Find the area of the part of the plane 5x + 2y + z = 10 that lies in the first octant.

Find the area of the surface. The part of the plane 5x+2y+z-10 that lies in the first octant given plane 1s 5x+2y+2=10 so, 2, 2. surface area is dA 30dA where dAis the area of the plane we have the constraints since given plane lies in the first octant, z 5x+2y+z 10 0 5x+2y-10 so, 0 sx S2. since z 2 0 →10-5x-2y20 so, 0 y 5--х. 4 10-5 therefore, surface area is 530