Area of portion of plane x+3y+2z=6 that lies in first octant.

How should i compute this problem?

Thanks

Write this as z = 3 - x/2 - 3y/2, with x,y such that x + 3y = 6 in the first quadrant.

So, the surface area equals
∫∫ √[1 + (z_x)^2 + (z_y)^2] dA
= ∫(y = 0 to 2) ∫(x = 0 to 6 - 3y) √[1 + (-1/2)^2 + (-3/2)^2] dx dy
= ∫(y = 0 to 2) (1/2)√10 (6 - 3y) dy
= (1/2)√10 (6y - 3y^2/2) {for y = 0 to 2}
= 3√10.

I hope this helps!