1. ## Line integral problem

Am having trouble with line integrals and can't manage to get the following problem correct;

$
\int\int_S y.dS
$

Where S is the part of the plane 2x + y + 3z = 6 for x=>0, y=>0, z=>0

I know that ;

And worked out the magnitude of the normal to be (14)^1/2, but i'm not sure if i'm parametrizing the plane correctly and i'm not sure of my limits of integration.
Any help would be greatly appreciated.

Cheers.

2. $y=6-2x-3z$

$\frac{\partial}{\partial x}y=-2....\frac{\partial}{\partial z}y=-3$

$ds=\sqrt{\left(\frac{dy}{dx}\right)^2+\left(\frac{ dy}{dz}\right)^2+1}dxdz$

$ds=\sqrt{14}dxdz$

$\int_{0}^{2}\int_{0}^{3-\dfrac{3z}{2}}(6-2x-3z)\sqrt{14}dxdz$

fixed thanks for calculas26

3. A couple of things here:

1. This is a surface integral not a line integral

2. dS = 14^(1/2) not 13^(1/2)

3. The inner integration limit should be 0 to 3 - 3z/2

4. the integration order is dx dz not dxdy

Monster what parameterization did you use? Hard to tell if its correct
without knowing what it is.