# Integrals 3

1. Jan 5, 2012

### bugatti79

1. The problem statement, all variables and given/known data

Evlute the surface integral

2. Relevant equations

f(x,y,z)=x+y+z where sigma is the parallelogram with parametric equations x=u+v, y=u-v and z=1+2u+v where 0 <=u<=2 and 0<=v<=1.

3. The attempt at a solution

I have no idea how to tackle this. Any suggestions?

2. Jan 5, 2012

### LCKurtz

You might begin by studying the material at:

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx

3. Jan 6, 2012

### bugatti79

I belive we have to use $\displaystyle \int \int _R f(x(u,v),y(u,v), z(u,v) || r_u \times r_v||dA$

I calculate the magnitude of the determinent to be \sqrt 2 hence

The surface integral becomes

$\displaystyle \sqrt{2} \int_{0}^{1} \int_{0}^{2} (4u+v+1) du dv$....?

4. Jan 6, 2012

### LCKurtz

I agree with everything except the $\sqrt 2$.

5. Jan 7, 2012

### bugatti79

since we have the determinant as 3i+1j-2k and the magnitude is

$\sqrt(3^2+(-2^2)+1)=\sqrt 14$ cheers

6. Jan 7, 2012

### LCKurtz

The $\sqrt {14}$ is correct, but 3i + j - 2k is not a determinant; it is a vector.