00:01
For this problem here to find the surface area of the region that's bounded by y equals negative x cubed over 12.
00:07
For x values in the closed interval negative 3 to 0 rotated about the x -exes.
00:12
Now the surface area here is given by s, which is equal to integral from a to b of 2 pi times y times the square root of 1 plus the square of d .y over dx and then dx.
00:30
If we're given y equals f of x as the function that bounds the region.
00:38
Now, d .y over dx is equal to negative 1 fourth of x squared.
00:48
So, d .y over dx squared is equal to 1 over 16 times x -rays to the fourth power.
00:58
So s is equal to the integral from negative 3 to z.
01:02
Of 2 pi times y which is negative x cubed over 12 times a square root of 1 plus 1 over 16 times x rays to the fourth power d x that's equal to negative pi over 6 integral from negative 2 to 0 of x cubed times the square root of 16 plus x raise to the fourth power over 16 square root of that, which will be 4, and then dx.
01:37
So this will give us negative pi over 24 integral from negative 3 to 0 of x cubed times 16 plus x -rays to the 4th power raised to a power of 1 half and then dx.
01:50
And then we apply substitution...