Question

Determine the surface area of the portion of 2x + 3y + 6z = 9 that is in the 1st octant.

Answer 1

We have given the portion,

3x + 2y + z = 9

The plane intersects the first octant in a triangle with the vertices $({2},0,0),(0,3,0),(0,0,6)$

since these are the intercepts with the positive x, y and z axes respectively. Thus this is the

surface area of the part of the surface $z= 6-3x-2y$ over the region $0 \le x \le {2}$ , $0 \le y\le 3 - \dfrac{3x}{2}$

The derivatives of the function $6-3x-2y$ in the x and y directions are -3 and 2, and so the surface area is

$\int_{0}^{2} \int_{0}^{3-\dfrac{3x}{2}} \sqrt{1+(-3)^2+(-2)^2}dydx = 3 \sqrt{14}$