Evaluating a Surface Integral: A Parallelogram

[bugatti79]

Homework Statement

Evlute the surface integral

Homework 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.

The Attempt at a Solution

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

 

[LCKurtz]

Homework Statement

Evlute the surface integral

Homework 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.

The Attempt at a Solution

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

You might begin by studying the material at:

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

 

[bugatti79]

You might begin by studying the material at:

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

I believe 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

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

 

[LCKurtz]

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

 

[bugatti79]

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

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

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

 

[LCKurtz]

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

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

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