00:01
So we have given here the surface s equation z is equal to 1 minus x minus y.
00:10
Now, and in first octant.
00:14
So first we need to find the normal vector n.
00:19
First we need to parametrization with respect to u and v.
00:23
So let's say our x is equal to u and y is equal to v.
00:30
So from here our z is equal to we need to write here 1 minus u minus v.
00:41
So these are two parameter u and v.
00:45
So now we need to find the derivative that is a del r upon del u.
00:53
That is equal to, we need to find here, 1, 0 and here minus of 1.
01:02
And now next we need to find our derivative with respect to v.
01:07
So it become dr upon del v is equal to 0, 1, minus of 1.
01:18
Now we need to find the normalized vector n to obtain the unit normal vector n.
01:25
So normalize the vector n by dividing the magnitude.
01:29
So the magnitude of n is equal to, we need to write here under root 1 square plus 1 square plus 1 square.
01:41
So that is equal to we have here under root 3.
01:44
So therefore, unit normal vector.
01:48
So therefore, unit normal vector is, is n is equal to 1 upon under root 3, 1 upon under root 3, and 1 upon under root 3.
02:03
Now, and to find the normal vector, we need to multiply here the cross multiplication of del r upon del u multiply with del r upon del v.
02:17
So from here, our unit vector become 1, 0, minus of 1 and cross multiply with 0, 1, minus of 1.
02:29
So after cross multiplication, it becomes that is 1, 1, 1.
02:35
So unit normal vector we obtained by this formula.
02:39
Now we need to calculate the product of f dot n.
02:47
So to calculate the dot product of f and n, we need to write our f function here.
02:55
X comma y comma z that is equal to 6zi minus 3j plus yk...