kosh squared x dx and then from here we can write this as the integral from minus l and 2 to l and 2 of pi times the quantity 1 plus kosh of 2x d x which is equal to pi times x plus 1 half cinch of 2x evaluated, of course, at our limits of integration.
02:47
So we get that this is pi times l .2 plus 1 half times cinch of 2 l .2 minus negative l .2 plus 1 half times cinch of 2 l .2, minus negative l2 plus 1 half times cinch.
03:20
Of minus 2ln2.
03:29
And then we get that this is pi times ln2 plus 1 half cinch of 2ln2 plus ln2 and then this minus can come out in front of cinch because cinch is an odd function and then we also have a minus sign here...