00:01
To calculate this surface integral, we're going to take the surface area of f of x, y, z, d of us.
00:14
And we are given that x is equal to u cosine v, y is equal to u sine v, and z is equal to u .s, and z is equal to u.
00:25
In order to represent this as a vector.
00:29
We'll write this as this is equal to u cosine v, u sine v, and u.
00:40
To derive this with respect to u and respect to v, we'll give cosine v, sine v, 1 for u, and we'll give negative u sine v, sine v, you cosine v, 0 for v.
01:06
From here we wish to come up with the cross product of both them, and we want to take the magnitude of that.
01:13
So to get the cross product, that is going to be the partial with respect to u, cross with the partial with respect to x, and that will be equal to the i components, j components, and the k components, which will be cosine v, sine v, and 1 for u.
01:40
And it'll be negative u sine v, u, u cosine v, and 0 for v.
01:50
I'm just going to move my ijks to better line up with those terms.
01:58
So you will have i, j, and k.
02:02
Okay.
02:06
Solving this, we'll give r u cross our v, which will be equal to negative u cosine v i plus u sine of vj plus parentheses u squared v, minus, sorry, plus u sine squared v, the minus sign should be attached to this j term minus u -sign v of j...