00:01
For this problem, we want to find the surface area of the region that's bounded by the function 2 square root of 4 minus y in the closed interval 0 to 15 over 4, and that is revolved about the y -axis.
00:12
So for this, the surface area formula that we're going to use is s, which is equal to the integral from a to b of 2 pi times our function g of y times the square root of 1 plus the square of g prime of y dy.
00:30
So first you want to get g prime of y.
00:32
So since g prime of y equals 2 times a half of 4 minus y raised to negative 1 half times negative 1, which is the same as negative 1 over square root of 4 minus y, then our s equals the integral from 0 to 15 over 4 of 2 pi times 2 square root of 4 minus y times the square root of 1 plus the square of negative 1 over square root of 4 minus y dy.
01:07
Simplifying that, we should get 4 pi integral from 0 to 15 over 4 of the square root of 4 minus y times the square root of 1 plus 1 over 4 minus y dy.
01:21
That can be simplified further into 4 pi integral from 0 to 15 over 4 of the square root of 4 minus y times the square root of 4 minus y plus 1 over 4 minus y dy.
01:34
Now we can cancel out this square root of 4 minus y here and this denominator here, which then give us 4 pi integral from 0 to 15 over 4 of the square root 5 minus y dy...