Parametrized Surfaces: Evaluating and Integrating f(x,y,z) = yz with u and v

[Chandasouk]

Express f(x,y,z) = yz in terms of u and v and evaluate \int\int_S f(x,y,z)dS

This is supposed to be simple but I really don't know how to do this.

I rewrote f(x,y,z) = yz as x = g(y,z) so then \Phi(y,z) = (y,z, x)

Tx = (0,0,1) and Ty=(1,0,0) and their corss product, n, is <0,0,-1>

Am I even doing this right? if so where do I go from here?

 

[lanedance]

can you explain the integral a bit more, is this an integral over a surface, if so what is the surface? if i guess what you are trying to do you first need parametrise the surface in terms of u & v giving
\textbf{x}^T = (x(u,v), y(u,v), z(u,v))^T

\int_S f(x,y,z) dS = \int \int f(\textbf{x}(u,v)) \left| \frac{\partial \textbf{x}}{\partial u} \times \frac{\partial \textbf{x}}{\partial v} \right|du dv

see below for more
http://en.wikipedia.org/wiki/Surface_integral

 

[HallsofIvy]

In order for this problem to make any sense at all, you would have to be given formulas for u and v in terms of x and y (or vice-versa) as well as what region S is. I see none of that here.

 

[Chandasouk]

This was the third part of a problem. The complete problem is this

Show that \Phi(u,v) = (2u+1, u-v, 3u+v) parametrizes the plane 2x-y-z=2. Then:

a) Calculate Tu, Tv, and n(u,v)

B) Find the area of S = \Phi(D), where D = {(u,v): 0\lequ\leq2,
0\leqv\leq1

c)Express f(x,y,z) = yz in terms of u and v and evaluate \int\int_S f(x,y,z)dS

I already did A and B. The answers are Tu = (2,1,3), Tv = (0,-1,1), and n(u,v) = <4,-2,-2>

and Area(S) = 4*sqrt(6)