00:01
We are looking to calculate the surface integral of x squared z squared for the cone given that z squared is equal to x squared plus y squared now the first thing i immediately want to see from this is that this looks a lot like polar that we have r squared is equal to x squared plus y squared and this tells us we should probably be using the equation at the surface area of the integral of x y z f of x, y, z is equal to the integral of f of r, u .v, multiplied by the magnitude of the cross product for the u partial and the v partial da.
01:00
Now, since we know that z square is equal to x square plus y squared, if we think of z in terms of r, we can write the parametric representation for the cone.
01:12
As v sine u, v cosine u, and v.
01:27
And we are given at the bounds for this are also 1 to 3 for z, so v will be from 1 to 3.
01:34
Now we want to get our u and v partials for this, are of course being what's right at the top here.
01:44
So our u partial is going to be v cosine u of un.
01:51
Minus v sine u for j plus 0 k and then our v partial is going to be equal to sine u for i plus k and now to get the cross product of this r u crossed with r v that is equal to our v that is equal to v.
02:30
Cosine u sine u sine u v sine u cosine u and zero and one for the i j and k components solving this out we'll give i times negative v sine u cosine u 1 minus j of v cosine u 0 sine u 1 plus k of k of v cosine u and cosine u and this tells us that r u crossed with r v is equal to negative v sine u i minus v cosine u j plus v k plus v k and the magnitude of r of the r partial of r cross with the v partial of r is going to be equal to the square root of all the terms square so magnitude of ru crossed with rv is equal to the square root of v squared sine squared u plus v squared cosine squared of u plus v squared.
04:31
And this is equal to v times the square root of two.
04:39
So now plugging back into our original equation, we have x squared, z squared, which we said x squared is equal to v sine u so v sine u squared times z squared which we say is equal to v so v squared multiplied by the magnitude of the cross products so multiplied by v square root 2 d a and we are given the parametization which is that u is from 0 to 2 pi 0 u to 2 pi and v to 2 pi and v as we said earlier, that we came up with, is from 1 to 3.
05:27
1 to v to 3.
05:30
Let me just write these a little more distinctively for you, so they don't look at the same letter...