Find the area of the part of the plane 3x 2y 6 that les in the first octant. Part1ofs The surface Sis the graph of a function zx. y) over a region Din the xy-plane. The surface area of S can be calculated by A(S) -L V-()-() dA. The plane 3r+ 2y+6 is the graph of the function z fx, y)-6- -2y. We, therefore, have - and - 2 Part 2 ofS Therefore. Since we want the area of the section of the plane that les abeve the first octant. then the region D will be the triangular region bounded by the axis, the y-axis, and the line formed by the intersection of the plane with the plane z-0. Substituting z0 into 3x + 2y 6 and solving for y, we find that this line of intersection has the equation Semt Shyocnot.come bac

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