Surface Integral

[mammothrob]

Can someone please check if I set my integral up correctly, because I keep getting a negative answer for my surface integral. I have evaluated it and got the same answer more than once. Im not sure what my flaw is.


Integrate g(x,y,z)=xyz over the portion of the plane 2x+2y+z=2 that lies in the first octant.

My plan is project the surface S on the yz-plane
So im using the following evaluation formula for surface integrals.

\(\displaystyle I = \iint\limits_{yz} {g(k(}y,z),y,z)\sqrt {k_y ^2 (y,z) + k_z ^2 (y,z) + 1} dA\)

so I found

\(\displaystyle X = k(y,z) = \frac{1}
{2}(2 - z - 2y)\)

thus;
\(\displaystyle k_y (y,z) = - 1\)
\(\displaystyle k_z (y,z) = - \frac{1}
{2}\)

I know the intagrand in my formula is now
\(\displaystyle \sqrt {k_y ^2 (y,z) + k_z ^2 (y,z) + 1} = \frac{3}
{2}\)

so I piece my formula togeather like this
\(\displaystyle \int\limits_0^1 {\int\limits_0^{2 - 2y} {\frac{1}
{2}} } (2 - z - 2y)yz\frac{3}
{2}dzdy\)

I got
.
.
-9/240 = I
.
.
I got my integral limits from the following:

The plane in the first octant it looks like a triangle with verticies (1,0,0) (0,0,2) and (0,1,2)

So in the first octant y goes from [0,1]
and
z goes from
0 to letting x=0 in my x=k(y,z) and solving for z.

 

[Unco]

Your integral is fine, but your evaluating is lacking - you should get 1/20.