00:01
We're given a vector field f and an oriented surface s, and we're asked to find the flux of f across s.
00:09
F is the vector field yi plus z minus y, j, plus x, k, and s is the surface of the tetrahedron, with vertices 0 ,000, 1 -0 ,0, and 0 -0 -1.
00:28
So if you can envision the tetrahedron in your head lying in the first octant, well, the surface s consists of four surfaces actually, one for each of the faces of the tetrahedron.
00:50
So we have s1, which is the triangular face with vertices 1 ,0 ,010, and 0 .01.
01:07
So this is the face that would be most visible if you were to draw this.
01:19
We have s2, which is the triangular face of the tetrahedron in the xy plane.
01:30
This has vertices 0 -00 -0 -0 -0 -0 -0 -0.
01:35
So the underneath, s3, this will be the face of the tetrahedron in the x, z plane, and this has vertices 0 -0 -0.
01:48
00, 00, and 001.
01:53
So this is essentially lying behind the tetrahedron on the left.
02:00
And we have a face s4.
02:03
This is the face in the yz plane.
02:09
This has the vertices 0 -0 -0 -1 -0 -0 -0 -0 -0 -0.
02:16
And this essentially lies behind the tetrahedron on the right.
02:23
Now on s1, s1, this is the portion of the plane z equals 1 minus x minus y, but x lies between 0 and 1, and y lies between the line y equals 0 for these values of x and between 1 minus x.
03:00
If you look at the projection into the xy plane.
03:04
And because this is on top of the tetrahedron, so to speak, this has an upward orientation.
03:17
Therefore, it follows up the flux of f across s1.
03:28
This is going to be the integral from x equals 0 to 1, integral from y equals 0 to 1 minus x of.
03:37
And because the surface described by a function of x and y, we have the opposite of the second the first component of our vector field, which is y.
03:50
So this is negative y, times the partial derivative of z with respect to x, which is negative 1, plus the opposite of the second component of f, which is z minus y.
04:07
So this is minus z minus y, times the partial derivative of z with respect to y, which is negative 1 plus the third component of f which is x the y d x and in this next step i'm going to simplify so we get the integral from 0 x equals 0 to x equals 1 integral from y equals 0 to 1 minus x and let's see these ys will cancel out and so i get x my plus 2 d y d y d x that's not a 2, my mistake.
04:59
That's a z.
05:03
Sorry, z plus x, d, y, dx.
05:05
And next i'll make the substitution for z in terms of x and y.
05:10
And so this becomes the integral from 0 to 1, 0 to 1 minus x of 1 minus y, dy, d .y, dx.
05:23
Taking the antiderivative with respect to y, i get integral from x equals 0 to 1 of y minus 1 half y squared from 0 to 1 minus x d x and substituting for y i get 1 minus x minus 1 half times 1 minus x squared so 1 minus x minus 1 1 1ā2 plus x minus 1ā2 plus x squared and this simplifies to 1 half times the integral from 0 to 1 of 1 minus x squared.
06:17
So the term of order 1 cancels out.
06:22
Taking the antiderivative with respect to x, we get 1 half times x minus 1 3rd, x cubed from 0 to 1...